For a symmetrical binomial distribution with n=8, find P(X 1) - Vidyadip

Question

For a symmetrical binomial distribution with $n=8$, find $P(X \leq 1)$

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Answer

Here, it is given that binomial distribution is symmetric.
$\therefore \mathrm{P}=\mathrm{q}=\frac{1}{2}, \mathrm{n}=8$
Putting, $n=8, P=\frac{1}{2}$ and $q=\frac{1}{2}$ in
$p(x)={ }^n C_x p^x q^{n-x}$
$ p(x)={ }^8 C_x\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{8-x}$
$ P(X \leq 1)=P(X=0,1)$
$ =[p(0)+p(1)]$
$ =\left[{ }^8 C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^8+{ }^8 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7\right]$
$ =\left[\left(\frac{1}{2}\right)^8+8\left(\frac{1}{2}\right)^8\right]$
$ =\left[\frac{1}{256}+\frac{8}{256}\right]$
$ =\frac{9}{256}$
Hence, $P[X \leq 1]$ obtained is $\frac{9}{256}$.

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