Question

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

Answer 1

Here, random variable X = 0, 1, 2

P(4) = `1/10`

`"P"(bar4) = 1 - 1/10 = 9/10`

P(X = 0) = `"P"(bar4)*"P"(bar4)`

= `9/10 xx 9/10`

= `81/100`

P(X = 1) = `"P"(bar4)*"P"(bar4) + "P"(4)*"P"(bar4)`

= `9/10 xx 1/10 + 1/10 xx 9/10`

= `18/100`

P(X = 2) = P(4).P(4)

= `1/10 xx 1/10`

= `1/100`

X	0	1	2
P(X)	`81/100`	`18/100`	`1/100`

We know that V(X) = E(X²) – [E(X)]²

E(X) = `0 xx 81/100 + 1 xx 18/100 + 2/100`

= `20/100`

= `1/5`

E(X²) = `0 xx 81/100 + 1 xx 18/100 + 4 xx 1/100`

= `22/100`

= `11/50`

∴ Var(X) = `11/50 - (1/5)^2`

= `11/50 - 1/25`

= `9/50`

= 0.18

Hence, the required variance = 0.18.