Therefore,
P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)
= 48
48
144
52
52
169
×
=
P(X = 1) = P(ace and non-ace or non-ace and ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace) P(non-ace) + P (non-ace) P(ace)
= 4
48
48
4
24
52
52
52
52
169
×
+
×
=
and
P(X = 2) = P (ace and ace)
= 4
4
1
52
52
169
 
 
Thus, the required probability distribution is
X
0
1
2
P(X)
144
169
24
169
1
169
Example 25 Find the probability distribution of number of doublets in three throws of
a pair of dice © NCERT
not to be republished
 562
MATHEMATICS
Solution Let X denote the number of doublets