Problem 15 · 2019 AMC 8
Medium
Counting & Probability
conditional-probabilityproportion
On a beach 50 people are wearing sunglasses and 35 people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is 25. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?
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Answer: B — 7/25.
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Hint 1 of 2
Both probabilities are about the same group — the people wearing both. So nail down that one number first; it's the bridge between the two questions.
Still stuck? Show hint 2 →
Hint 2 of 2
"2/5 of cap-wearers also wear sunglasses" means both = (2/5) of 35. Once you have a head-count for both, each probability is just both ÷ (the group you're conditioning on).
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Approach: find the both-count, then re-divide
- The given probability is a fraction of the 35 cap-wearers: people wearing both = (2/5) × 35 = 14.
- The reversed question keeps the same 14 in both, now as a fraction of the 50 sunglasses-wearers: P(cap | sunglasses) = 14 / 50 = 7/25.
- Why this transfers: a conditional probability is always (overlap) ÷ (the condition group). Convert one conditional into the raw overlap count, and you can flip it onto any other group — the overlap doesn't change, only the denominator does.
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