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Probability

Section: Mathematics

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Question : 41 of 92

Marks: +1, -0

A random variable X follows a binomial distribution in which the difference between its mean and variance is 1 . If

2 P(x=2)=3 P(x=1)

n^2 P(x>1)=

[AP EAPCET 21 May 2025 S1]

13
11
15
12

Solution:

Given, mean - variance

=1

n p - n p q = 1

n p (1-q) = 1

n p^2 = 1 \cdots (i)

and given

2 P(X=2)=3 P(X=1)

2 \cdot {}^{n}C_{2} p^2 q^{n-2} = 3 \cdot {}^{n}C_{1} p q^{n-1}

2 \cdot \frac{n(n-1)}{2} p = 3 \cdot n \cdot q

(n-1) p = 3 q = 3(1-p)

p(n+2) = 3 \cdots (ii)

Solving Eqs. (i) and (ii), we get

p = \frac{1}{2}, q = \frac{1}{2} \text{ and } n = 4

Now,

P(X>1)=1-P(X=0)-P(X=1)

= 1 - 4 \left(\frac{1}{2}\right)^4 - \left(\frac{1}{2}\right)^4

= 1 - \frac{5}{16} = \frac{11}{16}

\because n^2 P(X>1) = 16 \cdot \frac{11}{16} = 11

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