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A box contains 4 red chips and 2 blue chips. If two chips ar

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

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New post 29 Oct 2009, 01:15
3
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A
B
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D
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Difficulty:

  5% (low)

Question Stats:

92% (01:26) correct 8% (01:35) wrong based on 326 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 1-same color??
1-Both red + 1-both 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  [#permalink]

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New post 18 Jan 2011, 09:55
4
3
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  [#permalink]

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New post 29 Oct 2009, 01:40
2
1
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: A box contains 4 red chips and 2 blue chips. If two chips ar  [#permalink]

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New post 29 Oct 2009, 02:25
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  [#permalink]

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New post Updated on: 14 Oct 2014, 12:29
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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

Originally posted by mvictor on 14 Oct 2014, 12:23.
Last edited by mvictor on 14 Oct 2014, 12:29, edited 3 times in total.
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Re: A box contains 4 red chips and 2 blue chips. If two chips ar  [#permalink]

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New post 14 Oct 2014, 12:39
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  [#permalink]

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New post 14 Oct 2014, 12:56
mvictor wrote:
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


Yes, that's correct.
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Re: A box contains 4 red chips and 2 blue chips. If two chips ar  [#permalink]

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New post 24 Mar 2017, 08:25
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  [#permalink]

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New post 17 Jul 2018, 18:41
niks18 pushpitkc chetan2u pikolo2510

What 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  [#permalink]

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New post 17 Jul 2018, 18:59
adkikani wrote:
niks18 pushpitkc chetan2u pikolo2510

What 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 [#permalink] 17 Jul 2018, 18:59
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