Question

Two biased dice are thrown together. For the first die P(6) = `1/2`, the other scores being equally likely while for the second die, P(1) = `2/5` and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.

Answer 1

Given that: for the first die, P(6) = `1/2`

And `"P"(bar6) = 1 - 1/2 = 1/2`

⇒ P(1) + P(2) + P(3) + P(4) + P(5) = `1/2`

But P(1) = P(2) = P(3) = P(4) = P(5)

∴ 5.P(1) = `1/2`

⇒ P(1) = `1/10` and `"P"(bar1) = 1 - 1/10 = 9/10`

For the second die, P(1) = `2/5` and `"P"(bar1) = 1 - 2/5 = 3/5`

Let X be the number of one’s seen

∴ X = 0, 1, 2

⇒ P(X = 0) = `"P"(bar1)."P"(bar1)`

= `9/10 * 3/5`

= `27/50`

= 0.54

P(X = 1) = `"P"(bar1)*"P"(1) + "P"(1)*"P"(bar1)`

= `9/10*2/5 + 1/10*3/5`

= `(18 + 3)/50`

= `21/50`

= 0.42

P(X = 2) = `"P"(1)_"I" * "P"(1)_"II"`

= `1/10*2/5`

= `2/50`

= 0.04

Hence, the required probability distribution is

X	0	1	2
P(X)	0.54	0.42	0.04