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07 Jan 2014, 04:43
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
83% (02:25) correct
17%
(02:42)
wrong
based on 511
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The only contents of a container are 4 blue disks and 8 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
A. 21/55
B. 28/55
C. 34/55
D. 5/8
E. 139/220
07 Jan 2014, 04:51
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sanjoo
APPROACH #1:
\(P(BGG) = \frac{C^1_4*C^2_8}{C^3_{12}}=\frac{4*28}{220}=\frac{28}{55}\).
APPROACH #2:
\(P(BGG) =\frac{3!}{2!}*\frac{4}{12}*\frac{8}{11}*\frac{7}{10}=\frac{28}{55}\). We need to multiply by 3!/2! because BGG scenario can occur in several ways: BGG, GBG, GGB (permutations of 3 letters BGG).
Answer: B.
Hope it's clear.
General Discussion
bharatdivvela
Joined: 17 Oct 2013
Last visit: 02 Jan 2024
Posts: 34
Given Kudos: 21
Schools: HEC Dec"18
GMAT Date: 02-04-2014
07 Jan 2014, 10:30
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sanjoo
Prob of selecting one blue and two green is (4*8*7/12*11*10)
one blue and two green can be arranged in 3!/2! ways
hence the prob becomes 3*4*8*7/12*11*10=28/55
