Let X denote the number obtained on the throw Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6 © NCERT
not to be republished
 568
MATHEMATICS
Also
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1
6
Therefore, the Probability distribution of  X  is
X
1
2
3
4
5
6
P(X)
1
6
1
6
1
6
1
6
1
6
1
6
Now
E(X) =
1
(
)
n
i
i
i
x p x
  
=
1
1
1
1
1
1
21
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+ ×
+ ×
+ ×
+ ×
+ ×
=
Also
E(X2) =
2
2
2
2
2
2
1
1
1
1
1
1
91
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+
×
+
×
+
×
+
×
+
×
=
Thus,
Var (X) = E (X2) – (E(X))2
=
2
91
21
91
441
6
6
6
36
⎛
⎞
−
=
−
⎜
⎟
⎝
⎠
  
35
12
 
Example 29 Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards Find the mean, variance and standard deviation
of the number of kings