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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Aicharya · Accepted Answer

We have $n p=4$ and $n p q=2$ $	herefore \mathrm{q}=\frac{2}{4}=\frac{1}{2} \quad \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} 	ext { and } \mathrm{n}\left(\frac{1}{2}ight)=4 \Rightarrow \mathrm{n}=8$ $	herefore P(x=2)={ }^8 C_2\left(\frac{1}{2}ight)^2\left(\frac{1}{2}ight)^6=\frac{8 !}{2 ! 6 !}\left(\frac{1}{2}ight)^8=\frac{8 	imes 7}{2 	imes 2^3 	imes 2^5}=\frac{7}{64}=\frac{28}{256}$

If the mean and variance of a binomial distribution are 4 and 2 respectively, then probability of getting 2…

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