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TAGS: Manager Joined: 12 Jan 2016
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There are 4 blue disks [#permalink]
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Question Stats: 80% (01:07) correct 20% (01:38) wrong based on 5 sessions
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 GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4810
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 118

Kudos [?]: 1899  , given: 397

Re: There are 4 blue disks [#permalink]
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Expert's post
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.

_________________

Sandy
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Manager  Joined: 01 Nov 2018
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Re: There are 4 blue disks [#permalink]
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 Re: There are 4 blue disks   [#permalink] 09 Jan 2019, 00:31
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