Problem 3 · AMC 8 Stretch
Stretch
Counting & Probability
work-backwardcomplementary-countingorganizing-data
Imagine a tiny 'year' with only 5 days, and a group of 3 people who each have a birthday on a random one of those 5 days. What is the probability that at least two of them share a birthday? (This is the same idea as the famous '30 students, 365 days' birthday problem, just with small numbers you can actually compute.)
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Answer: \(P(\text{shared})=1-0.48=0.52\)
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Hint 1 of 4
Finding 'at least two share' directly is messy (there are several ways it could happen). Ask the EASIER opposite question first: what is the probability that all 3 birthdays are different?
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Hint 2 of 4
Line the people up. Person 1 can be born on any day. Person 2 must avoid person 1's day. Person 3 must avoid both used days.
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Hint 3 of 4
All different: \(\frac{5}{5}\times\frac{4}{5}\times\frac{3}{5}\). (Each new person has fewer 'free' days.)
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Approach: Complementary counting — find P(all different) and subtract from 1
- Computing 'at least two share' head-on is messy, so ask the easier opposite question: what is the chance all 3 birthdays are different? Then subtract from 1.
- Line the people up. Person 1: any of the 5 days works, \(\frac{5}{5}\). Person 2: must avoid person 1's day, so 4 of 5 are safe, \(\frac{4}{5}\). Person 3: must avoid both used days, so 3 of 5 are safe, \(\frac{3}{5}\).
- Multiply: \(P(\text{all different})=\frac{5}{5}\times\frac{4}{5}\times\frac{3}{5}=\frac{60}{125}=\frac{12}{25}=0.48\).
- Since 'all different' and 'at least one shared' are opposite events that together are certain, \(P(\text{at least two share})=1-0.48=0.52\). Even with only 3 people a shared birthday is slightly more likely than not — the same reason the real 30-people, 365-day problem comes out to about 70%.
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