Now
P(X = 2) = P({(1,1)}) 
361
 
P(X = 3) = P({(1,2), (2,1)}) 
362
 
P(X = 4) = P({(1,3), (2,2), (3,1)}) 
363
 
P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)})
364
 
P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)})
365
 
P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)})
366
 
P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)})
365
 
© NCERT
not to be republished
PROBABILITY      565
P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})
364
 
P(X = 10) = P({(4,6), (5,5), (6,4)})
363
 
P(X = 11) = P({(5,6), (6,5)})
362
 
P(X = 12) = P({(6,6)}) 
361
 
The probability distribution of X is
X or xi
2
3
4
5
6
7
8
9
10
11
12
P(X) or pi
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Therefore,
μ = E(X) =
1
1
2
3
4
2
3
4
5
36
36
36
36
n
i
i
i
x p
=
= ×
+ ×
+ ×
+ ×
∑
5
6
5
6
7
8
36
36
36
 
 
 
 
 
 
4
3
2
1
9
10
11
12
36
36
36
36
 
 
 
 
 
 
 
 
= 2
6
12
20
30
42
40
36
30
22
12
36
 
 
 
 
 
 
 
 
 
 
 = 7
Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7 13 6 3  Variance of a random variable
The mean of a random variable does not give us information about the variability in the
values of the random variable