Aspire Faculty ID #17262 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

The probability mass function of Random variable $X$ is:

$P(X=x)=(0.6)^x(0.4)^{1-x}, ; x=0,1$

The variance of $X$ is:

Solution

For $x=0$:

$P(0)=(0.6)^0(0.4)^1=0.4$

For $x=1$:

$P(1)=(0.6)^1(0.4)^0=0.6$

So $X$ follows Bernoulli distribution with

$p=0.6$

Variance of Bernoulli:

$\mathrm{Var}(X)=p(1-p)$

$=0.6(0.4)$

$=0.24$

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