A bag contains 3 red and 2 blue counters. One counter is drawn and not replaced. A second counter is then drawn. What is the probability that both counters are red?

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Question

A bag contains 3 red and 2 blue counters. One counter is drawn and not replaced. A second counter is then drawn. What is the probability that both counters are red?

Accepted Answer

3/10 — P(red first) = 3/5. After removing a red, 2 red and 2 blue remain, so P(red second | red first) = 2/4 = 1/2. Multiplying gives 3/5 × 1/2 = 3/10. Option B incorrectly uses replacement, and D ignores probability entirely.

Answer

1/2

Answer

3/10

Answer

9/25

Answer

6/25

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