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Tamil Nadu Board of Secondary EducationHSC Science Class 12

In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable - Mathematics

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Question

In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable

Sum

Solution

n = 5, X ~ B(n, p)

Given, P(X = 1) = 0.4096

P(X = 2) = 0.2048

P(X = x) = ⁿCx px qn-x, x = 0, 1, 2, …. n

⇒ 5C1 pq4 = 0.4096

⇒ 5C2 p2q3 = 0.2048

5pq4 = 0.4096  .......(1)

10p2q3 = 0.2048  .......(2)

(1) divided by (2),

⇒ `(5"pq"^4)/(10"p"^2"q"^3)` = 2

`"p"/"q"` = 4

q = 4p

q = 4(1 – q)

q = 4 – 4q

5q = 4

q = `4/5`

p = 1 – q = `1/5`

Mean = np

= `5 xx 1/5` = 1

Variance = npq

= `1 xx 4/5 = 4/5`

Distribution

P(X = x) = `""^5"C"_x (1/5)^x (4/5)^(5-x)`

x = 0, 1, 2, 3, 4, 5

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Theoretical Distributions: Some Special Discrete Distributions
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Chapter 11: Probability Distributions - Exercise 11.5 [Page 218]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 11 Probability Distributions
Exercise 11.5 | Q 9 | Page 218

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