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(D) Let $P(X=1)$ be the probability of one success and $P(X=2)$ be the probability of two successes. The probability mass function for a Binomial dist

Vedclass · Accepted Answer

$\frac{4}{625}$

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(D) Let $P(X=1)$ be the probability of one success and $P(X=2)$ be the probability of two successes. The probability mass function for a Binomial distribution is $P(X=k) = { }^n C_k p^k q^{n-k}$. Given $n=5$,we have: $P(X=1) = { }^5 C_1 p^1 q^4 = 5pq^4 = 0.4096$ ...$(i)$ $P(X=2) = { }^5 C_2 p^2 q^3 = 10p^2 q^3 = 0.2048$ ...(ii) Dividing equation $(i)$ by (ii): $\frac{5pq^4}{10p^2q^3} = \frac{0.4096}{0.2048} = 2$ $\frac{q}{2p} = 2 \Rightarrow q = 4p$. Since $p+q=1$,we have $p+4p=1 \Rightarrow 5p=1 \Rightarrow p=\frac{1}{5}$ and $q=\frac{4}{5}$. Now,the probability of getting exactly $4$ successes is: $P(X=4) = { }^5 C_4 p^4 q^1 = 5 	imes (\frac{1}{5})^4 	imes (\frac{4}{5}) = 5 	imes \frac{1}{625} 	imes \frac{4}{5} = \frac{4}{625}$.

In a Binomial distribution consisting of $5$ independent trials,probabilities of exactly $1$ and $2$ successes are $0.4096$ and $0.2048$ respectively,then the probability of getting exactly $4$ successes is

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