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**Answer**

**Variance of Bernoulli Distribution**

The expected value of Bernoulli Distribution is *p*

$Var\left(x\right)=E\left[{x}^{2}\right]-[E{\left[x\right]]}^{2}$

$E\left[x\right]=p$

$E\left[{x}^{2}\right]=\sum _{x=0}^{1}{x}^{2}P(x=x)$

$={0}^{2}\times {p}^{0}{(1-p)}^{1-0}+{1}^{2}\times {p}^{1}{(1-p)}^{1-1}$

$=0+p$

$=p$

$Var\left(x\right)=E\left[{x}^{2}\right]-[E{\left[x\right]]}^{2}$

$Var\left(x\right)=p-{\left[p\right]}^{2}$

$=p[1-p]=pq$

$=pq$

Thus the Variance of Bernoulli Distribution is given by; $pq$

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