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A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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29 Oct 2009, 02:15

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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?

A. 1/2 B. 8/15 C. 7/12 D. 2/3 E. 7/10

............................................................ 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)

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.

Re: A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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18 Jan 2011, 10:55

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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: A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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25 Oct 2013, 07:45

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A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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14 Oct 2014, 13:23

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can someone explain if the method used by me is correct?

we have 6 chips, and out of which we have to select 2, hence 6C2=15 we have 4 red chips - 4C1=4 2 blue chips - 2C1=2 then we multiply the combinations 4x2 and get 8 putting all together, we have the probability of selecting one red and one blue - 8/15

Last edited by mvictor on 14 Oct 2014, 13:29, edited 3 times in total.

can someone explain if the method used by me is correct?

we have 6 chips, and out of which we have to select 2, hence 6C2=15 we have 4 red chips - 4C1=4 2 blue chips - 2C1=2 then we multiply the combinations 4x2 and get 8 putting all together, we have the probability of selecting one red and one blue - 8/15

Re: A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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09 Dec 2015, 14:55

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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30 Dec 2016, 22:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A box contains 4 red chips and 2 blue chips. If two chips ar [#permalink]

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24 Mar 2017, 09:25

Quote:

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?

(Just for my practice) I am considering this problem WITH replacement:

* One (conventional) way to solve is ([4][/6]x[2][/6] + [2][/6]x[4][/6]) = [4][/9];

* BUT, I am struggling hard to solve the same using Combinations: My approach [4C1 x 2C1][/6^2] demands a multiplication by 2 .