Category: Class 10

Students can download Maths Chapter 8 Statistics and Probability Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Additional Questions

I. Multiple Choice Questions

Question 1.
The range of the first 10 prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 is ______
(1) 28
(2) 26
(3) 29
(4) 27
Answer:
(4) 27
Hint:
Range = Largest value – Smallest value = 29 – 2 = 27

Question 2.
The least value in a collection of data is 14.1. If the range of the collection is 28.4, then the greatest value of the collection is _______
(1) 42.5
(2) 43.5
(3) 42.4
(4) 42.1
Answer:
(1) 42.5
Hint:
Given, S = 14.1; R = 28.4,
L = S + R = 28.4 + 14.1 = 42.5

Question 3.
The greatest value of a collection of data is 72 and the least value is 28. Then the coefficient of range is ______
(1) 44
(2) 0.72
(3) 0.44
(4) 0.28
Answer:
(3) 0.44
Hint:
Coefficient of range = \(\frac{L-S}{L+S}\)
= \(\frac{72-28}{72+28}=\frac{44}{100}\)
= 0.44

Question 4.
For a collection of 11 items, Σx = 132, then the arithmetic mean is _______
(1) 11
(2) 12
(3) 14
(4) 13
Answer:
(2) 12
Hint:
\(\bar{x}=\frac{\Sigma x}{n}=\frac{132}{11}=12\)

Question 5.
For any collection of n items, Σ(x – \(\bar{x}\)) = _____
(1) Σx
(2) \(\bar{x}\)
(3) n\(\bar{x}\)
(4) 0
Answer:
(4) 0
Hint:
We know that, For all collection of n items,
Σ(x – \(\bar{x}\)) = 0

Question 6.
For any collection of n items, (Σx) – \(\bar{x}\) = ________
(1) n\(\bar{x}\)
(2) (n – 2)\(\bar{x}\)
(3) (n – 1)\(\bar{x}\)
(4) 0
Answer:
(3) (n – 1)\(\bar{x}\)
Hint:

Question 7.
If t is the standard deviation of x, y, z, then the standard deviation of x + 5, y + 5, z + 5 is _______
(1) \(\frac{t}{3}\)
(2) t + 5
(3) t
(4) xyz
Answer:
(3) t
Hint:
The S.D. of distribution remains unchanged when each value is added or subtracted by the same quantity.

Question 8.
If the standard deviation of a set of data is 1.6, then the variance is _______
(1) 0.4
(2) 2.56
(3) 1.96
(4) 0.04
Answer:
(2) 2.56
Hint:
Variance = (S.D.)² = (1.6)² = 2.56

Question 9.
If the variance of a data is 12.25, then the S.D is _______
(1) 3.5
(2) 3
(3) 2.5
(4) 3.25
Answer:
(1) 3.5
Hint:
S.D = √Variance = √12.25 = 3.5

Question 10.
Variance of the first 11 natural numbers is ______
(1) √5
(2) √10
(3) 5√2
(4) 10
Answer:
(4) 10
Hint:
Variance = \(\frac{n^{2}-1}{12}=\frac{11^{2}-1}{12}=\frac{120}{12}=10\)

Question 11.
The variance of 10, 10, 10, 10, 10 is _______
(1) 10
(2) √10
(3) 5
(4) 0
Answer:
(4) 0
Hint:

Question 12.
If the variance of 14, 18, 22, 26, 30 is 32, then the variance of 28, 36, 44, 52, 60 is _______
(1) 64
(2) 128
(3) 32√2
(4) 32
Answer:
(2) 128
Hint:
Variance of the given numbers = 32;
S.D. = √32 = 4√2
Each data is multiplied by 2.
New S.D.= 4√2 × 2 = 8√2
Variance = (8√2)² = 128

Question 13.
The standard deviation of a collection of data is 2√2. If each value is multiplied by 3, then the standard deviation of the new data is ______
(1) √12
(2) 4√2
(3) 6√2
(4) 9√2
Answer:
(3) 6√2
Hint:
Given, S.D. = 2√2
Each value is multiplied by 3
New S.D. = 2√2 × 3 = 6√2

Question 14.
Given Σ(x – \(\bar{x}\))2 = 48, \(\bar{x}\) = 20 and n = 12. The coefficient of variation is ______
(1) 25
(2) 20
(3) 30
(4) 10
Answer:
(4) 10
Hint:

Question 15.
Mean and standard deviation of a data are 48 and 12 respectively. The coefficient of variation is _______
(1) 42
(2) 25
(3) 28
(4) 48
Answer:
(2) 25
Hint:
Coefficient of variation
C.V = \(\frac{\sigma}{\bar{x}} \times 100\) = \(\frac{12}{48} \times 100\) = 25

Question 16.
If Φ is an impossible event, then P(Φ) = ______
(1) 1
(2) \(\frac{1}{4}\)
(3) 0
(4) \(\frac{1}{2}\)
Answer:
(3) 0
Hint:
Probability of an impossible event is 0.

Question 17.
If S is the sample space of a random experiment, then P(S) = _______
(1) 0
(2) \(\frac{1}{8}\)
(3) \(\frac{1}{2}\)
(4) 1
Answer:
(4) 1
Hint:
Every event is a subset of S. Sample space contain all the possible element.
P(S) = 1

Question 18.
If p is the probability of an event A, then p satisfies _______
(1) 0 < p < 1
(2) 0 < p < 1
(3) 0 < p < 1
(4) 0 < p < 1
Answer:
(2) 0 < p < 1
Hint:
The Probability of an event is always greater than 0 and less than or equal to 1.

Question 19.
Let A and B be any two events and S be the corresponding sample space. Then P (\(\bar{A}\) ∩ B) = ______

(1) P(B) – P(A ∩ B)
(2) P(A ∩ B) – P(B)
(3) P(S)
(4) P[(A ∪ B)’]
Answer:
(1) P(B) – P(A ∩ B)
Hint:
P(\(\bar{A}\) ∩ B) means only B and not A

Question 20.
The probability that a student will score centum in mathematics is \(\frac{4}{5}\). The probability that he will not score centum is ______
(1) \(\frac{1}{5}\)
(2) \(\frac{2}{5}\)
(3) \(\frac{3}{5}\)
(4) \(\frac{4}{5}\)
Answer:
(1) \(\frac{1}{5}\)
Hint:
P(\(\bar{A}\)) = 1 – P(A) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)

Question 21.
If A and B are two events such that P(A) = 0.25, P(B) = 0.05 and P(A ∩ B) = 0.14, then P(A ∪ B) = _____
(1) 0.61
(2) 0.16
(3) 0.14
(4) 0.6
Answer:
(2) 0.16
Hint:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.25 + 0.05 – 0.14 = 0.16

Question 22.
There are 6 defective items in a sample of 20 items. One item is drawn at random. The that it is probability a non-defective item is ________
(1) \(\frac{1}{10}\)
(2) 0
(3) \(\frac{3}{10}\)
(4) \(\frac{2}{3}\)
Answer:
(1) \(\frac{1}{10}\)
Hint:
Non-defective item = 20 – 6 = 14
Probability of non-defective items = \(\frac{14}{20}\) = \(\frac{7}{10}\)

Question 23.
If A and B are mutually exclusive events and S is the sample space such that P(A) = \(\frac{1}{3}\) P(B) and S = A ∪ B, then P(A) = _____
(1) \(\frac{1}{4}\)
(2) \(\frac{1}{2}\)
(3) \(\frac{3}{4}\)
(4) \(\frac{3}{8}\)
Answer:
(1) \(\frac{1}{4}\)
Hint:
P(A) = \(\frac{1}{3}\) P(B)
P(B) = 3 P(A)
P(A ∪ B) = P(A) + P(B)
[A and B are mutually exclusive]
P(A ∪ B) = P(A) + 3 P(A)
1 = 4P(A) [But P(A ∪ B) = 1]
P(A) = \(\frac{1}{4}\)

Question 24.
The probabilities of three mutually exclusive events A, B and C are given by \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{5}{12}\). Then P(A ∪ B ∪ C) is _______
(1) \(\frac{19}{12}\)
(2) \(\frac{11}{12}\)
(3) \(\frac{7}{12}\)
(4) 1
Answer:
(4) 1
Hint:
P (A ∪ B ∪ C) = P(A) + P(B) + P(C)
\(\frac{1}{3}+\frac{1}{4}=\frac{5}{12}=\frac{4+3+5}{12}=\frac{12}{12}=1\)

Question 25.
If P(A) = 0.25, P(B) = 0.50, P(A ∩ B) = 0.14 then P(neither A nor B) = ______
(1) 0.39
(2) 0.25
(3) 0.11
(4) 0.24
Answer:
(1) 0.39
Hint:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.25 + 0.50 – 0.14
= 0.61
P (neither A nor B) = P(\(\bar{A}\) ∩ \(\bar{B}\))
= 1 – P(A ∪ B)
= 1 – 0.61
= 0.39

Question 26.
A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random, the probability that it is not red is _______
(1) \(\frac{5}{12}\)
(2) \(\frac{4}{12}\)
(3) \(\frac{3}{12}\)
(4) \(\frac{3}{4}\)
Answer:
(4) \(\frac{3}{4}\)
Hint:
P(\(\bar{R}\)) = 1 – P(R)
\(=1-\frac{3}{12}=\frac{9}{12}=\frac{3}{4}\)

Question 27.
Two dice are thrown simultaneously. The probability of getting a doublet is ________
(1) \(\frac{1}{36}\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{2}{3}\)
Answer:
(3) \(\frac{1}{6}\)
Hint:
n(S) = 36
Let A be the event of getting a doublet
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n(A) = 6
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

Question 28.
A fair die is thrown once. The probability of getting a prime or composite number is _______
(1) 1
(2) 0
(3) \(\frac{5}{6}\)
(4) \(\frac{1}{6}\)
Answer:
(3) \(\frac{5}{6}\)
Hint:
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
The required probability = \(\frac{5}{6}\)
[Science is neither prime not a composite number]

Question 29.
Probability of getting 3 heads or 3 tails in tossing a coin 3 times is _______
(1) \(\frac{1}{8}\)
(2) \(\frac{1}{4}\)
(3) \(\frac{3}{8}\)
(4) \(\frac{1}{2}\)
Answer:
(2) \(\frac{1}{4}\)
Hint:
S= {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
A = {HHH, TTT}; n(A) = 2
The required probability = \(\frac{2}{8}\) = \(\frac{1}{4}\)

Question 30.
A card is drawn from a pack of 52 cards at random. The probability of getting neither an ace nor a king card is _____
(1) \(\frac{2}{13}\)
(2) \(\frac{11}{13}\)
(3) \(\frac{4}{13}\)
(4) \(\frac{8}{13}\)
Answer:
(2) \(\frac{11}{13}\)
Hint:
n(S) = 52
Number of ace cards = 4
number of king cards = 4
n(non-ace and non-king cards) = 52 – 8 = 44
P(neither an ace nor a king) = \(\frac{44}{52}\) = \(\frac{11}{13}\)

Question 31.
The probability that a leap year will have 53 Fridays or 53 Saturdays is ______
(1) \(\frac{2}{7}\)
(2) \(\frac{1}{7}\)
(3) \(\frac{4}{7}\)
(4) \(\frac{3}{7}\)
Answer:
(4) \(\frac{3}{7}\)
Hint:
Leap year contains 52 weeks and 2 days
Sample space = {(sun, mon), (mon, tue), (tue, wed), (wed, thu), (thu, fri), (fri, sat), (sat, sun)}
n(S) = 7
The required probability = \(\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{3}{7}\)

Question 32.
The probability that a non-leap year will have 53 Sundays and 53 Mondays is ________
(1) \(\frac{1}{7}\)
(2) \(\frac{2}{7}\)
(3) \(\frac{3}{7}\)
(4) 0
Answer:
(4) 0
Hint:
Non leap year contains 52 weeks and one day
Sample space (S) = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
The required probability = \(\frac{1}{7}+\frac{1}{7}-\frac{2}{7}\)
= \(\frac{2}{7}-\frac{2}{7}\)
= 0

Question 33.
The probability of selecting a queen of hearts when a card is drawn from a pack of 52 playing cards is ______
(1) \(\frac{1}{52}\)
(2) \(\frac{16}{52}\)
(3) \(\frac{1}{13}\)
(4) \(\frac{1}{26}\)
Answer:
(1) \(\frac{1}{52}\)
Hint:
n(S) = 52 [1 queen of hearts in 52 cards]
n(A) = 1
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{1}{52}\)

Question 34.
Probability of sure event is ______
(1) 1
(2) 0
(3) 100
(4) 0.1
Answer:
(1) 1

Question 35.
The outcome of a random experiment result in either success or failure. If the probability of success is twice the probability of failure, then the probability of success is ______
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) 1
(4) 0
Answer:
(2) \(\frac{2}{3}\)
Hint:
n(A ∪ B) = 1; P(\(\bar{A}\)) = 1 – P(A)
Given P(A) = 2P(\(\bar{A}\))
P(A) = 2 [1 – P(A)] = 2 – 2 P(A)
3 P(A) = 2
P(A) = \(\frac{2}{3}\)

II. Answer the following questions.

Question 1.
Find the range and the coefficient of range of the following data.

Answer:
Largest value (L) = 690
Smallest value (S) = 610
Range R = L – S = 690 – 610 = 80
Coefficient of range = \(\frac{L-S}{L+S}=\frac{690-610}{690+610}=\frac{80}{1300}\) = 0. 06

Question 2.
Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them
(ii) 5 will come up at both dice
Answer:
S = {(1, 1)(1, 2)(1, 3)(1, 4)(1, 5)(1, 6), (2, 1)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6), (3, 1)(3, 2)(3, 3)(3, 4)(3, 5)(3, 6), (4, 1)(4, 2)(4, 3)(4, 4)(4, 5)(4, 6), (5, 1)(5, 2)(5, 3)(5, 4)(5, 5)(5, 6), (6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)}
n(S) = 36
(i) Let A be the event of getting 5 on either of them.
A = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 5)}
n(A) = 11
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{11}{36}\)
Probability that 5 will not come up on either of them = 1 – P (A)
\(=1-\frac{11}{36}=\frac{36-11}{36}=\frac{25}{36}\)
(ii) Let B be the event of getting 5 will come up at both dice
B = {(5, 5)}
n(B) = 1
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{36}\)

Question 3.
The king, Queen and Jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards, find the probability of getting
(i) a card of clubs
(ii) a queen of diamond
Answer:
Sample space (S) = (52 – 3) = 49
n (S) = 49
(i) Let A be the event of getting a card of clubs.
n(A) = (13 – 3) = 10
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{49}\)
(ii) Let B be the event of getting a queen of diamond
n(B) = 1
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{49}\)

Question 4.
The standard deviation of 20 observations is √5. If each observation is multiplied by 2, find the standard deviation and variance of the resulting observations.
Answer:
Given a standard deviation of 20 observations = √5
Each observation is multiplied by 2 then,
New standard deviation = 2 × √5 = 2√5
New variance = (2√5)² = 20

Question 5.
Calculate the variance standard deviation of the following data 38, 40, 34, 31, 28, 26, 34.
Answer:
Arrange the given data in ascending order we get, 26, 28, 31, 34, 38, 40
Assumed mean = 34


Variance = 22
Standard deviation(σ) = √Variance = √22 = 4.69

Question 6.
Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the times and the sum of the squares of all items.
Answer:
The mean of 100 items = 48
The sum of 100 items (Σx) = 100 × 48 = 4800
standard deviation (σ²) = 10

Sum of the squares of all items (Σx²) = 240400

Question 7.
If n = 10, \(\bar{x}\) = 12 and Σx² = 1530, then calculate the coefficient of variation.
Answer:

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \Rightarrow \frac{3}{12} \times 100=25\)

Question 8.
If the coefficient of variation of a collection of data is 57 and its standard deviation is 6, 84, then find the mean.
Answer:
Given the coefficient of variation = 57
Standard deviation (σ) = 6.84

Arithmetic mean = \(\bar{x}\) = 12

Question 9.
Find the standard deviation and the variance of first 23 natural numbers?
Answer:
Standard deviation of first “n” natural numbers = \(\sqrt{\frac{n^{2}-1}{12}}\)
Standard deviation of first “23” natural numbers = \(\sqrt{\frac{23^{2}-1}{12}}\)
\(=\sqrt{\frac{529-1}{12}}=\sqrt{\frac{528}{12}}=\sqrt{44}\)
= 6.63

Question 10.
Find the coefficient of variation of the following data: 18, 20, 15, 12, 25.
Answer:
Let us calculate the A.M of the given data.


The coefficient of variation is 24.6

Question 11.
Three rotten eggs are mixed with 12 good ones. One egg is chosen at random. What is the probability of choosing a rotten egg?
Answer:
Number of good eggs = 12
Number of rotton eggs = 3
Total number of eggs = 12 + 3 = 15
Sample space n(S) = 15
Let A be the event of choosing a rotten egg
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{15}=\frac{1}{5}\)
The Probability is \(\frac{1}{5}\)

Question 12.
Two coins are tossed together. What is the probability of getting at most one head?
Answer:
Sample space (S) = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Let A be the event of getting atmost one head
A = {HT, TH, TT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
The probability is \(\frac{3}{4}\)

Question 13.
A number is selected at random from integers 1 to 100. Find the probability that it is
(i) a perfect square
(ii) not a perfect cube.
Answer:
Sample space (S) = {1, 2, 3, …,100}
n(S) = 100
(i) Let A be the event of getting a perfect square.
A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(A) = 10
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{100}=\frac{1}{10}\)
(ii) Let B be the event of getting a perfect cube.
B = {1, 8, 27, 64}
n(B) = 4
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{4}{100}=\frac{1}{25}\)
The Probability that the selected number is not a perfect cube is
P(\(\bar{B}\)) = 1 – P(B)
P(\(\bar{B}\)) = 1 – \(\frac{1}{25}\) = \(\frac{24}{25}\)

Question 14.
Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.
Answer:
Sample space (S) = {(1, 1, 1) (1, 1, 2) (1, 1, 3) ….(6, 6, 6)}
n(S) = 63 = 216
Let A be the event of getting the same number on all the three dice.
A = {(1, 1, 1) (2, 2, 2) (3, 3, 3) (4, 4, 4) (5, 5, 5) (6, 6, 6)}
n(A) = 6
P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{216}=\frac{1}{36}\)
The probability is \(\frac{1}{36}\)

Question 15.
If P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{10}\), P(A∪B) = 1, Find
(i) P(A ∩ B)
(ii) P(A’ ∪ B’)
Answer:

III. Answer the following questions.

Question 1.
The mean of the following frequency distribution is 53 and the sum of all frequencies is 100. compute the missing frequencies f₁ and f₂.

Answer:


f₂ = 29
Substitute the value of f₂ = 29 in (1)
f₁ + 29 = 47
⇒ f₁ = 47 – 29 = 18
The value of f₁ = 18 and f₂ = 29

Question 2.
Calculate the standard deviation of the following data.

Answer:
Assumed mean (A) = 13

Standard deviation = 6.32

Question 3.
The time (in seconds) taken by a group to walk across a pedestrian crossing is given in the table below.

Calculate the variance and standard deviation of the data.
Answer:
Assumed mean (A) = 17.5, c = 5


Standard deviation (σ) = √36.76 = 6.063
Variance 36. 76, Standard deviation = 6.06

Question 4.
The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.
Answer:
Given, mean of 20 items (\(\bar{x}\)) = 10
and standard deviation (σ) = 2



(i) Corrected mean = 10. 2
(ii) Corrected Standard deviaton = 1.99

Question 5.
Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.
Answer:
Arrange in ascending order, we get 18, 20, 24, 26, 32
Assumed mean = 24

Question 6.
The marks scored by two students A, B in a class are given below.

Who is more consistent?
Answer:
Student = A
\(\bar{x}\) = 60




From student A and B the coefficient of variation for A is less than the coefficient of variation for B.
Student A is more Consistent.

Question 7.
If for distribution Σ(x – 7) = 3, Σ(x – 7)² = 57 and total number of item is 20. find the mean and standard deviation.
Answer:


(i) Arithmetic mean = 7. 15
(ii) Standard deviation = 1.68

Question 8.
Two unbiased dice are rolled once. Find the probability of getting
(i) a sum 8
(ii) a doublet
(iii) a sum greater than 8
Answer:
When two dice are thrown, the sample space is
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 6 × 6 = 36
(i) Let A be the event of getting a sum 8
A= {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(A) = 5
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{5}{36}\)
(ii) Let B be the event of getting a doublet.
B = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(B) = 6
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum greater than 8
C = {(3, 6) (4, 5) (4, 6) (5, 4) (5, 5) (5, 6) (6, 3) (6, 4) (6, 5) (6, 6)}
n(C) = 10
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{10}{36}=\frac{5}{18}\)

Question 9.
A die is thrown twice. Find the probability that atleast one of the two throws conies up with the number 5 (use addition theorem).
Answer:
In rolling a die twice, the size of the sample space, n(S) = 36
Let A be the event of getting 5 in the first throw.
A = {(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)}
Thus, n(A) = 6, and P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}\)
Let B be the event of getting 5 in the second throw.
B = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)}
Thus, n(B) = 6, and P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{6}{36}\)
A and B are not mutually exclusive events, since A ∩ B = {(5, 5)}
n(A ∩ B) = 1 and P(A ∩ B) = \(\frac{1}{36}\)
By addition theorem,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(=\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}\)

Question 10.
Let A, B, C be any three mutually exclusive and exhaustive events such that P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B). Find P(A).
Answer:
Let P(A) = p
Now, P(B) = \(\frac{3}{2}\), P(A) = \(\frac{3}{2}\) p
Also, P(C) = \(\frac{1}{2}\) P(B)
= \(\frac{1}{2}\) (\(\frac{3}{2}\) p) = (\(\frac{3}{4}\) p)
Given that A, B and C are mutually exclusive and exhaustive events.
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) and S = A ∪ B ∪ C
Now, P(S) = 1
That is, P(A) + P(B) + P(C) = 1
p + \(\frac{3}{2}\) p + \(\frac{3}{4}\) p = 1
⇒ 4p + 6p + 3p = 4
Thus, p = \(\frac{4}{13}\)
Hence, P(A) = \(\frac{4}{13}\)

Question 11.
A bag contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.
Answer:
Sample space (S) = (50 + 150) = 200
n(S) = 200
Number of rusted bolts = \(\frac{1}{2}\) (50) = 25
Number of rusted nuts = \(\frac{1}{2}\) (150) = 75
Let A be the event of getting rusted bolts and nuts.
n (A) = 25 + 75 = 100
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{100}{200}\)
Let B be the event of getting a bolt.
n(B) = 50
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{50}{200}\)
Number of bolts which are rusted n(A ∩ B) = 25

Question 12.
Two dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event of getting the sum is divisible by 3
A = {(1, 2) (2, 1) (1, 5) (5, 1) (2, 4) (4, 2) (3, 3) (3, 6) (6, 3) (4, 5) (5, 4) (6, 6)}
n(A) = 12
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{12}{36}\)
Let B be the event of getting a sum is divisible by 4.
B = {(1, 3) (2, 2) (2, 6) (3, 1) (3, 5) (4, 4) (5, 3) (6, 2) (6, 6)}
n(B) = 9
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{9}{36}\)

Question 13.
In a class, 40% of the students participated in Mathematics-quiz, 30% in Science -quiz and 10% in both the quiz programmes. If a students is selected at random from the class, find the probability that the students participated in Mathematics or science or both quiz programmes.
Answer:
Sample space (S) = 100
n(S) = 100
Let A be the event of getting a number of students participated in mathematics-quiz programme
n(A) = 40
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{40}{100}\)
Let B be the event getting a number of students participated in science-quiz programme
n(B) = 30

\(=\frac{40+30-10}{100}=\frac{60}{100}=\frac{3}{5}\)
The required probability = \(\frac{3}{5}\)

Question 14.
A two digit number is formed with the digits 2, 5, 9 (repetition is allowed). Find the probability that the number is divisible by 2 or 5.
Answer:
Sample space (S) = {22, 25, 29, 55, 52, 59, 99, 92, 95}
n(S) = 9
Let A be the event of getting number is divisible by 2.
A = {22, 52, 92}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{9}\)
Let B be the event of getting number is divisible by 5.
B = {25, 95, 55}
n(B) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{9}\)
If A and B are mutually exclusive.
P(A ∪ B) = P(A) + P(B)
P(A ∪ B) = \(\frac{3}{9}+\frac{3}{9}=\frac{6}{9}=\frac{2}{3}\)
The required probability is \(\frac{2}{3}\)

Question 15.
The probability that A, B and C can solve a problem are \(\frac{4}{5}\), \(\frac{2}{3}\) and \(\frac{3}{7}\) respectively. The probability of the problem being solved by A and B is \(\frac{8}{15}\), B and C is \(\frac{2}{7}\), A and C is \(\frac{12}{35}\). The probability of the problem being solved by all the three is \(\frac{8}{35}\). Find the probability that the problem can be solved by atleast one of them.
Answer:
Given P(A) = \(\frac{4}{5}\)
P(B) = \(\frac{2}{3}\)
P(C) = \(\frac{3}{7}\)
P(A ∩ B) = \(\frac{8}{15}\)
P(B ∩ C) = \(\frac{2}{7}\)
P(A ∩ C) = \(\frac{12}{35}\)
P(A ∩ B ∩ c) = \(\frac{8}{35}\)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

The probability of the problem can be solved by at least one of them = \(\frac{101}{105}\)

Students can download Maths Chapter 8 Statistics and Probability Unit Exercise 8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Unit Exercise 8

Question 1.
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂.

Answer:
Arithmetic Mean (\(\bar{x}\)) = 62.8
Sum of all the frequencies (Σf_i) = 50
Let the missing frequencies be f₁ and f₂


Substitute of the value of f₂ in (1)
f₁ + 12 = 20
⇒ f₁ = 20 – 12 = 8
The Missing frequency is 8 and 12.

Question 2.
The diameter of circles (in mm) drawn in the design are given below.

Calculate the standard deviation.
Answer:
Assumed mean = 42. 5

Question 3.
The frequency distribution is given below

In table k is a positive integer, has a variance of 160. Determine the value of k.
Answer:
Assumed mean = 3k



The value of k = 7

Question 4.
The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?
Solution:
F° = (C° × 1.8) + 32
\(\begin{array}{l}{\sigma_{c}=5^{\circ} \mathrm{C}} \\ {\sigma_{\mathrm{F}}=\left(1.8 \times 5^{\circ} \mathrm{C}\right) \cdot 9^{\circ} \mathrm{F}}\end{array}\)
Adding value to data doesn’t affect standard deviation.
New variance = \(\sigma_{\mathrm{F}}^{2}=81^{\circ} \mathrm{F}\)

Question 5.
If for a distribution, Σ (x – 5) = 3, Σ (x – 5)² = 43, and total number of observations is 18, find the mean and standard deviation.
Answer:


(i) Arithmetic mean (\(\bar{x}\)) = 5.17
(ii) Standard deviation (σ) = 1.53

Question 6.
Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Answer:
Coefficients of variation of prices in city A.
Arrange in ascending order we get, 16, 19, 20, 22, 23
Assumed mean = 20

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100\)
= \(\frac{2.45}{20} \times 100\)
= 12. 25%
Coefficient of variation = 12.25%
Coefficients of variation of prices in city B.
Arrange in ascending order we get, 10, 12, 15, 18, 20
Assumed mean = 15


Prices in city A is more stable (since 12.25 < 24.6 %)

Question 7.
If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Answer:
Range of the data (R) = 20
L – S = 20 ……(1)
Coefficient of range = 0.2
Coefficient of range = \(\frac{L-S}{L+S}\)
0.2 = \(\frac{20}{L+S}\)

substituting the value of L = 60 in (2)
60 + S = 100
S = 100 – 60 = 40
Largest value = 60
Smallest value = 40

Question 8.
If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Answer:
Sample space = {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6), (3, 1), (3, 2), (3, 3),(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1),(6, 2), (6, 3), (6, 4),(6, 5), (6, 6)}
n(S) = 36
(i) Let A be the event of getting product of face value 6.
A = {(1, 6), (2, 3), (3, 2) (6, 1)}
n(A) = 4
P (A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{4}{36}\)
(ii) Let B be the event of getting difference of face value is 5.
B = {(6, 1)}
n(B) = 1

The probability is \(\frac{1}{9}\)

Question 9.
In a two children family, find the probability that there is at least one girl in a family.
Answer:
Sample space (S) = {(Boy, Boy) (Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(S) = 4
Let A be the event of getting atleast one Girl
A = {(Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
Probability of atleast one girl in a family is \(\frac{3}{4}\)

Question 10.
A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Answer:
Let the number of black balls be “x”
Sample space (S) = x + 5
n(S) = x + 5
Let A be the event of drawing a black ball
n (A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{x}{x+5}\)
Let B be the event of getting a white ball
n(B) = 5
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{5}{x+5}\)
By the given condition,
\(\frac{x}{x+5}=2 \times\left(\frac{5}{x+5}\right)\)
⇒ \(\frac{x}{x+5}=\left(\frac{10}{x+5}\right)\)
⇒ 10x + 50 = x² + 5x
⇒ x² + 5x – 10x – 50 = 0
⇒ x² – 5x – 50 = 0
⇒ (x – 10)(x + 5) = 0
⇒ x = 10 or x = -5
Number of balls will not be negative.
Number of black balls = 10

Question 11.
The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Answer:
Let A be the event of getting student pass in English
Let B be the event of getting student pass in Tamil


Probability of passing the tamil examination is \(\frac{13}{20}\)

Question 12.
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Answer:
Total number of cards = 52
3 cards are removed
Remaining number of cards = 52 – 3 = 49
n(S) = 49
(i) Let A be the event of getting a diamond card.
n(A) = 13
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{13}{49}\)
(ii) Let B be the event of getting a queen card.
n(B) = (4 – 1) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{49}\)
(iii) Let C be the event of getting a spade card.
n(C) = (13 – 3) = 10
P(C) = \(\frac{n(C)}{n(S)}=\frac{10}{49}\)
(iv) Let D be the event of getting a 5 of heart card.
n(D) = 1
P(D) = \(\frac{n(\mathrm{D})}{n(\mathrm{S})}=\frac{1}{49}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.5

Multiple Choice Questions.

Question 1.
Which of the following is not a measure of dispersion?
(1) Range
(2) Standard deviation
(3) Arithmetic mean
(4) Variance
Answer:
(3) Arithmetic mean
Hint:
Measures of dispersion are,
(i) Range
(ii) Mean deviation
(iii) Quartile deviation
(iv) Standard deviation
(v) Variance
(vi) coefficient of variation

Question 2.
The range of the data 8, 8, 8, 8, 8.. . 8 is
(1) 0
(2) 1
(3) 8
(4) 3
Solution:
(1) 0
Hint
R = L – S = 8 – 8 = 0

Question 3.
The sum of all deviations of the data from its mean is _______
(1) always positive
(2) always negative
(3) zero
(4) non-zero integer
Answer:
(3) zero

Question 4.
The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is
(1) 40000
(2) 160900
(3) 160000
(4) 30000
Solution:
(2) 160900
Hint:
σ = 3

Question 5.
Variance of first 20 natural numbers is ______
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25
Hint:
Variance of 20 natural numbers is

Question 6.
The standard deviation of a data Is 3. If each value is multiplied by 5 then the new variance is
(1) 3
(2) 15
(3) 5
(4) 225
Solution:
σ = 3. 1f each is multiplied by 5. The new standard variation is also multiplied by 3.
∴ The new S.D = 5 × 3 = 15
Variance = 15² = 225

Question 7.
If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is ________
(1) 3p + 5
(2) 3p
(3) p + 5
(4) 9p + 15
Answer:
(2) 3p
Hint:
(i) Each value is added by any constant there is no change in standard deviation.
(ii) Each value is multiplied by 3 standard deviations also multiplied by 3.
The standard deviation is 3p.

Question 8.
If the mean and coefficient of variation of a data are 4 and 87.5% then the standard
deviation is
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Solution:
(1) 3.5
Hint:
\(\bar{x}\) = 4, coefficient of variation is = 87. 5%

Question 9.
Which of the following is incorrect?
(1) P(A) > 1
(2) 0 ≤ P(A) ≤ 1
(3) P(Φ) = 0
(4) P(A) + P(\(\bar{A}\)) = 1
Answer:
(1) P(A) > 1
Hint:
Probability is always less than one or equal to one.

Question 10.
The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is _________
(1) \(\frac{q}{p+q+r}\)
(2) \(\frac{p}{p+q+r}\)
(3) \(\frac{p+q}{p+q+r}\)
(4) \(\frac{p+r}{p+q+r}\)
Answer:
(1) \(\frac{q}{p+q+r}\)
Hint:
Sample spaces = p + q + r
Let A be the event of getting red
n(A) = p
P(A) = \(\frac{q}{p+q+r}\)

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is _______
(1) \(\frac{3}{10}\)
(2) \(\frac{7}{10}\)
(3) \(\frac{3}{9}\)
(4) \(\frac{7}{9}\)
Answer:
(2) \(\frac{7}{10}\)
Hint:
Here n(S)= 10 (given digit at imit place. It has two digit)
n(A) = 7
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{7}{10}\)

Question 12.
The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is.
(1) 2
(2) 1
(3) 3
(4) 1.5
Solution:
(2) 1
Hint:
Probability of getting a job = \(\frac{x}{3}\)
Probability of not getting a job = 1 – \(\frac{x}{3}\)

Question 13.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac{1}{9}\), then the number of tickets bought by Kamalam is _______
(1) 5
(2) 10
(3) 15
(4) 20
Answer:
(3) 15
Hint:
n(S) = 135
P(A) = \(\frac{1}{9}\)

Question 14.
If a letter is chosen at random from the English alphabets {a, b, …, z}, then the probability that the letter chosen precedes x.
(1) \(\frac{12}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{23}{26}\)
(4) \(\frac{3}{26}\)
Solution:
(3) \(\frac{23}{26}\)
Hint:
n(S) = 26
Let A denote the letter chosen precedes x
A= {a, b, c, d, …, x}
n(A) = 23
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{23}{26}\)

Question 15.
A purse contains 10 notes of ₹ 2000, 15 notes of ₹ 500, and 25 notes of ₹ 200. One note is drawn at random. What is the probability that the note is either a ₹ 500 note or ₹ 200 note?
(1) \(\frac{1}{5}\)
(2) \(\frac{3}{10}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{5}\)
Answer:
(4) \(\frac{4}{5}\)
Hint:
Sample space (S) = 10 + 15 + 25 = 50
n(S) = 50
Let A be the event of getting ₹ 500 note
n (A) = 15
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{15}{50}\)
Let B be the event of getting ₹ 200 note
n (B) = 25
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{25}{50}\)
Probability of the note is either a ₹ 500 note or ₹ 200 note
P(A) + P(B) = \(\frac{15}{50}+\frac{25}{50}\) = \(\frac{40}{50}\) = \(\frac{4}{5}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.4

Question 1.
If P (A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A ∪ B) = \(\frac{1}{3}\), then find P(A ∩ B).
Answer:
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)
\(\frac{1}{3}=\frac{2}{3}+\frac{2}{5}\) – P (A ∩ B)

Question 2.
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B)=016. Find (i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Solution:
(a) P(A) = 0.42 ;
P(B) = 0.48
P(A∩B) = 0.16
(i) P(not A) = P(\(\overline{\mathbf{A}}\)) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = P(\(\overline{\mathbf{B}}\)) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.42 + 0.48 – 0.16
= 0.74

Question 3.
If A and B are two mutually exclusive events of a random experiment and P (not A) = 0.45, P (A ∪ B) = 0.65, then find P(B).
Answer:
P(not A) = 0.45
1 – P (A) = 0.45
P (A) = 1 – 0.45 = 0.55
P(A ∪ B) = P (A) + P (B)
0. 65 = 0.55 + P(B)
0. 65 – 0.55 = P(B)
0.10 = P (B)
P(B) = 0.1

Question 4.
The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(\(\overline{\mathbf{A}}\)) + P(\(\overline{\mathbf{B}}\)).
Solution:
P(A∪B) = 0.6
P(A∩B) = 0.2
P(A) + P(B) = [1 – P(A∪B)] + [1 – P(A∩B)] = [1 – 0.6] + [1 – 0.2]
= 0.4 + 0.8 = 1.2

Question 5.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen.
Answer:
Here P(A) = 0.5, P (B) = 0.3
P(A ∪ B) = P (A) + P(B) [A and B are mutually exclusive]
= 0.5 + 0.3
= 0.8
Probability of neither A nor [P(A ∪ B)’] = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (S) = 36
Let A be the event of getting an even number on the first time
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (A) = 18
\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\)
(ii) Let B be the event of getting a total of face sum 8.
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{36}\)
A ∩ B = {(2, 6) (4, 4) (6, 2)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)

The required probability = \(\frac{5}{9}\)

Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen.
Answer:
n(S) = 52
Let A be the event of getting a red king
n(A) = 2
\(P(A)=\frac{n(A)}{n(S)}=\frac{2}{52}\)
Let B be the event of getting a black Queen king
n(B) = 2
\(P^{\prime}(B)=\frac{n(B)}{n(S)}=\frac{2}{52}\)
It A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
\(=\frac{2}{52}+\frac{2}{52}=\frac{4}{52}=\frac{1}{13}\)
The required probability is \(\frac{1}{13}\)

Question 8.
A box contains cards numbered 3, 5, 7, 9,… 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Answer:
Sample space = {3, 5, 7, 9,…….,35, 37}
n(S) = 18
Let A be the event of getting a multiple of 7
A = {7, 21, 35}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{18}\)
Let B be the event of getting a prime number
B = {3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

Probability of getting a multiple of 7 or a prime number = \(\frac{13}{18}\)

Question 9.
Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting atmost 2 tails.
A = {HTT, THT, TTH, HHT, HTH, THH, HHH}
n(A) = 7

Probability of getting atmost two tails or atleast 2 heads = \(\frac{7}{8}\)

Question 10.
The probability that a person will get an electrification contract is \(\frac{3}{5}\) and the probability that he will not get plumbing contract is \(\frac{5}{8}\). The probability of getting atleast one contract is \(\frac{5}{7}\). What is the probability that he will get both?
Answer:
Let A and B represent the event of getting electrification control and plumbing contract.


Probability of getting both the job is \(\frac{73}{280}\)

Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Answer:
Total number of people in a town is 8000.
n(S) = 8000
Total number of females = 3000
Let A be the event of getting number of females
n(A) = 3000
\(P(A)=\frac{n(A)}{n(S)}=\frac{3000}{8000}\)
Number of people over 50 years = 1300
Let B be the event of getting number of people over 50 years.
n(B) = 1300
\(P(B)=\frac{n(B)}{n(S)}=\frac{1300}{8000}\)
Given 30% of the females are over 50 years.


Proability of getting either a female or over 50 years = \(\frac{17}{40}\)

Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}\)
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3



The probability is 1.

Question 13.
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P (A ∩ B) = \(\frac{1}{6}\), P(B ∩ C) = \(\frac{1}{4}\), P(A ∩ C) = \(\frac{1}{8}\), P(A ∪ B ∪ C) = \(\frac{9}{10}\) and P (A ∩ B ∩ C) = \(\frac{1}{15}\), then find P(A), P(B) and P(C)?
Answer:
By the given condition,
P(B) = 2 P(A), P(C) = 3 P(A)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

Question 14.
In a class of 35, students are numbered from 1 to 35. The ratio of boys and girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Answer:
Sample space (S) = {1, 2, 3,… ,35}
n(S) = 35
Total number of students = 35
Number of boys = \(\frac{4}{7}\) × 35 = 20 [Boys Numbers = {1, 2, 3,…, 20}]
Number of girls = \(\frac{3}{7}\) × 35 = 15 [Girls Numbers = { 21, 22,…, 35}]
Let A be the event of getting a boy role number with prime number
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\frac{8}{35}\)
Let B be the event of getting girls roll number with composite number.
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}
n(B) = 12
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}\) = \(\frac{12}{35}\)
Let C be the event of getting an even roll number.
C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}
n(C) = 17


Probability of getting roll number is \(\frac{29}{35}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.3

Question 1.
Write the sample space for tossing three coins using tree diagram.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Question 3.
If A is an event of a random experiment such that P(A) : P(\(\bar{A}\)) = 17 : 15 and n(s) = 640 then find (i) P(\(\bar{A}\))
(ii) n(A)
Answer:

Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting consecutive tails
A = {HTT, TTH, TTT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}\)

Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Solution:
22² = 484
31² = 961
23² = 529
32² = 1024
23, 24, 25, 26, 27, 28, 29, 30, 31 has squares below 500 × 1000.
(i) P(first player wins a prize) = \(\frac{9}{1000}\)
(ii) P(second player ins if first has won) = \(\frac{8}{999}\)

Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Answer:
Sample space = 12 + x
n(S) = x + 12
(i) Let A be the event of getting a red ball
n(A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\left(\frac{x}{x+12}\right)\)
(ii) 8 more red balls are added
Sample space = x + 12 + 8 = x + 20
Number of red balls = x + 8
Probability of drawing red ball = \(\frac{x+8}{x+20}\)
By the given condition
\(\frac{x+8}{x+20}=2\left(\frac{x}{x+12}\right)\)
(x + 8)(x + 12) = 2x(x + 20)
x² + 20x + 96 = 2x² + 40x
x² + 20x – 96 = 0
(x + 24)(x – 4) = 0
x = -24 (or) x = 4
The value of x = 4 (Number of balls will not be negative)
The probability of getting red balls = \(\left(\frac{x}{x+12}\right)=\frac{4}{16}=\frac{1}{4}\)

Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Answer:
(i) Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event of getting doublet
A = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting a product is a prime number.
B = {(1, 2) (1, 3) (1, 5) (2, 1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum is a prime number
C = {(1, 1) (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2), (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)}
n(C) = 15
\(P(C)=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting a sum is 1
n(D) = 0
\(P(D)=\frac{n(D)}{n(S)}=\frac{0}{36}=0\)
Probability of getting a sum is 1 is 0

Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Solution:
Possible outcomes = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
No. of possible outcomes = 2 × 2 × 2 = 8
(i) Prob(all heads) = \(\frac{1}{8}\)
(ii) Atleast one tail = {(THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
Prob(atleast one tail) = \(\frac{7}{8}\)
(iii) Atmost one head = {(HTT), (THT), (TTH), (TTT)}
∴ Prob(atmost one head) = \(\frac{4}{8}=\frac{1}{2}\)
(iv) Atmost two tail = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT)}
∴ Prob(atmost two tail) = \(\frac{7}{8}\)

Question 9.
Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Answer:
1st dice A = {1, 2, 3, 4, 5, 6}
2nd dice B = {1, 1, 2, 2, 3, 3}
Sample Space (S) = {(1, 1), (1, 1), (1, 2), (1, 2), (1, 3), (1, 3), (2, 1), (2, 1), (2, 2), (2, 2), (2, 3), (2, 3), (3, 1), (3, 1), (3, 2), (3, 2), (3, 3), (3, 3), (4, 1), (4, 1), (4, 2), (4, 2), (4, 3), (4, 3), (5, 1), (5, 1), (5, 2), (5, 2), (5, 3), (5, 3),(6, 1), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3)}
n(S) = 36
(i) Let A₁ be the event of getting sum is 2
A1 = {(1, 1) (1, 1)}
n(A₁) = 2
\(P\left(A_{1}\right)=\frac{n\left(A_{1}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)
(ii) Let A₂ be the event of getting a sum is 3.
A₂ = {(1, 2) (1, 2) (2, 1) (2, 1)}
n(A₂) = 4
\(P\left(A_{2}\right)=\frac{4}{36}=\frac{1}{9}\)
(iii) Let A₃ be the event of getting a sum is 4.
A₃ = {(1, 3) (1, 3) (2, 2) (2, 2) (3, 1) (3, 1)}
n(A₃) = 6
\(P\left(A_{3}\right)=\frac{6}{36}=\frac{1}{6}\)
(iv) Let A₄ be the event of getting a sum is 5.
A₄ = {(2, 3) (2, 3) (3, 2) (3, 2) (4, 1) (4, 1)}
n(A₄) = 6
\(P\left(A_{4}\right)=\frac{6}{36}=\frac{1}{6}\)
(v) Let A₅ be the event of getting a sum is 6.
A₅ = {(3, 3) (3, 3) (4, 2) (4, 2) (5, 1) (5, 1)}
n(A₅) = 6
\(P\left(A_{5}\right)=\frac{6}{36}=\frac{1}{6}\)
(vi) Let A₆ be the event of getting a sum is 7.
A₆ = {(4, 3) (4, 3) (5, 2) (5, 2) (6, 1) (6, 1)}
n(A₆) = 6
\(P\left(A_{6}\right)=\frac{6}{36}=\frac{1}{6}\)
(vii) Let A₇ be the event of getting a sum is 8.
A₇ = {(5, 3) (5, 3) (6, 2) (6, 2)}
n(A₇) = 4
\(P\left(A_{7}\right)=\frac{4}{36}=\frac{1}{9}\)
(viii) Let A₈ be the event of getting a sum is 9.
A₈ = {(6, 3) (6, 3)}
n(A₈) = 2
\(P\left(A_{8}\right)=\frac{2}{36}=\frac{1}{18}\)

Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Answer:
Sample space (S) = 5 + 6 + 7 + 8
n(S) = 26
(i) Let A be the event of getting a white ball
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(P(A)=\frac{6}{26}=\frac{3}{13}\)
(ii) Let A be the event of getting a black ball
n(A) = 8
\(P(A)=\frac{n(A)}{n(S)}=\frac{8}{26}\)
Let B be the event of getting a red ball
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{26}\)
Probability of getting black or red ball
P(A ∪ B) = P (A) + P (B)
= \(\frac{8}{26}+\frac{5}{26}=\frac{13}{26}=\frac{1}{2}\)
(iii) Not white probability of getting white ball
P(A) = \(\frac{3}{13}\) from (i)
Probability of not getting white ball P(\(\bar{A}\)) = 1 – P(A)
\(1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}\)
(iv) Probability of getting a white ball.
P(A) = \(\frac{6}{26}\) (from 1)
Let B be the event of getting a black ball
n(B) = 8
\(P(B)=\frac{n(B)}{n(S)}=\frac{8}{26}\)
P(A ∪ B) = P(A) + P(B) = \(\frac{6}{26}+\frac{8}{26}=\frac{14}{26}\)
Probability of neither white nor black P(A ∪ B)’ = 1 – P(A ∪ B)
= \(1-\frac{14}{26}\)
= \(\frac{26-14}{26}=\frac{12}{26}=\frac{6}{13}\)

Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is \(\frac{3}{8}\) then, find the number of defective bulbs.
Answer:
Let the number of defective bulbs be “x”
Sample space (S) = 20 + x
n(S) = 20 + x
Let A be the event of getting to be defective
n(A) = x
\(P(A)=\frac{n(A)}{n(S)}\)
⇒ \(\frac{3}{8}=\frac{x}{20+x}\)
⇒ 8x = 3(20 + x) = (60 + 3x)
⇒ 8x – 3x = 60
⇒ 5x = 60
⇒ x = \(\frac{60}{5}\)
⇒ x = 12
Number of defective bulbs = 12

Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Answer:
King diamond + Queen diamonds = 1 + 1 = 2 …….(1)
Queen hearts + Jack of hearts = 1 + 1 = 2 …….(2)
Jack spade + King of spades =1 + 1 = 2 …….(3)
Remaining number of cards = 52 – (6)
n(S) = 46
(i) Let A be the event of getting a clavor
n (A) = 13
\(P(A)=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting a queen of red card
n(B) = 2
But the above two cards are removed from (1) and (2)
n(B) = 0
\(P(B)=\frac{n(B)}{n(S)}=\frac{0}{46}=0\)
(iii) Let B be the event of getting a king of black card
n(B) = (2 – 1) [from (3) one black card is removed]
n (B) = 1
\(P(B)=\frac{n(B)}{n(S)}=\frac{1}{46}\)

Question 13.
Some boys are playing a game, in which the stone thrown by them landing in a circular region given in the figure is considered as win and landing other than the circular region is considered as a loss. What is the probability to win the game?
Answer:

Area of a rectangle = l × b sq. feet = 3 × 4 sq. feet = 12 sq. feet
sample space (S) = 12
n(S) = 12
Let A be the event of getting the stone landing in a circular region
n(A) = Area of a circle
= πr²
= π × 1 × 1 (radius of a circle = 1 feet)
= π

Probability to win the game = \(\frac{11}{42}\) (or) \(\frac{157}{600}\)

Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Answer:
Sample space (S) = 6 × 6 = 36
n(S) = 36
[priya and Amuthan are visiting a particular shop in any one of 6 days is 6 × 6 = 36]
(i) Let A be the event of getting both are shopping on the same day
A = {(Mon, Mon) (Tue, Tue) (Wed, Wed) (Thu, Thu) (Fri, Fri) (Sat, Sat)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of shopping in different days.
n(B) = 36 – 6 = 30
\(P(B)=\frac{n(B)}{n(S)}\)
\(=\frac{30}{36}=\frac{5}{6}\)
(iii) Let C be the event of shopping consecutive days
C = {(Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat)}
n(C) = 5
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}\) = \(\frac{5}{36}\)

Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Answer:
Sample space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting double entry fee (only getting 3 heads)
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting her entry fee (one or two heads to show)
n(B) = Probability of one head + Probability of 2 head
= \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)
(iii) To loss the entry means not getting the head (only tail)
n(C) = 1
\(P(C)=\frac{n(C)}{n(S)}=\frac{1}{8}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2

Question 1.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (\(\bar{x}\)) = 12.5
Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \%\)
= \(\frac{6.5}{12.5} \times 100 \%=52 \%\)
Coefficient of variation = 52%

Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer:
Standard deviation (σ) = 1.2
Coefficient of variation = 25.6

Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Answer:
Mean (\(\bar{x}\)) = 15
Co efficient of variation = 48

Question 4.
If n = 5, \(\bar{x}\) = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:

Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Answer:
Arrange in ascending order we get 24, 26, 29, 31, 33, 37
Assumed mean = 29

Question 6.
The time taken (in minutes) to complete homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Answer:
Arrange in ascending order we get, 38, 40, 43, 44, 46, 47, 49, 53.
Assumed mean = 46


= \(\frac{453}{45}\)%
= 10.066
Coefficient of variation = 10.07%

Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation of 4.6 and 2.4 respectively. Who is more consistent in performance?
Answer:
Total marks scored by sathya = 460
Total marks scored by vidhya = 480
Number of subjects = 5
Mean marks of sathya = \(\frac{460}{5}\)
\(\bar{x}\) = 92%
Given standard deviation, (σ) = 4.6

Vidhya coefficient of variation is less than Sathya.
Vidhya is more consistent.

Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.

Which of the three subjects shows the highest variation and which shows the lowest variation in marks?
Answer:
(i) Mathematics:
Mean (\(\bar{x}\)) = 56
Standard deviation (σ) = 12
Coefficient variation (CV₁) = \(\frac{\sigma}{\bar{x}} \times 100=\frac{12}{56} \times 100\)

Science shows the highest variation
Social science shows the lowest variation

Question 9.
The temperature of two cities A and B in the winter season are given below.

Find which city is more consistent in temperature changes?
Answer:
(i) city A:
Assumed mean = 22




C.V of city A < C.V of city B
City A is more consistent in temperature change.

Students can download Maths Chapter 7 Mensuration Unit Exercise 7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Unit Exercise 7

Question 1.
The barrel of a fountain-pen cylindrical in shape is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one-fifth of a litre?
Answer:

Question 2.
A hemispherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemispherical tank = 1.75 m
Volume of the tank = \(\frac{2}{3} \pi r^{3}\) cu.units

Time taken = \(\frac{11229.17}{7}\) = 1604.17 seconds = 26.74 minutes = 27 minutes (approximately)

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Answer:
Radius of a cone = Radius of a hemisphere = r unit

Height of a cone = r units
(height of the cone = radius of a hemisphere)
Maximum volume of the cone

Question 4.
An oil funnel of the tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion by 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Answer:

Total height of oil funnel = 22 cm
Height of the cylindrical portion = 10 cm
Height of the frustum (h) = 22 – 10 = 12 cm
Radius of the cylindrical portion = 4 cm
Radius of the bottom of the frustum = 4 cm
Top radius of the funnel (frustum) = \(\frac{18}{2}\) = 9 cm
Area of the tin sheet required = C.S.A of the frustum + C.S.A of the cylinder
= π (R + r) l + 2πrh sq. units.
= [π(9 + 4) \(\sqrt{12^{2}+(9-4)^{2}}\) + 2π × 4 × 10] cm²
= π [13 × \(\sqrt{144+25}\) + 25 + 80] cm²
= \(\frac{22}{7}\) [13 × 13 + 80] cm²
= \(\frac{22}{7}\) [169 + 80] cm²
= \(\frac{22}{7}\) × 249 cm²
= 782.57 cm²
Area of sheet required to make the funnel = 782.57 cm²

Question 5.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer:
Radius of the cylinder = \(\frac{4.5}{2}\) cm
Height of the cylinder = 10 cm
Volume of the cylinder = πr²h cu. units


Number of coins = 450

Question 6.
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder.
Answer:
External radius of the hollow cylinder R = 4.3 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be “x”
Volume of the solid cylinder = Volume of the hollow cylinder

Diameter of the solid cylinder = 2 × 2.4 = 4.8 cm

Question 7.
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m.
Answer:
Slant height of a frustum (l) = 4 m
Perimeter of the top part = 18 m
2πR = 18


Cost of painting = ₹ 100 × 68 = ₹ 6800

Question 8.
A hemispherical hollow bowl has material of volume cubic \(\frac{436 \pi}{3}\) cubic cm. Its external diameter is 14 cm. Find its thickness.
Answer:
External radius of a hemisphere (R) = 7 cm
Volume of a hemi-spherical bowl = \(\frac{436 \pi}{3}\) cm³

Internal radius = 5 cm
Thickness of the hemisphere = (7 – 5) cm = 2 cm

Question 9.
The volume of a cone is 1005\(\frac{5}{7}\) cu. cm. The area of its base is 201\(\frac{1}{7}\) sq. cm. Find the slant height of the cone.
Answer:
Area of the base of a cone = 201\(\frac{1}{7}\) sq. cm

Question 10.
A metallic sheet in the form of a sector of a circle of radius 21 cm has a central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Answer:

Radius of a cone (r) = 21 cm
Central angle (θ) = 216°
Let “R” be the radius of a cone
Circumference of the base of a cone = arc length of the sector
2πR = \(\frac{\theta}{360} \times 2 \pi r\)

Students can download Maths Chapter 7 Mensuration Ex 7.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.5

Multiple Choice Questions

Question 1.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is
(1) 60π cm²
(2) 68π cm²
(3) 120π cm²
(4) 136π cm²
Solution:
(4) 13671 cm²
Hint:
Here, h = 15 cm, r = 8 cm

C.S.A of a cone = πrl sq. units. = π × 8 × 17 = 136π cm³

Question 2.
If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is
(1) 4πr² sq. units
(2) 6πr² sq. units
(3) 3πr² sq. units
(4) 8πr² sq. units
Answer:
(1) 4πr² sq. units
Hint:
When you joined two hemispheres together, the solid sphere is formed
C.S.A of the new solid = C.S.A of a sphere = 4πr² sq. units.

Question 3.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Solution:
(1) 12 cm
Hint:
Here r = 5 cm and l = 13 cm

Question 4.
If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is _________
(1) 1 : 2
(2) 1 : 4
(3) 1 : 6
(4) 1 : 8
Answer:
(2) 1 : 4
Hint:
Let the radius of the cylinder be “r” and the height be “h”
Radius of the new cylinder = \(\frac{r}{2}\) (Height will be same)
Volume of the new cylinder : Volume of the original cylinder
= \(\pi r_{1}^{2} h: \pi r_{2}^{2} h\) (πh is same)
= \(r_{1}^{2}: r_{2}^{2}\)
= \(\left(\frac{r}{2}\right)^{2}: r^{2}\)
= \(\frac{r^{2}}{4}: r^{2}=\frac{1}{4}: 1\)
= 1 : 4

Question 5.
The total surface area of a cylinder whose radius is \(\frac{1}{3}\) of its height is _______
(1) \(\frac{9 \pi h^{2}}{8}\) sq. units
(2) 24πh² sq.units
(3) \(\frac{8 \pi h^{2}}{8}\) sq.units
(4) \(\frac{56 \pi h^{2}}{8}\) sq.units
Answer:
(3) \(\frac{8 \pi h^{2}}{8}\) sq.units
Hint:
Let the height of the cylinder be “h”
Radius of the cylinder = \(\frac{1}{3}\) h
T.S.A of the cylinder = 2πr(h + r)

Question 6.
In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is
(1) 5600π cm³
(2) 11200π cm³
(3) 56π cm³
(4) 3600π cm³
Solution:
(2) 112007π cm³
Hint:
Here, let the external radius be “R” and the internal radius be “r”
R + r = 14 ……(1)
Width (R – r) = 4 ……(2)
Height of the hollow cylinder = 20 cm
Volume of the hollow cylinder = πh × (R² – r²)
= πh(R + r) (R – r)
= π × 20 (14) × 4
= π × 1120
= 1120π cm³

Question 7.
If the radius of the base of a cone is tripled and the height is doubled then the volume is ______
(1) made 6 times
(2) made 18 times
(3) made 12 times
(4) unchanged
Answer:
(2) made 18 times
Hint:
Radius of a cone = r
Height of a cone = h
Volume of the cone = \(\frac{1}{3}\) πr²h cu. units
When the radius is increased three-time (tripled) and the height is doubled
Radius is 3r and the height is 2h
Volume of the new cone

Volume is increased 18 times.

Question 8.
The total surface area of a hemisphere is how many times the square of its radius.
(1) π
(2) 4π
(3) 3π
(4) 2π
Solution:
(3) 3π
Hint:
T.S.A of the hemisphere = 3πr²
The square of the radius is 3π times.

Question 9.
A solid sphere of radius x cm is melted and cast into a shape of a solid cone of the same radius. The height of the cone is _______
(1) 3x cm
(2) x cm
(3) 4x cm
(4) 2x cm
Answer:
(3) 4x cm
Hint:
Radius of a sphere = Radius of a cone = x cm
Volume of a cone = Volume of a sphere

Question 10.
A frustum of a right circular cone is of height 16cm with radii of its ends as 8cm and 20cm. Then, the volume of the frustum is
(1) 3328π cm³
(2) 3228π cm³
(3) 3240πcm³
(4) 3340π cm³
Solution:
(1) 3328π cm³
Hint:
Here, h = 16 cm, r = 8 cm, R = 20 cm
Volume of the frustum

Question 11.
A shuttlecock used for playing badminton has the shape of the combination of ______
(1) a cylinder and a sphere
(2) a hemisphere and a cone
(3) a sphere and a cone
(4) frustum of a cone and a hemisphere
Answer:
(4) frustum of a cone and a hemisphere
Hint:

Question 12.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is _______
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer:
(1) 2 : 1
Hint:
Volume of the first sphere : Volume of second sphere = 8 : 1

Question 13.
The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is ________
(1) \(\frac{4}{3} \pi\)
(2) \(\frac{10}{3} \pi\)
(3) 5π
(4) \(\frac{20}{3} \pi\)
Answer:
(1) \(\frac{4}{3} \pi\)
Hint:
Radius of the sphere = 1 cm
Volume of the Sphere = \(\frac{4}{3}\) πr³ cu. units
= \(\frac{4}{3}\) × π × 1 × 1 × 1 cm³
= \(\frac{4}{3}\) π cm³

Question 14.
The height and radius of the cone of which the frustum is a part are h₁ units and r₁ units respectively. Height of the frustum is h₂ units and the radius of the smaller base is r₂ units. If h₂ : h₁ = 1 : 2 then r₂ : r₁ is ______
(1) 1 : 3
(2) 1 : 2
(3) 2 : 1
(4) 3 : 1
Answer:
(2) 1 : 2
Hint:
h₂ : h₁ = 1 : 2
h₁ : h₂ = 2 : 1
Ratio of their volumes

Volume is 2 : 1 the ratio of their radius also 2 : 1
r₁ : r₂ = 2 : 1 But r₂ : r₁ = 1 : 2

Question 15.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is
(1) 1 : 2 : 3
(2) 2 : 1 : 3
(3) 1 : 3 : 2
(4) 3 : 1 : 2
Solution:
(4) 3 : 1 : 2
Hint:
Volume of (cylinder : cone : sphere)

Students can download Maths Chapter 7 Mensuration Ex 7.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.4

Question 1.
An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.
Answer:
Sphere – Radius r₁ = 12 cm
Cylinder – Radius r₂ = 8 cm
h₂ = ?
Volume of cylinder = Volume of sphere melted

∴ Height of the cylinder made = 36 cm.

Question 2.
Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
Answer:
Length of the rectangular tank (l) = 50 m = 5000 cm
Width of the rectangular tank (b) = 44 m = 4400 cm
Level of water in the tank (h) = 21 cm
Volume of the tank = l × b × h cu. units = 5000 × 4400 × 21 cm³
Radius of the pipe (r) = 7 cm
Speed of the water = 15 km/hr.
(h) = 15000 × 100 cm / hr.
Volume of water flowing in one hour

Question 3.
A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius x r units. Find the height of water in the cylindrical flask.
Answer:
Radius of the conical flask = r units
Height of the conical flask = h units
Volume of the conical flask = \(\frac{1}{3} \pi r^{2} h\) cu.units
Radius of the cylindrical flask = x r units
Let the height of the cylindrical flask be “H” units
Volume of the cylindrical flask = Volume of the Conical flask

Height of the cylindrical flask = \(\frac{h}{3 x^{2}}\) units

Question 4.
A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
Answer:
Radius of a cone (V) = 7 cm
Height of a cone (h) = 8 cm
External radius of the hollow sphere (R) = 5 cm
Let the internal radius be “x”
Volume of the hollow sphere = Volume of the Cone


Internal diameter of the Hollowsphere = 2 × 3 = 6 cm.

Question 5.
Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m × 1.5 m × 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Answer:
Length of the cuboid tank (l) = 2 cm = 200 cm
Breadth of the cuboid tank (b) = 1.5 cm = 150 cm
Height of the tank (h) = 1 m = 100 cm
Volume of the cuboid = l × b × h cu. units
= 200 × 150 × 100 cm³
= 30,00,000 cm³
Radius of the tank (r) = 60 cm
Height of the tank (h) = 105 cm
Volume of the cylindrical tank = πr²h cu. units
= \(\frac{22}{7}\) × 60 × 60 × 105 cm³
= 22 × 60 × 60 × 15 cm³
= 1188000 cm³
Volume of water left in the sump = Volume of the sump – Volume of the tank
= 3000000 – 1188000 cm³
= 1812000 cm³

Question 6.
The internal and external diameter of a hollow hemispherical shell is 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder.
Answer:
Internal radius of the shell (r) = 3 cm
External radius of the shell (R) = 5 cm
Radius of the cylinder (r) = 7 cm
Let the height of the cylinder be “h”
Volume of the cylinder = Volume of the hemispherical shell

Height of the cylinder = 1.33 cm

Question 7.
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.
Answer:
Radius of a sphere (r) = 6 cm
External radius of the cylinder (R) = 5 cm
Height of the cylinder (h) = 32 cm
Let the internal radius of the cylinder be ‘x’
Volume of the hollow cylinder = Volume of a sphere
πh (R² – r²) = \(\frac{4}{3}\) πr³
π × 32 (5 + x) (5 – x) = \(\frac{4}{3}\) × π × 6 × 6 × 6
32 (25 – x²) = 4 × 2 × 6 × 6
25 – x² = 9
x² = 25 – 9 = 16
x = √16 = 4
Thickness of the cylinder = 5 – 4 = 1 cm.

Question 8.
A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Answer:
Let the height of the cylinder be “h”
radius is 50% more than the height


From (1) and (2) we get,
Volume of the cylinder = Volume of the hemisphere
It is possible to transfer the full quantity from the bowl into the cylindrical vessel.
100 % of the juice can be transferred.

Students can download Maths Chapter 7 Mensuration Ex 7.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.3

Question 1.
A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.
Answer:
Radius of a hemisphere = Radius of the cylinder

Question 2.
Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of the model that Nathan made.
Answer:
Radius of the cone = Radius of the cylinder

r = \(\frac{3}{2}\) cm
Height of the cone (H) = 2 cm
Height of the cylinder (h) = 12 – (2 + 2) cm = 8 cm
Volume of the model = Volume of the cylinder + Volume of 2 cones

Question 3.
From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm³.
Answer:
Radius of a cylinder = Radius of a cone r = 0.7 cm
Height of a cylinder = Height of a cone (h) = 2.4 cm

Volume of the remaining solid = Volume of the cylinder – Volume of a cone

Volume of the remaining soild = 2.46 cm³

Question 4.
A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.

Answer:
Radius of a cone = Radius of a hemisphere = Radius of a cylinder
r = 6 cm
Height of a cone (h) = 12 cm
Volume of the water displaced = Volume of the solid inside = Volume of the cone + Volume of the hemisphere

Volume of water displaced = 905. 14 cm³.

Question 5.
A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?
Answer:
Radius of a hemisphere = Radius of a Cylinder

r = \(\frac{3}{2}\) mm = 1.5 mm
Height of the cylinderical portion = 12 mm – (1.5 mm + 1.5 mm) = (12 – 3) mm = 9 mm
Volume of the capsule

Volume of the capsule = 77.8 cu. mm

Question 6.
As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

Answer:
Side of a cube (a) = 7 cm
Radius of a hemisphere (r) = \(\frac{7}{2}\) cm
Surface area of the solid = T.S.A of the cube + C.S.A of the hemisphere – Area of the base of the hemisphere

Question 7.
A right circular cylinder just encloses a sphere of radius r units. Calculate
(i) the surface area of the sphere
(ii) the curved surface area of the cylinder
(iii) the ratio of the areas obtained in (i) and (ii).
Answer:
(i) Surface area of sphere = 4πr² sq. units

Question 8.
A shuttlecock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttlecock is 7 cm. Find its external surface area.
Answer: