In a Binomial distribution,n=4 and 2 P(X=3)=3 | vedclass

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(D) The probability mass function for a Binomial distribution is given by $P(X=k) = {n \choose k} p^k q^{n-k}$,where $p+q=1$. Given $n=4$,the equation

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$\frac{4}{13}$

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(D) The probability mass function for a Binomial distribution is given by $P(X=k) = {n \choose k} p^k q^{n-k}$,where $p+q=1$. Given $n=4$,the equation $2 P(X=3) = 3 P(X=2)$ becomes: $2 	imes {4 \choose 3} p^3 q^1 = 3 	imes {4 \choose 2} p^2 q^2$ Since ${4 \choose 3} = 4$ and ${4 \choose 2} = 6$,we have: $2 	imes 4 	imes p^3 q = 3 	imes 6 	imes p^2 q^2$ $8 p^3 q = 18 p^2 q^2$ Dividing both sides by $2 p^2 q$ (assuming $p, q 
eq 0$): $4 p = 9 q$ Since $p = 1-q$,substitute this into the equation: $4(1-q) = 9q$ $4 - 4q = 9q$ $4 = 13q$ $q = \frac{4}{13}$

In a Binomial distribution,$n=4$ and $2 P(X=3)=3 P(X=2)$,then $q=$

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