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Container of 4 blue disks and 8 green disks

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by EbrahimHashem » Fri Apr 02, 2010 11:20 am
The only contents of a container are 4 are blue disks and 8 are green disks. If 3 disks are selected one after the other, and at random and without replacement from the container, what is the probability that 1 of the disks selected is blue, and 2 of the disks selected are green?


A) 21/55

B) 28/55

C) 34/55

D) 5/8

E) 139/220

Enjoy it :D
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by outreach » Fri Apr 02, 2010 11:45 am
1 blue ball can be selected in 4c1 ways
2 green ball can be selected in 8c2 ways
3 balls can be selected from 12 balls in 12c3 ways

total=(4c1*8c2)/12c3
=28/55
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by a_gmat » Fri Apr 02, 2010 12:04 pm
We need all cases for 1 blue and 2 green. Possible scenarios:

BGG = 4/12*8/11*7/10
GBG = 8/12*4/11*7/10
GGB = 8/12*7/11*4/10

add all 3 you get = 3*4*8*7/12*11*10 = 28/55 (B)
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by analyst218 » Fri Apr 02, 2010 12:10 pm
EbrahimHashem wrote:The only contents of a container are 4 are blue disks and 8 are green disks. If 3 disks are selected one after the other, and at random and without replacement from the container, what is the probability that 1 of the disks selected is blue, and 2 of the disks selected are green?


A) 21/55

B) 28/55

C) 34/55

D) 5/8

E) 139/220

Enjoy it :D
B.

BGG
GBG
GGB

28/165 x 3 = 28/55
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