Problem 44 · AMC 8 Stretch
Stretch
Counting & Probability
binomial-probabilityor-process-addsymmetry
A family has 5 children, each equally likely to be a boy or a girl. Find the probability that (a) at least 4 are boys; (b) at least 4 are girls.
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Answer: (a) 3/16; (b) 3/16
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Hint 1 of 3
There are \(2^5 = 32\) equally likely sequences. 'At least 4 boys' means exactly 4 OR exactly 5 boys.
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Hint 2 of 3
Count the sequences: exactly 5 boys (1 way) and exactly 4 boys (5 ways — which one child is the girl). Add, then divide by 32.
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Hint 3 of 3
For (b), boys and girls are equally likely, so by symmetry the 'at least 4 girls' answer is the same as 'at least 4 boys'.
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Approach: Binomial count, with a symmetry argument for part (b)
- There are \(2^5 = 32\) equally likely sequences.
- (a) At least 4 boys = exactly 4 OR exactly 5 boys. Exactly 5 boys: 1 sequence. Exactly 4 boys: 5 sequences (which single child is the girl). Total favorable: \(5 + 1 = 6\), so \(\frac{6}{32} = \frac{3}{16}\).
- (b) At least 4 girls: since boys and girls are equally likely, swapping the labels shows this has the SAME probability as part (a), so \(\frac{6}{32} = \frac{3}{16}\).
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