For a Binomial distribution,n=6,if 9 P(X=4)=P | vedclass

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(C) The probability mass function for a Binomial distribution is given by $P(X=k) = {}^nC_k p^k q^{n-k}$,where $p+q=1$. Given $n=6$,we have $P(X=4) =

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$\frac{3}{4}$

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(C) The probability mass function for a Binomial distribution is given by $P(X=k) = {}^nC_k p^k q^{n-k}$,where $p+q=1$. Given $n=6$,we have $P(X=4) = {}^6C_4 p^4 q^2$ and $P(X=2) = {}^6C_2 p^2 q^4$. According to the problem,$9 P(X=4) = P(X=2)$. Substituting the values: $9 	imes {}^6C_4 p^4 q^2 = {}^6C_2 p^2 q^4$. Since ${}^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$ and ${}^6C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$,the equation becomes: $9 	imes 15 p^4 q^2 = 15 p^2 q^4$. Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$): $9 p^2 = q^2$. Taking the square root of both sides: $3p = q$. Since $p+q=1$,we substitute $p = 1-q$: $3(1-q) = q \Rightarrow 3 - 3q = q \Rightarrow 4q = 3 \Rightarrow q = \frac{3}{4}$.

For a Binomial distribution,$n=6$,if $9 P(X=4)=P(X=2)$,then $q=$

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