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(C) Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure.Given $n = 5$ independent trials.The probability of $X$ succes

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$\frac{32}{625}$

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(C) Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure.Given $n = 5$ independent trials.The probability of $X$ successes is given by $P(X=k) = {}^nC_k p^k q^{n-k}$.We have $P(X=1) = 0.4096$ and $P(X=2) = 0.2048$.${}^5C_1 p^1 q^4 = 5pq^4 = 0.4096$ (Equation $1$).${}^5C_2 p^2 q^3 = 10p^2q^3 = 0.2048$ (Equation $2$).Dividing Equation $2$ by Equation $1$:$\frac{10p^2q^3}{5pq^4} = \frac{0.2048}{0.4096} = \frac{1}{2}$.$\frac{2p}{q} = \frac{1}{2} \Rightarrow 4p = q$.Since $q = 1 - p$,we have $4p = 1 - p \Rightarrow 5p = 1 \Rightarrow p = \frac{1}{5}$.Then $q = 1 - \frac{1}{5} = \frac{4}{5}$.Now,the probability of exactly $3$ successes is:$P(X=3) = {}^5C_3 p^3 q^2 = 10 	imes (\frac{1}{5})^3 	imes (\frac{4}{5})^2$.$P(X=3) = 10 	imes \frac{1}{125} 	imes \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}$.

Let in a Binomial distribution,consisting of $5$ independent trials,probabilities of exactly $1$ and $2$ successes be $0.4096$ and $0.2048$ respectively. Then the probability of getting exactly $3$ successes is equal to

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