If X B(7, P) is a binomial variate and P(X=3)=P(X=5), then P=

Correct Answer is : 5−152\; {5 - {15}}{2}25−15

Correct Answer is : 5−152\; {5 - {15}}{2}25−15

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TG EAMCET 2 May 2025 Shift 1 Paper

Question : 40 of 160

Marks: +1, -0

X \sim B(7, P)

P(X=3)=P(X=5)

P=

Solution:

Given

X \sim B(7, P)

is a binomial variate and

P(X=3)=P(X=5)

Using the binomial probability formula

P(X=K)=\binom{n}{K} P^K (1-P)^{n-K}

Here

n=7

So,

P(X=3)=\binom{7}{3} P^3 (1-P)^4

And,

P(X=5)=\binom{7}{5} P^5 (1-P)^2

\; \therefore \binom{7}{3} P^3 (1-P)^4 = \binom{7}{5} P^5 (1-P)^2

\; \Rightarrow 35 P^3 (1-P)^4 = 21 P^5 (1-P)^2

\; \Rightarrow 35 (1-P)^2 = 21 P^2, \; \text{ where } \; P \neq 0, P \neq 1

\; \Rightarrow 5 (1-P)^2 = 3 P^2

\; \Rightarrow 5 (1 - 2P + P^2) = 3 P^2

\; \Rightarrow 5 - 10P + 5P^2 - 3P^2 = 0

\; \Rightarrow 2 P^2 - 10 P + 5 = 0

\; \Rightarrow P = \frac{-(-10) \pm \sqrt{100 - 4 \times 2 \times 5}}{2 \times 2}

\; = \frac{10 \pm \sqrt{100-40}}{4}

\; = \frac{10 \pm 2 \sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2}

But

P = \frac{5 + \sqrt{15}}{2} > 1

, So it can't be a valid probability.

\therefore P = \frac{5 - \sqrt{15}}{2}

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