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I'm sorry, I do not seem to get a hold of this problem:
A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 apples are selected at random fron the basket, what is the probability that 2 of the apples will be red and 1 green?
I went this way:
(7/10)(6/9)(3/8)=7/40
the answer is posted as 21/40, could you explain what am I doing wrong and correct it so I won't make the same mistake?
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This need not be the case - you could select a green apple first, second or third. There are then two (basically equivalent) ways to look at how to get the correct answer:
1) Since there are three choices as to when the green apple is removed, multiply your answer by 3:
2) Since there are three choices as to when the green apple is removed, find the answer by adding the three relevant probabilities:
Last edited by Dross (2007-10-12 23:15:22)
Bad speling makes me [sic]
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You can also look at the problem this way.
Total number of ways of selecting 3 apples from 10 = [sup]10[/sup]C[sub]3[/sub] = 120
Total number of ways of selecting 2 red apples from 7 and 1 green one from 3 = [sup]7[/sup]C[sub]2[/sub]×[sup]3[/sup]C[sub]1[/sub] = 63
Hence probability of picking 2 red and 1 green apples from the bunch = 63⁄120 = 21⁄40.
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thank you so much guys.
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I think Jane's method is called Hypergeometric Distribution
http://www.mnlottery.com/hypergeo.html (The Wikipedia page is incomplete LoL)
Last edited by Identity (2007-10-13 01:32:25)
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