1.

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

Answer»

Let X denote the number of doublets. Possible doublets are 

(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6) 

Clearly, X can take the value 0, 1, 2, or 3. 

Probability of getting a doublet = 6/36 = 1/6

Probability of not getting a doublet = 1 - 1/6 = 5/6

Now P(X = 0) = P (no doublet) = 5/6 x 5/6 x 5/6 = 125/216

P(X = 1) = P (one doublet and two non-doublets) 

= 1/6 x 5/6 x 5/6 + 5/6 x 1/6 x 5/6 + 5/6 x 5/6 x 1/6

= 3(1/6 x 52/62) = 75/216

P(X = 2) = P (two doublets and one non-doublet)

= 1/6 x 1/6 x 5/6 + 5/6 x 1/6 x 1/6 + 1/6 x 5/6 x 1/6

= 3(1/62 x 5/6) = 15/216

and P(X = 3) = P (three doublets) = 1/6 x 1/6 x 1/6 = 1/216

Thus, the required probability distribution is

X0123
P(X)125/21675/21615/2161/216


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