# Probability Distribution Mean (Expectation), Variance :: Problems

 Problem Back to Problems Page
A number is chosen at random from the set 1, 2, 3, ... , 100 and another number is chosen at random from the set 1, 2, 3, ... , 50. What is the expected value of the sum and the expected value of the product?

 [Expectation: 25.5 ; Expectation(sum): 76 ; Expectation(product): 1,287.75]

 Solution

Let the variable representing a number from the first set be "x" and a number from the second set be be "y"

First Set

In the experiment of choosing a number from the set of 100 numbers, there are 100 elementary events i.e. the events of choosing 1, choosing 2,... , choosing 100

All these elementary events are equally likely (since any of the 100 numbers can appear on choosing a number) and mutually exclusive (since appear of one of the numbers prevents the appearance of the other numbers). They are exhaustive events since they form all possible events in the experiment.

Therefore probability of occurance of each elementary event i.e. getting each number is 1/100

The probability distribution of "x" would therefore be
x 1 2 ... ... 100
P(X = x)
 1 100
 1 100
 1 100
 1 100
 1 100

Calculations for Mean and Standard Deviations
x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
1
 1 100
 1 100
1
 1 100
2
 1 100
 2 100
4
 4 100
...
 1 100
...
...
 1 100
...
100
 1 100
 100 100
10,000
 10,000 100
Total 1
 16 7
 42 7
= 2.29 = 6
Σ px =
 1 100
+
 2 100
+   ...     ...
 100 100
=
 1 + 2 + ... + ... + 100 100

=
 100(100+1) 2
100
[Sum of the first "n" natural numbers =
 n(n+1) 2
=
 50 × 101 100
= 50.5
Σ px2 =
 1 100
+
 4 100
+   ...     ...
 10,000 100
=
 1 + 4 + ... + ... + 10,000 100
=
 12 + 22 + ... + ... + 1002 100

=
 100(100+1)(2 × 100 + 1) 6
100
[Sum of the squares of the first "n" natural numbers =
 n(n+1)(2n+1) 6
=
 100(101)(201) 6
×
 1 100
=
 101 × 37 2
= 1,868.5
⇒ Expectation of the distribution
 ⇒ E (x) = Σ px = 50.5

Second Set

In the experiment of choosing a number from the set of 50 numbers, there are 50 elementary events i.e. the events of choosing 1, choosing 2,... , choosing 50

All these elementary events are equally likely (since any of the 50 numbers can appear on choosing a number) and mutually exclusive (since appear of one of the numbers prevents the appearance of the other numbers). They are exhaustive events since they form all possible events in the experiment.

Therefore probability of occurance of each elementary event i.e. getting each number is 1/50

The probability distribution of "y" would therefore be
y 1 2 ... ... 50
P(X = y)
 1 50
 1 50
 1 50
 1 50
 1 50

Calculations for Mean and Standard Deviations
y P (X = y) py
[y × P (X = y)]
y2 py2
[y2 × P (X = y)]
1
 1 50
 1 50
1
 1 50
2
 1 50
 2 50
4
 4 50
...
 1 50
...
...
 1 50
...
50
 1 50
 50 50
2,500
 2,500 50
Total 1
 16 7
 42 7
= 2.29 = 6
Σ py =
 1 50
+
 2 50
+   ...     ...
 50 50
=
 1 + 2 + ... + ... + 50 50

=
 50(50+1) 2
50
[Sum of the first "n" natural numbers =
 n(n+1) 2
=
 25 × 51 50
= 25.5
Σ py2 =
 1 50
+
 4 50
+   ...     ...
 2,500 50
=
 1 + 4 + ... + ... + 2,500 50
=
 12 + 22 + ... + ... + 502 50

=
 50(50+1)(2 × 50 + 1) 6
50
[Sum of the squares of the first "n" natural numbers =
 n(n+1)(2n+1) 6
=
 50(51)(101) 6
×
 1 50
=
 17 × 101 2
= 858.5

⇒ Expectation of the distribution
 ⇒ E (y) = Σ py = 25.5

Expectation of the sum of the two variables
 ⇒ E(x + y) = E (x) + E(y) = 50.5 + 25.5 = 76

Expectation of the product of the two variables
 ⇒ E(x × y) = E (x) × E(y) = 50.5 × 25.5 = 1,287.75

 Credit : Vijayalakshmi Desu