A biased die is such that P(4)`=1/10` and other scores being equally likely. The die tossed twice. If X is the number of four seen, then find the variance of the random variable X.

Explanation

since, X= Number of fours seen
On tossing twi die, X=0,1,2.
Also, `P_(4)=1/10andP_(("not" 4))=9/10`
So, `P(X=0)=P(("not" 4))cdotP_(("not" 4))=9/10cdot9/10=81/100`
P(X=1)=`P(("not" 4))cdotP_((4))+P_(4)cdotP_(("not"4))=9/10cdot1/10+1/10cdot9/10=18/100`
P(X=2)=`P_((4))cdotP_((4))=1/10cdot1/10=1/100`
Thus, we get folllowing table

`therefore` Var (X)=`E(X^(2))-[E(X)]^(2)=sumX^(2)P(X)-[sumXP(X)]^(2)`
`=[0+18/100+4/100]-[0+18/100+2/100]^(2)`
`=22/100-(20/100)^(2)=11/50-1/25`
`=(11-2)/50=9/50=18/100=0.18`