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#1 2007-10-12 22:51:06

simone
Member
Registered: 2007-10-05
Posts: 19

probability

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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#2 2007-10-12 23:14:07

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: probability

By calculating
, what you're actually calculating is the probability of removing two red apples from the bag, and THEN removing one green apple. (because by using the fraction 3/8, you're assuming there are 8 apples left in the bag when you select one of the three green apples)

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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#3 2007-10-12 23:26:02

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: probability

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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#4 2007-10-13 01:09:17

simone
Member
Registered: 2007-10-05
Posts: 19

Re: probability

thank you so much guys.kissroflol

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#5 2007-10-13 01:25:01

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: probability

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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