Question

Find the probability distribution of the number of doublets in three throws of a pair of dice

Answer 1

Let X denotes the number of doublets in three throws of a pair of dice.

∴ Possible values of X are 0, 1, 2 and 3.

In a toss of pair of dice, possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

∴ Probability of getting a doublet = p

= `6/36`

= `1/6`

∴ Probability of not getting a doublet = q

= `1 - 1/6`

= `5/6`

∴ P(X = 0) = P(no doublet)

= q × q × q

= `5/6 xx 5/6 xx 5/6`

= `125/216`

P(X = 1) = P(one doublet)

= pqq + qpq + qqp

= `1/6 xx 5/6 xx 5/6 + 5/6 xx 1/6 xx 5/6  + 5/6 xx 5/6 xx 1/6`

= `75/216`

P(X = 2) = P(two doublets)

= ppq + pqp + qpp

= `1/6 xx 1/6 xx 5/6 + 1/6 xx 5/6 xx 1/6 + 5/6 xx 1/6 xx 1/6`

= `15/216`

P(X = 3) = P(three doublets)

= p × p × p

= `1/6 xx 1/6 xx 1/6`

= `1/216`

∴ Probability distribution of X is as follows:

X	0	1	2	3
P(X = x)	`125/216`	`75/216`	`15/216`	`1/216`