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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:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 1

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:
Given that die is marked ‘ 1 ’ on one face, ‘3’ on two of its faces and ‘5’ on remaining three faces. i.e., {1, 3, 3, 5, 5, 5} in a single die.
When it is thrown twice, the number of sample points is 36, in which the sum of faces numbers are 2, ,4, 6, 8 and 10 are the value of random variable ‘X’.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 2

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

(i) Probability mass function:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 3

(ii) The Cumulative distribution function:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 4

(iii) (4 ≤ 10) = P(X = 4) + P(X = 6) + P(X = 8)
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 5

(iv) P(X ≥ 6) = P (X = 6) + P (X = 8) + P (X = 10)
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 6

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 which denotes the number of girl children in the family of 4 children and X takes the values of 0, 1, 2, 3, 4.
Probability of child being a boy = P (B) = \(\frac{1}{2}\)
Probability of child being a girl = P (G) = \(\frac{1}{2}\)
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 7
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 77

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 8
Find
(i) the value of k
(ii) cumulative distribution function
(iii) P(X ≥ 1).
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 9

Question 5.
The cumulative distribution function of a discrete random variable is given by
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 10
Find the (i) the probability mass function
(ii) P(X < 1)
(iii) P(X ≥ 2)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 11
For x = -1, f(x) = 0.15 – 0 = 0.15
For x = 0, f(x) = 0.35 – 0.15 = 0.20
For x = 1, f(x) = 0.60 – 0.35 = 0.25
For x = 2, f(x) = 0.85 – 0.60 = 0.25
For x = 3, f(x) = 1 – 0.85 = 0.15
(i) Probability mass function table
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 12
(ii) P (X < 1) = P(X = -1) + P(X = 0) = 0.15 + 0.20 = 0.35
(iii) P (X ≥ 2) = P (X = 2) + P (X = 3) = 0.25 + 0.15 = 0.40

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

Question 6.
A random variable X has the following probability mass function.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 13
Solution:
Given probability mass function
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 14
(i) We know that \(\Sigma P_{i}\) = 1
i.e., k2 + 2k2 + 3k2 + 2k + 3k = 1
6k2 + 5k = 1
6k2 + 5k – 1 = 0
(k + 1) (6k – 1) = 0
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 16

(ii) P (2 ≤ X < 5)
= P (X = 2) + P (X = 3) + P (X = 4)
= 2k2 + 3k2 + 2k = 5k2 + 2k
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 165

(iii) P (3 < X) = P (X > 3)
= P (X = 4) + P (X = 5) = 2k + 3k = 5k = 5/6

Question 7.
The cumulative distribution function of a discrete random variable is given by.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 18
Find (i) the probability mass function
(ii) P(X < 3) and
(iii) P(X ≥ 2).
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 19
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 20

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

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

Question 1.
Find the probability mass function, and the cumulative distribution function for getting ‘3’s when two dice are thrown.
Solution:
Two dice are thrown. Let X be the random variable of getting number of ‘3’s.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 21
Cumulative distribution function:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 22

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 23

Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 24

Question 3.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 25
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 26

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

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 28
(i) P(0.5 < X < 0.75)
(ii) P(X ≤ 0.5)
(iii) P(X > 0.75)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 288

Question 5.
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 29
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 30

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