Problem 43 · AMC 8 Stretch
Core
Counting & Probability
binomial-probabilityor-process-add
A family has 4 children, each equally likely to be a girl or a boy. Find the probability that (a) exactly 3 are girls; (b) none are girls; (c) at least 2 are girls.
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Answer: (a) 1/4; (b) 1/16; (c) 11/16
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Hint 1 of 3
There are \(2^4 = 16\) equally likely boy/girl sequences for the 4 children.
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Hint 2 of 3
For each part, count how many of the 16 sequences fit, then divide by 16.
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Hint 3 of 3
(a) choose which 3 of 4 are girls: 4 ways. (b) all boys: 1 way. (c) 'at least 2 girls' means 2, 3, or 4 girls: count \(6 + 4 + 1\).
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Approach: Favorable sequences over 16 equally likely sequences
- There are \(2^4 = 16\) equally likely sequences for the 4 children.
- (a) Exactly 3 girls: 4 ways to pick which 3 are girls, so \(\frac{4}{16} = \frac{1}{4}\).
- (b) No girls (all boys): just 1 sequence, so \(\frac{1}{16}\).
- (c) At least 2 girls means 2, 3, or 4 girls: counts \(6 + 4 + 1 = 11\), so \(\frac{11}{16}\).
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