The Professor's Response
Hello professor,
I also need help with question number 45 from the mixed review from chapter 4. It starts like this "Your campus newspaper decides to run a raffle..."
See the professor's answer below.
Hi Katherine,
Here is the question: Your campus newspaper decides to run a raffle that has relatively good odds. The raffle involves paying $5 to draw a single colored marble from a bag. The bag contains 2 red, 6 white, 18 blue, and 54 yellow marbles. If the participant selects a red marble, he/she is given $25. If a white marble is drawn, he/she gets $7. If a blue marble is selected, he/she gets $5, and if a yellow marble is selected, he/she is given $4. What is the expected value for a single raffle drawing?
To answer this problem, you might want to set up a table, but before we do that let's mention some important ideas. The game requires us to pay $5 just to play, so we start out at a loss. This means if we select a red marble, we only net $20 in profit. That is because we paid $5 to have the right to select a marble. If we draw a yellow marble, we end up losing a dollar because we are only paid $4 for the yellow marble, yet it cost us $5 to play. Lastly, there are a total of 80 marbles in the bag. This means that the probability of drawing a red marble = # of red marbles/ total number of marbles = 2 / 80. These ideas should help you understand the table below.
| Outcome | X (Net $ Amount Lost or Gained) | P(X) |
| red | 20 | 2/80 |
| white | 2 | 6/80 |
| blue | 0 | 18/80 |
| yellow | -1 | 54/80 |
Then, we multiply the x column by the P(x) column and add the results to find the mean:
| Outcome (marble color) | X (Net $ Amount Lost or Gained) | P(X) | X*P(X) |
| red | 20 | 2/80 | 40/80 |
| white | 2 | 6/80 | 12/80 |
| blue | 0 | 18/80 | 0 |
| yellow | -1 | 54/80 | -54/80 |
| -2/80 = -0.025 |
I hope that helps,
Professor McGuckian