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(A) In a Binomial distribution with $n=5$ trials,the probability of $k$ successes is given by $P(X=k) = {}^{5}C_{k} \cdot p^{k} \cdot q^{n-k}$,where $

Vedclass · Accepted Answer

$\frac{32}{625}$

Vedclass · Answer

(A) In a Binomial distribution with $n=5$ trials,the probability of $k$ successes is given by $P(X=k) = {}^{5}C_{k} \cdot p^{k} \cdot q^{n-k}$,where $p+q=1$.Given $P(X=1) = {}^{5}C_{1} \cdot p \cdot q^{4} = 5pq^{4} = 0.4096$.Given $P(X=2) = {}^{5}C_{2} \cdot p^{2} \cdot q^{3} = 10p^{2}q^{3} = 0.2048$.Dividing the two equations: $\frac{10p^{2}q^{3}}{5pq^{4}} = \frac{0.2048}{0.4096} \Rightarrow \frac{2p}{q} = 0.5 \Rightarrow q = 4p$.Since $p+q=1$,we have $p+4p=1 \Rightarrow 5p=1 \Rightarrow p = \frac{1}{5} = 0.2$ and $q = \frac{4}{5} = 0.8$.Now,the probability of exactly $3$ successes is $P(X=3) = {}^{5}C_{3} \cdot p^{3} \cdot q^{2}$.$P(X=3) = 10 \cdot (\frac{1}{5})^{3} \cdot (\frac{4}{5})^{2} = 10 \cdot \frac{1}{125} \cdot \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}$.

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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