If X follows the Binomial distribution with p | vedclass

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(A) The probability mass function of a Binomial distribution is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.Given $n=6$,we have $P(X=4) = {}^{6}

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$1/4$

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(A) The probability mass function of a Binomial distribution is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.Given $n=6$,we have $P(X=4) = {}^{6}C_{4} p^{4} (1-p)^{2}$ and $P(X=2) = {}^{6}C_{2} p^{2} (1-p)^{4}$.According to the problem,$9 P(X=4) = P(X=2)$.Substituting the values,we get $9 \cdot {}^{6}C_{4} p^{4} (1-p)^{2} = {}^{6}C_{2} p^{2} (1-p)^{4}$.Since ${}^{6}C_{4} = {}^{6}C_{2} = 15$,the equation simplifies to $9 p^{4} (1-p)^{2} = p^{2} (1-p)^{4}$.Dividing both sides by $p^{2} (1-p)^{2}$ (assuming $p 
eq 0, 1$),we get $9 p^{2} = (1-p)^{2}$.Taking the square root of both sides,$3p = 1-p$ (since $p$ must be positive).$4p = 1$,which gives $p = 1/4$.

If $X$ follows the Binomial distribution with parameters $n=6$ and $p$ and $9 P(X=4) = P(X=2)$,then $p$ is

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