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There are 4 blue disks

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There are 4 blue disks [#permalink] New post 19 Sep 2016, 07:18
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There are 4 blue disks,6 green disk and nothing else in the container.If 4 disk are to be selected one after the another at random and without replacement
from the container.What is the probability that 2 selected are blue and 2 selected are green?

A) \(\frac{1}{7}\)

B) \(\frac{5}{14}\)

C) \(\frac{3}{7}\)

D) \(\frac{25}{56}\)

E) \(\frac{12}{25}\)



when we solve the above it should be (4/10)*(3/9)*(6/8)*(5/7)=1/14 which is not even present in the option.Plzz explain.
[Reveal] Spoiler: OA

Last edited by Carcass on 20 Sep 2016, 07:38, edited 1 time in total.
Edited by Carcass
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Re: There are 4 blue disks [#permalink] New post 19 Sep 2016, 07:51
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Sonalika42 wrote:

when we solve the above it should be (4/10)*(3/9)*(6/8)*(5/7)=1/14 which is not even present in the option.Plzz explain.


Hi,

you assume it is blue blue green green, but there are 5 other possible orders:

bgbg
bggb
gbgb
ggbb
gbbg

You can observe, though, that the probabilities of these orders appearing are equal, hence you can arrive at the correct answer by multiplying 1/14 by 6.

Hence option C is correct.

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Re: There are 4 blue disks [#permalink] New post 09 Jan 2019, 00:31
Expert's post
There are two easy ways to answer this question.

1st way.
just multiplying probabilities.
P(B)*P(G)*P(B)*P(G)= (4/10)*(3/9)*(6/8)*(5/7)= 1/14
P(B)*P(B)*P(G)*P(G)= 1/14 as well, so every probability = 1/14

4 selections will be made, 2 Green ,and 2 Blue, use a shortcut for finding the total combinations of 2 blue and 2 gren and just do 4c2 =6
thus the total probability of getting 2 blue and 2 green is 1/14*6= 6/14= 3/7= C

The 2nd way is quicker and you know some combinatorics.

The total combinations of 4 disks out of 10 is 10c4= 210
The total combinations of 2 out of 6 green disks is 6c4= 15
The total combinations of 2 out of 4 blue disks is 4c2= 6

(15*6)/210= 90/210= 3/7

C is the answer
Re: There are 4 blue disks   [#permalink] 09 Jan 2019, 00:31
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There are 4 blue disks

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