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11 Dec 2014, 06:57
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (02:00) correct
34%
(02:14)
wrong
based on 545
sessions
History
Date
Time
Result
Not Attempted Yet
A computer wholesaler sells eight different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?
A) 15/56
B) 3/28
C) 1/28
D) 1/56
E) 1/168
Source: Chili Hot GMAT
A) 15/56
B) 3/28
C) 1/28
D) 1/56
E) 1/168
Source: Chili Hot GMAT
Direct probability:
2/8*1/7*1=1/28
we have 3!/2!*1!=3 such cases
(1/28)*3=3/28
Reverse probability:
Not desirable cases when we have all three not highest price or only one in highest price
6/8*5/7*4/6=5/14
2/8*6/7*5/6=(5/28)*3 cases=15/28
1 - (5/14 + 15/28) = 1 - 25/28=3/28
Combination approach:
denominator (all possible cases): 8C3=8!/3!*5!=56
numerator (desirable cases): 2C2*6C1=6
6/56=3/28
Reverse combination:
denimonator (all cases): 8C3=56
numerator (undesirable cases): 6C3 + (2C1*6C2) = 50
1 - 50/56=6/56=3/28
B
in hope to get Kudos from Bunuel) and you, guys
2/8*1/7*1=1/28
we have 3!/2!*1!=3 such cases
(1/28)*3=3/28
Reverse probability:
Not desirable cases when we have all three not highest price or only one in highest price
6/8*5/7*4/6=5/14
2/8*6/7*5/6=(5/28)*3 cases=15/28
1 - (5/14 + 15/28) = 1 - 25/28=3/28
Combination approach:
denominator (all possible cases): 8C3=8!/3!*5!=56
numerator (desirable cases): 2C2*6C1=6
6/56=3/28
Reverse combination:
denimonator (all cases): 8C3=56
numerator (undesirable cases): 6C3 + (2C1*6C2) = 50
1 - 50/56=6/56=3/28
B
in hope to get Kudos from Bunuel) and you, guys
General Discussion
11 Dec 2014, 11:23
Kudos
Bookmarks
Bunuel
8!/3!*5!=56
1( 1 way of picking top 2 comps in pricing list) ∗ 6 ( ypu can pick third comp in 6 ways as there are 6 more left)=6
and then favourite outcomes / total number of outcomes => 6/56 = 3/28, so B?
