Example 1: Find the missing probability in the following distribution and also find $E(X)$?
$x$

0

1

2 
3

4 
$p(x)$

$\frac{3}{8}$ 
$\frac{1}{4}$ 
 
$\frac{3}{16}$ 
$\frac{1}{16}$

Solution: Let the missing probability be $k$
For a probability distribution , $\sum$ $p(x)$ = 1
Therefore
$\frac{3}{8}$ +
$\frac{1}{4}$ + $k$ +
$\frac{3}{16}$ +
$\frac{1}{16}$ = 1
$k$ +
$\frac{14}{16}$ = 1
$k$ = 1 
$\frac{14}{16}$
$k$ =
$\frac{1}{8}$Therefore, the probability distribution is:
$x$

0 
1 
2

3  4

$p(x)$ 
$\frac{3}{8}$

$\frac{1}{4}$ 
$\frac{1}{8}$ 
$\frac{3}{16}$  $\frac{1}{16}$

Thus $E(x)$ = $\sum$ $x \times p(x)$
= 0 $\times$
$\frac{3}{8}$ + 1 $\times$
$\frac{1}{4}$ + 2 $\times$
$\frac{1}{8}$ + 3 $\times$
$\frac{3}{16}$ + 4 $\times$
$\frac{1}{16}$
=
$\frac{21}{16}$Example 2: In a lottery there are 1000 tickets costing Rs 1 each. There is one first prize worth Rs. 100/ two second prizes worth Rs 20/ each and ten third prizes worth Rs 10/ each. Find the expected loss in buying one ticket?
Solution: Let $X$ : Amount that one ticket fetches after deducting the purchase cost
$X$ is a random variable which takes the values 99, 19, 9 and  1 with respective probabilities.
p(99) = P[1st Prize] =
$\frac{1}{1000}$= 0.001
p(19) = P [2nd Prize] =
$\frac{2}{1000}$ = .002
p(9) = P[3rd Prize] =
$\frac{10}{1000}$= 0.01
p( 1) = P[No Prize] =
$\frac{987}{1000}$= 0.987
Thus expectation of the net amount
$E(X)$ = $\sum$ $x . p(x)$
= 99 $\times$ 0.001 + 19 $\times$ 0.002 + 9 $\times$ 0.01 + (1) $\times$ 0.987
=  0.76
= Rs. 0.76 (loss)