Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2

Question 1.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Solution:
Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n (S) = 8
Let X be the random variable denoting the number of heads
X = {0, 1, 2, 3}
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 1
Probability mass function
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 2

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Question 2.
A six sided die is marked ‘1’ on one face, ‘3’ on two of its faces and ‘5’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find
(i) the probability mass function
(ii) the cumulative distribution function
(iii) P(4 ≤ X < 10)
(iv) P(X ≥ 6)
Solution:
Let X be the random variable denotes the total score in two thrown of a die.
Sample space S
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 3
n (S) = 36
X = {2, 4, 6, 8, 10}
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 4

(i) Probability mass function
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 5

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

(ii) Cumulative distribution function
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 6
= \(\frac { 36 }{ 36 }\)
= 1
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 7
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 8

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Question 3.
Find the probability mass function and cumulative distribution function of number of girl child in families with 4 children, assuming equal probabilities for boys and girls.
Solution:
Let X be the random variable denotes number of girl child among 4 children
X = {0, 1, 2, 3, 4}
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 9
(i) Probability mass function
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 10

(ii) Cumulative distribution
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 11
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 12

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Question 4.
Suppose a discrete random variable can only take the values 0, 1 and 2. The probability mass function is defined by
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 13
Find (i) the value of k .
(ii) cumulative distribution function
(iii) P(X ≥ 1)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 14
(i) Given f is a probability mass function
\(\sum_{x}\) f(x) = 1
Probability mass function is
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 15

(ii) Cumulative distribution function
F(0) = P(x ≤ 0)
= P(x = 0)
= \(\frac { 1 }{ 8 }\)
F(1) = P(x ≤ 1)
= P(x = 0) + P(x = 1)
= \(\frac { 1 }{ 8 }\) + \(\frac { 2 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
F(2) = P(x ≤ 2)
= P(x = 0) + P(x = 1) + P(x = 2)
= \(\frac { 1 }{ 8 }\) + \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) = 1
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 16

(iii) P(X ≥ 1)
= P(X = 1) + P(X = 2)
= \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) + \(\frac { 7 }{ 8 }\)

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Question 5.
The cumulative distribution function of a discrete random variable is given by
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 17
Find
(i) the probability mass function
(ii) P(X < 1)
(iii) P(X ≥ 2)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 18
f(-1) = P(X = -1) = F(-1) – F(-1) = 0.15 – 0
= 0.15
f(0) = P(X = 0) = F(0) – F(-1) = 0.35 – 0.15
= 0.20
f(1) = P(X = 1) = F(1) – F(0) = 0.60 – 0.35
= 0.25
f(2) = P(X = 2) = F(2) – F(1) = 0.85 – 0.60
= 0.25
f(3) = P(X = 3) = F(3) – F(2) = 1 – 0.85
= 0.15
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 19
(ii) P(X < 1)
P(X < 1) = P(X = -1) + P(X = 0)
= 0.15 + 0.20
= 0.35

(iii) P(X ≥ 2)
P(X ≥ 2) = P(X = 2) + P(X = 3)
= 0.25 + 0.15
= 0.40

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Question 6.
A random variable X has the following probability mass function.
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 20
Find
(i) the value of k
(ii) P(2 ≤ X < 5)
(iii) P(3 < X)
Solution:
(i) Given f(x) in a probability mass function
\(\sum_{x}\) f(x) = 1
k² + 2k² + 3k² + 2k + 3k = 1
6k² + 5k = 1
6k² + 5k – 1 = 0
(k + 1) (6k – 1) = 0
k = \(\frac { 1 }{ 6 }\)
(k ≠ -1 neglecting negative terms)
Probability mass function
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 21

(ii) P(2 ≤ X < 5)
P(2 ≤ X < 5) = P(X = 2) + P(X = 3) + P(X = 4)
= \(\frac { 2 }{ 36 }\) + \(\frac { 3 }{ 36 }\) + \(\frac { 2 }{ 6 }\)
= \(\frac { 2+3+12 }{ 36 }\) + \(\frac { 17 }{ 36 }\)

(iii) P(X > 3) = P(X = 4) + P(X = 5)
= \(\frac { 2 }{ 6 }\) + \(\frac { 3 }{ 6 }\)
= \(\frac { 5 }{ 6 }\)

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

Question 7.
The cumulative distribution function of a discrete random variable is given by
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 22
Find
(i) the probability mass function
(ii) P(X < 3) and (iii) P(X ≥ 2)
Solution:
(i) Probability mass function
Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2 23

(ii) P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 5 }\)
= \(\frac { 4 }{ 5 }\)

(ii) P(X ≥ 2)
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)
= \(\frac { 1 }{ 5 }\) + \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 10 }\)
= \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Samacheer Kalvi 12th Maths Guide Chapter 11 Probability Distributions Ex 11.2

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