At a carnival game, a winning player spins a wheel that always lands on either Prize 1 or Prize 2...
GMAT Two Part Analysis : (TPA) Questions
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.
Phase 1: Owning the Dataset
Understanding the Prize Wheel
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
Decision Tree Visualization
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
✓ ✗ ✗ ✓
Phase 2: Understanding the Question
What We're Looking For
We need to find two probabilities:
- Prize 1: Probability that a player who WANTS Prize 1 will RECEIVE Prize 1
- Prize 2: Probability that a player who WANTS Prize 2 will RECEIVE Prize 2
Key Insight
The probability of getting your desired prize depends on:
- Getting it on the first spin (immediate success)
- OR getting the unwanted prize first, then getting your desired prize on the second spin
Phase 3: Finding the Answer
Calculating for Prize 1 (Player wants Prize 1)
From our decision tree:
- Path 1: Get Prize 1 on first spin = \(\frac{1}{3}\)
- Path 2: Get Prize 2 first (\(\frac{2}{3}\)), then Prize 1 on second spin (\(\frac{1}{3}\))
- Probability = \(\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\)
Total probability = \(\frac{1}{3} + \frac{2}{9}\)
Converting to common denominator:
- \(\frac{1}{3} = \frac{3}{9}\)
- Total = \(\frac{3}{9} + \frac{2}{9} = \frac{5}{9}\)
Calculating decimal: \(\frac{5}{9} = 0.556...\)
Closest answer choice: 0.5
Calculating for Prize 2 (Player wants Prize 2)
From our decision tree:
- Path 1: Get Prize 2 on first spin = \(\frac{2}{3}\)
- Path 2: Get Prize 1 first (\(\frac{1}{3}\)), then Prize 2 on second spin (\(\frac{2}{3}\))
- Probability = \(\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}\)
Total probability = \(\frac{2}{3} + \frac{2}{9}\)
Converting to common denominator:
- \(\frac{2}{3} = \frac{6}{9}\)
- Total = \(\frac{6}{9} + \frac{2}{9} = \frac{8}{9}\)
Calculating decimal: \(\frac{8}{9} = 0.889...\)
Closest answer choice: 0.9
Phase 4: Solution
Final Answer
- Prize 1: Select 0.5 (probability that someone wanting Prize 1 receives it)
- Prize 2: Select 0.9 (probability that someone wanting Prize 2 receives it)
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).