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| Math Central | Quandaries & Queries |
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Question from Brian: Two participants are alternatively selecting eggs from a basket of 6 hard-boiled eggs and 6 uncooked eggs. Thank you Brian |
Hi Brian,
At the start there are 12 eggs in the basket, 6 hard-boiled and 6 uncooked so the probability that the first egg selected is hard-boiled is $\frac{6}{12}.$ Now there are 11 eggs in the basket, 5 hard-boiled and 6 uncooked so the probability that the second egg selected is uncooked given that the first egg was hard-boiled is $\frac{6}{11}.$ Thus the probability that the first egg selected is hard-boiled AND the second egg selected is uncooked given that the first egg was hard-boiled is $\frac{6}{12} \times \frac{6}{11}.$
The pattern continues, the probability that the third egg selected is hard-boiled given that first egg selected is hard-boiled AND the second egg selected is uncooked given that the first egg was hard-boiled is $ \frac{6}{12} \times \frac{6}{11} \times \frac{5}{10}.$
Continue for 8 eggs.
Penny
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