The probability of success p for the Binomial | vedclass

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(D) For a Binomial distribution,$P(X=k) = {}^nC_k p^k q^{n-k}$,where $q = 1-p$.Given $n=6$,the relation is $4 P(X=4) = P(X=2)$.Substituting the formul

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$\frac{1}{3}$

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(D) For a Binomial distribution,$P(X=k) = {}^nC_k p^k q^{n-k}$,where $q = 1-p$.Given $n=6$,the relation is $4 P(X=4) = P(X=2)$.Substituting the formula: $4 \cdot {}^6C_4 p^4 q^{6-4} = {}^6C_2 p^2 q^{6-2}$.$4 \cdot {}^6C_4 p^4 q^2 = {}^6C_2 p^2 q^4$.Since ${}^6C_4 = {}^6C_2 = 15$,we have:$4 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$.Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$):$4 p^2 = q^2$.Taking the square root on both sides: $2p = q$.Since $p + q = 1$,we substitute $q = 2p$:$p + 2p = 1 \Rightarrow 3p = 1 \Rightarrow p = \frac{1}{3}$.

The probability of success $p$ for the Binomial distribution satisfying the relation $4 P(X=4) = P(X=2)$ with parameter $n=6$ is

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