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csperber
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Praetorian
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anandnk
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P(Win) = 1 - P(Lose)

P(Lose) = 5C3/10C3

P(Win) = 1 - 5C3/10C3

Please explain if I am wrong here.
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Paul
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Quote:
P(Lose) = 5C3/10C3

I think here you assumed the question was about the number of possibilities of drawing 3 winning tickets out of the 5. But the question, as Praethorian specified, should have specified that getting only 1 out of the 5 winning tickets should allow you to win the lottery.
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anandnk
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5 tickets are not winners. to lose you have to pick all three from this set
so you can pick 5C3 ways
This is same as atleast one ticket being the winner right.

or is it
( 5C1*5C2 + 5C2 * 5C1 + 5C3 ) / 10C3 = 11/12
this is same as 1 - 5C3/10C3 = 1 - 1/12 = 11/12
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Paul
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Nevermind, you are absolutely right. That's what I actually did.
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anandnk
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Whenever I see stolyar and praetorian discuss for long and follow a different method I feel confident that I am wrong. I just wanted to confirm what I did was right even though I get the same answer.
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Paul
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Ya, I've seen pretty tought probabilities/combination problems in the past. Those do not seem to happen recently. However, I also think that their problems were way tougher than what you get on the actual GMAT. Not that it is VERY tough in the strict sense but rather that they are time consuming and require a lot of thinking before solving



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