© NCERT
not to be republished
 562
MATHEMATICS
Solution 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
1
36
6
 
 
Probability of not getting a doublet 
1
5
1
6
6
 
 
 
Now
P(X = 0) = P (no doublet) = 5
5
5
125
6
6
6
216
 
 
 
P(X = 1) = P (one doublet and two non-doublets)
= 1
5
5
5
1
5
5
5
1
6
6
6
6
6
6
6
6
6
 
 
 
 
 
 
 
 
=
2
2
1
5
75
3 6
216
6
 
 
 
 
 
 
 
 
P(X = 2) = P (two doublets and one non-doublet)
=
2
1
1
5
1
5
1
5
1
1
1
5
15
3
6
6
6
6
6
6
6
6
6
6
216
 6
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
and
P(X = 3) = P (three doublets)
= 1
1
1
1
6
6
6
216
 
 
 
Thus,  the required probability distribution is
X
0
1
2
3
P(X)
125
216
75
216
15
216
1
216
Verification Sum of the probabilities
1
n
i
i
p
=∑
 = 125
75
15
1
216
216
216
216
 
 
 
= 125
75
15
1
216
1
216
216
 
 
 
 
© NCERT
not to be republished
PROBABILITY      563
Example 26 Let X denote the number of hours you study during a randomly selected
school day The probability that X can take the values x, has the following form, where
k is some unknown constant