Problem 33 · AMC 8 Stretch
Core
Counting & Probability
and-process-multiplylogical-reasoning
A box has 7 marbles: 3 red and 4 blue. Two are drawn one after another. Find the probability both are red if (a) the first is put back before the second draw; (b) the first is not put back.
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Answer: (a) 9/49; (b) 1/7
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Hint 1 of 3
Two draws form an AND process. Ask whether the first draw changes the box for the second.
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Hint 2 of 3
With replacement the chance stays \(3/7\) each draw; without replacement, after one red is gone there are 2 reds left out of 6.
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Hint 3 of 3
(a) \((3/7) \times (3/7)\). (b) \((3/7) \times (2/6)\).
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Approach: AND process (with and without replacement)
- (a) With replacement (independent): each draw is red with probability \(\frac{3}{7}\), so \(\frac{3}{7} \times \frac{3}{7} = \frac{9}{49}\).
- (b) Without replacement (dependent): after one red is removed, 2 reds remain among 6 marbles, so \(\frac{3}{7} \times \frac{2}{6} = \frac{3}{7} \times \frac{1}{3} = \frac{1}{7}\).
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