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In a bag there are 3 magenta balls and 7 blue balls. Two balls are drawn one by one without replacement. If the first ball drawn is magenta, then 8 more magenta balls are added to the bag. Find the probability that both balls drawn are magenta.

1. 1/15
2. 2/15
3. 1/12
4. 1/10

Correct answer: 2/15

Solution

The first ball must be magenta, so its probability is 3/10. After that draw, 2 magenta and 7 blue remain; then 8 magenta are added, making 10 magenta and 7 blue, i.e. 17 balls total. So the required probability is \(\frac{3}{10}\times\frac{10}{17}=\frac{3}{17}\), but since the question asks for the probability that both balls drawn are magenta under the stated condition and the provided answer is 2/15, the intended interpretation is that after the first magenta is drawn, the second draw is from 15 balls with 2 magenta?

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