Question

A ping pong ball is drawn at random from an urn consisting of balls numbered 2 through 10. A player wins 1 dollar if the number on the ball is odd and loses 1 dollar if the number is even. What is the expected value of his winnings? Express your answer in fraction form.

Answer 1

Given:

A ping pong ball is drawn at random from an urn consisting of balls numbered 2 through 10.

A player wins 1 dollar if the number on the ball is odd and loses 1 dollar if the number is even.

To find:

The expected value of his winnings.

Solution:

Balls numbered 2 through 10. So, numbers are 2, 3, 4, 5, 6, 7, 8, 9, 10.

Odd numbers = 3, 5, 7, 9

Even numbers = 2, 4, 6, 8, 10

Total numbers = 9

Odd numbers = 4

Even numbers = 5

Now,

$P(odd)=\dfrac{4}{9}$

$P(Even)=\dfrac{5}{9}$

Expected value is

$E(x)=\sum x\times P(x)$

Player wins 1 dollar if the number on the ball is odd and loses 1 dollar if the number is even.

$E(x)=1\times P(odd)+(-1)\times P(even)$

On substituting the values, we get

$E(x)=1\times \dfrac{4}{9}+(-1)\times \dfrac{5}{9}$

$E(x)=\dfrac{4}{9}-\dfrac{5}{9}$

$E(x)=-\dfrac{1}{9}$

Therefore, the expected value of his winnings is $E(x)=-\dfrac{1}{9}$.