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?

Answer 1

Answer:

The expected value is defined as:

EV = ∑xₙ*pₙ

Where xₙ represents the n-th event, and pₙ is the probability of that event.

Now, the events are:

The possible outcomes are: {2, 3, 4, 5, 6, 7, 8, 9, 10} 9 in total.

Winning $1, when the number is odd. The outcomes in this case are: {3. 5. 7. 9}

4 out of 9, then the probability is p = 4/9.

Lossing $1 when the number is even. The outcomes in this case are {2, 4, 6, 8, 10}

5 out of 9, the probability is p = 5/9.

Then the expected value is:

EV = (4/9)*$1 - (5/9)*$1 = (-1/9)*$1  = -$0.11...