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There are 10 new action figures. Jim has 3 of them. What is

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There are 10 new action figures. Jim has 3 of them. What is [#permalink] New post 06 Jan 2004, 06:12
There are 10 new action figures. Jim has 3 of them. What is the probability that the first two action figures from the group of 10 will NOT be any of the 3 that Jim has?
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 [#permalink] New post 06 Jan 2004, 09:04
assuming you cannot pick the same figure twice, 7/10 * 6/9, 42/90

assuming you can pick the same figure twice, 7/10 * 7/10, 49/100

Maybe I misunderstand the question?
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 [#permalink] New post 07 Jan 2004, 06:32
Total ways to chose 3 actions figures is 10C3 = 120
3 actions figures can be arranged in 3! ways = 6
For each of these combinations the thrid figure can be one amoung 8
so we have 6*8= 48 ways
probability that any two of the action figures are 1 and 2 = 48/120 = 2/5

probability that the set of three does not contain two of them = 1-2/5 = 3/5

Please post the official answer.

Hi Geethu,
Explain how you gor 5/8. May then the problem will become clear.
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 [#permalink] New post 07 Jan 2004, 06:58
I thought the answer should be 1-(10C2/10C3).
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 [#permalink] New post 07 Jan 2004, 07:02
You are not accounting for third action figure ( it can have 8 possible combinations right ? )
  [#permalink] 07 Jan 2004, 07:02
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There are 10 new action figures. Jim has 3 of them. What is

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