I need to be explained about this question. Any help will be kind!

Question

A ping pong ball is drawn at random from an urn consisting of balls numbered 4 through 9. A player wins $1 if the number on the ball is odd and loses $1.5 if the number is even. 

Let x be the amount of money a player will win/lose when playing this game, where x is negative when the player loses money. 

a. Construct the probability table for this game. Round your answers to two decimal places.

b) What is the expected value of player's winnings? Round to the nearest hundreth. 

Answer 1

Since there are 6 numbers, the the probability of an even = probability of an odd.

3/6 = 1/2 or 0.5

So the expected value is the sum of the payout for an outcome times the probability of that outcome.

Pr(even)=Pr(odd)=0.5

Expected Value = $1*(0.5) +(-$1.5)(0.5) = - $0.25

Answer 2

Rinzi

There are two possible events here - odd number or even number.

There are 6 possible outcomes (4,5,6,7,8,9).

Probability (odd) = 3/6 = 1/2 = 0.5 because there are 3 odd outcomes out of 6

Probability (even) = 3/6 = 1/2 =0.5 because there are 3 even outcomes out of 6

Expected winnings = P(odd) x winnings when odd + P(even) x winnings when even

= 0.5 * 1 - 0.5 * 1.5 = - 0.25

What this means is if the player play a large number of times, on average he will lose $ 0.25 per game. So if he plays 10000 times he would expect to lose $ 2500 approximately.

Mike