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If the mean and variance of a binomial distribution are 4 and 2 respectively, then probability of getting 2 heads is
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Verified Answer
The correct answer is:
$\frac{28}{256}$
We have $n p=4$ and $n p q=2$
$$
\therefore \mathrm{q}=\frac{2}{4}=\frac{1}{2} \quad \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \text { and } \mathrm{n}\left(\frac{1}{2}\right)=4 \Rightarrow \mathrm{n}=8
$$
$\therefore$
$$
P(x=2)={ }^8 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{8 !}{2 ! 6 !}\left(\frac{1}{2}\right)^8=\frac{8 \times 7}{2 \times 2^3 \times 2^5}=\frac{7}{64}=\frac{28}{256}
$$
$$
\therefore \mathrm{q}=\frac{2}{4}=\frac{1}{2} \quad \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \text { and } \mathrm{n}\left(\frac{1}{2}\right)=4 \Rightarrow \mathrm{n}=8
$$
$\therefore$
$$
P(x=2)={ }^8 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{8 !}{2 ! 6 !}\left(\frac{1}{2}\right)^8=\frac{8 \times 7}{2 \times 2^3 \times 2^5}=\frac{7}{64}=\frac{28}{256}
$$
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