If X sim B(4, p) and 2 P(X=3)=3 P(X=2),then t | vedclass

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(A) Given $X \sim B(4, p)$,where $n=4$. The probability mass function is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.Given the condition $2 P(X=3) = 3 P(X=2)

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$\frac{9}{13}$

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(A) Given $X \sim B(4, p)$,where $n=4$. The probability mass function is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.Given the condition $2 P(X=3) = 3 P(X=2)$.Substituting the values: $2 	imes ({}^{4}C_{3} p^{3} q^{1}) = 3 	imes ({}^{4}C_{2} p^{2} q^{2})$.Calculating combinations: $2 	imes (4 p^{3} q) = 3 	imes (6 p^{2} q^{2})$.Simplifying: $8 p^{3} q = 18 p^{2} q^{2}$.Dividing both sides by $2 p^{2} q$ (assuming $p, q 
eq 0$): $4 p = 9 q$.Since $q = 1 - p$,we have $4 p = 9(1 - p)$.$4 p = 9 - 9 p$.$13 p = 9$.Therefore,$p = \frac{9}{13}$.

If $X \sim B(4, p)$ and $2 P(X=3)=3 P(X=2)$,then the value of $p$ is:

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