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A box contains 4 red chips and 2 blue chips. If two chips ar
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29 Oct 2009, 02:15
Question Stats:
89% (01:07) correct 11% (00:50) wrong based on 317 sessions
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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 1same color?? 1Both red + 1both blue (4/6)(3/6)+(2/6)(1/6)
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Re: A box contains 4 red chips and 2 blue chips. If two chips ar
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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




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Re: A box contains 4 red chips and 2 blue chips. If two chips ar
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29 Oct 2009, 02:40
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 1same color?? 1Both red + 1both 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/ 5You 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: A box contains 4 red chips and 2 blue chips. If two chips ar
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29 Oct 2009, 03:25
thank you..I'll go jump off the balcony now..!!grumble grumble..Silly silly!!
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A box contains 4 red chips and 2 blue chips. If two chips ar
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Updated on: 14 Oct 2014, 13:29
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
Originally posted by mvictor on 14 Oct 2014, 13:23.
Last edited by mvictor on 14 Oct 2014, 13:29, edited 3 times in total.



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Re: A box contains 4 red chips and 2 blue chips. If two chips ar
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14 Oct 2014, 13:39
Bunuel, can u pls specify if my method is good to use for solving this problem



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Re: A box contains 4 red chips and 2 blue chips. If two chips ar
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14 Oct 2014, 13:56



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Re: A box contains 4 red chips and 2 blue chips. If two chips ar
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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 . Anyone to help? TIA !



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A box contains 4 red chips and 2 blue chips. If two chips ar
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17 Jul 2018, 19:41
niks18 pushpitkc chetan2u pikolo2510What is incorrect in below approach? Probability = (Favorable outcomes) / (total no of outcomes) P (red) = 1/4 P (blue) = 1/2 Combined probability = P(r)* P(b) = 1/8
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Re: A box contains 4 red chips and 2 blue chips. If two chips ar
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17 Jul 2018, 19:59
adkikani wrote: niks18 pushpitkc chetan2u pikolo2510What is incorrect in below approach? Probability = (Favorable outcomes) / (total no of outcomes) P (red) = 1/4 P (blue) = 1/2 Combined probability = P(r)* P(b) = 1/8 P(r) is 4/6... Since 4 reds are there so favourable outcomes and 6 total outcomes.. Since it is without replacement, only 5 are left, so total outcomes =7 and blue are 2 so P(b)=2/5 If blue first then 2/6 and red 4/5
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Re: A box contains 4 red chips and 2 blue chips. If two chips ar &nbs
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