Problem 41 · AMC 8 Stretch
Stretch
Counting & Probability
binomial-probability
Find the probability of getting (a) a three or a five in exactly 2 out of 4 rolls of a fair die; (b) exactly 2 even numbers in 4 rolls of a fair die.
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Answer: (a) 8/27; (b) 3/8
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Hint 1 of 3
Find the single-roll success chance first: 'three or five' is \(2/6 = 1/3\); 'even' is \(3/6 = 1/2\).
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Hint 2 of 3
For each, count the patterns (which 2 of the 4 rolls are the successes) and multiply by (success chance)^2 times (failure chance)^2.
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Hint 3 of 3
There are 6 ways to choose which 2 of 4 rolls succeed. (a) \(p = 1/3\); (b) \(p = 1/2\).
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Approach: Binomial probability with 6 patterns of 2-of-4
- There are 6 ways to choose which 2 of the 4 rolls are the successes.
- (a) Success = three or five, \(p = \frac{2}{6} = \frac{1}{3}\), failure \(q = \frac{2}{3}\): \(6 \times \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 = 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{24}{81} = \frac{8}{27}\).
- (b) Success = even, \(p = \frac{1}{2}\): \(6 \times \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8}\).
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