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4 red chips and 2 blue chips

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4 red chips and 2 blue chips [#permalink] New post 29 Oct 2009, 02:15
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Guys please tell me where am i going wrong:

A box contains 4 red chips and 2 blue chips. If two chips are selected at random without replacement, what is the probability that the chips are different colors?


............................................................
My solution:
Got the correct answer by the following method.
(4/6)(2/5)+(2/6)(4/5)= 8/15>>OA

My querie:
Why are we not getting the answer when we do 1-same color??
1-Both red + 1-both blue
(4/6)(3/6)+(2/6)(1/6)
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Re: 4 red chips and 2 blue chips [#permalink] New post 29 Oct 2009, 02:40
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tejal777 wrote:
Guys please tell me where am i going wrong:

A box contains 4 red chips and 2 blue chips. If two chips are selected at random without replacement, what is the probability that the chips are different colors?


............................................................
My solution:
Got the correct answer by the following method.
(4/6)(2/5)+(2/6)(4/5)= 8/15>>OA

My querie:
Why are we not getting the answer when we do 1-same color??
1-Both red + 1-both blue
(4/6)(3/6)+(2/6)(1/6)


First solution is right.

As for the second one: you've made a little mistake you've avoided when solving for the different colors.

We surely can get the answer for same color by exactly the same method:

P(both same color)=4/6*3/5+2/6*1/5

You just forgot that when taking the first chip there are 5 left, so the chances of getting is out of 5, not out of 6.

Hope it's clear.
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Re: 4 red chips and 2 blue chips [#permalink] New post 29 Oct 2009, 03:25
thank you..I'll go jump off the balcony now..!!grumble grumble..Silly silly!!
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Re: 4 red chips and 2 blue chips [#permalink] New post 18 Jan 2011, 10:55
Total selection ways: 6C2=6!/2!4!=15
Selecting one blue chip out of two: 2C1=2!/1!1!=2
Selecting one red chip out of four: 4C1=4!/1!3!=4
Thus, (2C1*4C1)/6C2=(2*4)/15=8/15
Re: 4 red chips and 2 blue chips   [#permalink] 18 Jan 2011, 10:55
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