Question

Find $Var\left(X\right)$ if$P(X=a)=p=1-P(X=b)$

Short answer

We have found to be$Var\left(X\right)=(a-b{)}^2(1-p)p$

Step-by-step solution

Given information

Given in the question is,

$P(X=a)=p=1-P(X=b)$

Substitution

The random variable $X$can consider only two values, $a$with probability $p$and $b$with the probability $1-p$.

Using the meaning of the mean, we have that

$E\left(X\right)=a\times p+b\times (1-p)=:\mu$

Calculation

Currently, operating the definition of the variance, we have that

$Var\left(X\right)=E\left((X-\mu {)}^2\right)=(a-\mu {)}^2·p+(b-\mu {)}^2·(1-p)$

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

Substitute the given expression,

$=(a-b{)}^2(1-p{)}^{2}p+(b-a{)}^2{p}^2(1-p)$

role="math" localid="1646669147074" $=(a-b{)}^2(1-p\left)p\right[1-p+p]$

We get,

$=(a-b{)}^2(1-p)p$.

Final answer

The solution of the$Var\left(X\right)$is found to be$(a-b{)}^2(1-p)p$.