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## 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).
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).
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.
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.
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.
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.
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.
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?
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?
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.
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)?
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.
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 $$\frac{29}{35}$$

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