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

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

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New post 29 Oct 2009, 02:15
4
12
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

91% (01:27) correct 9% (01:35) wrong based on 391 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 are selected  [#permalink]

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New post 18 Jan 2011, 10:55
4
6
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 are selected  [#permalink]

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New post 29 Oct 2009, 02:40
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 are selected  [#permalink]

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

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New post Updated on: 14 Oct 2014, 13:29
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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, 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 are selected  [#permalink]

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New post 14 Oct 2014, 13:39
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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 are selected  [#permalink]

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New post 14 Oct 2014, 13:56
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Re: A box contains 4 red chips and 2 blue chips. If two chips are selected  [#permalink]

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New post 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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Re: A box contains 4 red chips and 2 blue chips. If two chips are selected  [#permalink]

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New post 17 Jul 2018, 19:41
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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 are selected  [#permalink]

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New post 17 Jul 2018, 19:59
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 are selected  [#permalink]

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New post 07 Jan 2019, 09:26
Question...

Why is (4/6)(2/5) + (2/6)(4/5) correct answer?

The way I am interpreting this question is that whatever you draw on the first attempt does not matter.

If you get RED chip on the first draw, you just need to draw blue chip on the second draw, which is probability of 2/5.
If you get BLUE chip on the 1st draw, you just need to draw red chip on the 2nd draw, which is probability of 4/5.

Because what you draw on your 1st pick doesn't matter, wouldn't we just add (2/5) + (4/5)?

Please help!!
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Re: A box contains 4 red chips and 2 blue chips. If two chips are selected  [#permalink]

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New post 18 Jan 2019, 01:51
ylee6262
I find the direct approach confusing. Do it the other way and work out how to have all the same colour

Chance all red 4/6·3/5=2/5=6/15
Chance all blue 2/6·1/5=1/15

15/15-6/15-1/15=8/15
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Re: A box contains 4 red chips and 2 blue chips. If two chips are selected   [#permalink] 18 Jan 2019, 01:51
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