The random variable X can take only the values 0,1,2. If
`P(X=0)=P(X=1)=p and E(X^(2))=E[X]`,
then find valu of p.

Explanation

Since, X=0,1,2 and P(X) at X =0 and 1 is p, let at X=2,P(X) is x.
`rArr P+P+x=1`
`rArrx=1-2p`
We get, the following distribution.

`therefore E[X]=sumXP(X)`
`=0cdotp+1cdotp+2(1-2p)`
and `E(X^(2))=sumX^(2)P(X)`
`=0cdotP+1cdotp+4cdot(1-2p)`
`=p+4-8p=4=-7p`
Also, given that`E(X^(2))=E[X]`
`rArr4-7p=2-3p`
`rArr4p=2rArrp=1/2`