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At a carnival game, a winning player spins a wheel that always lands on either Prize 1 or Prize 2 to determine which of the two prizes he or she wins. The probability that the prize wheel indicates Prize 2 is double the probability that it indicates Prize 1. If a player does not want the prize that the prize wheel first indicates, then he or she may spin the wheel again. In such cases, the player must accept whichever prize the prize wheel indicates on the second spin.
Select for Prize 1 the number nearest to the probability that a winning player who wants Prize 1 will receive Prize 1 after one or two spins of the prize wheel, and select for Prize 2 the number nearest to the probability that a winning player who wants Prize 2 will receive Prize 2 after one or two spins of the prize wheel. Make only two selections, one in each column.
0.1
0.3
0.5
0.7
0.9
Let's create a visual representation of our prize wheel and the spinning process:
Prize Wheel:
Prize 1: |----| (1 part)
Prize 2: |--------| (2 parts)
Total: 3 parts
Probabilities:
P(Prize 1) = 1/3
P(Prize 2) = 2/3
Player wants Prize 1: Player wants Prize 2:
START START
| |
First Spin First Spin
/ \ / \
1/3 2/3 1/3 2/3
Prize 1 Prize 2 Prize 1 Prize 2
✓ | | ✓
KEEP Re-spin Re-spin KEEP
/ \ / \
1/3 2/3 1/3 2/3
Prize 1 Prize 2 Prize 1 Prize 2
✓ ✗ ✗ ✓
We need to find two probabilities:
The probability of getting your desired prize depends on:
From our decision tree:
Total probability = \(\frac{1}{3} + \frac{2}{9}\)
Converting to common denominator:
Calculating decimal: \(\frac{5}{9} = 0.556...\)
Closest answer choice: 0.5
From our decision tree:
Total probability = \(\frac{2}{3} + \frac{2}{9}\)
Converting to common denominator:
Calculating decimal: \(\frac{8}{9} = 0.889...\)
Closest answer choice: 0.9
These answers make intuitive sense: Since Prize 2 appears twice as often as Prize 1, players wanting the more common prize (Prize 2) have a higher chance of success than those wanting the rarer prize (Prize 1).