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We have a tree with 100 apples. There are 10 red apples and the rest are green. Lisa is picking apples at random. When she pick the 3rd red apple she stop. What is the probability that Lisa has exactly 7 green apples (in addition to the 3 red apples she picked)?

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Hint: she has to pick 2 red and 7 green, then pick a red. How many ways are there to do that? How many ways to pick 10 apples?

Added: to pick an unordered 2 red and 7 green, you need to choose 2 out of 10 red and choose 7 of 90 green. Divide this by an unordered choice of 9 of 100

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  • $\begingroup$ i thought about it too but when she pick a red apple the probability of picking red apple next time will changed, same with green $\endgroup$ Dec 6 '13 at 15:10
  • $\begingroup$ But you can just look at the chance that the first nine are what you want. You pick 9 out of the 100 and ask the chance they have the correct color distribution. $\endgroup$ Dec 6 '13 at 15:16
  • $\begingroup$ pick 2 red: (10/100)*(9/99)* pick 7 green:(90/98)*(89/97)....*(84/92) sorting 9 apples in a line : 9! choosing the last red apple: (8/91) now what? $\endgroup$ Dec 6 '13 at 15:26
  • $\begingroup$ We are counting unordered. Pick two red is ${10 \choose 2}$ and seven green is ${90 \choose 7}$ ways, so the chance of that is $\frac{{10 \choose 2}{90\choose 7}}{{100 \choose 9}}\approx 0.1767$ Then the chance of a red (from what is left) is $\frac 8{91}$ $\endgroup$ Dec 6 '13 at 17:41

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