A pair of dice is rolled twice. Find the probability distribution of no. of doublets.

Explanation

Let X denote the no. of doublets. Possible doublets are
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Clearly X can take value 0, 1, or 2
Probability of getting a doublet `=6/36=1/6`
Probability of not getting a doublet `=1-1/6=5/6`
P(X=0)=P(no doublet)`=5/6xx5/6=25/36`
P(X=1)=P(one doublet and two non doublets)
`=1/6xx5/6+5/6xx1/6=10/36`
P(X=2) = (two doublets and one non doublet)
`=1/6xx1/6=1/36`
Thus, the required probability distribution is