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Probabilities

by anastasios1984 » Mon Mar 26, 2012 3:12 am
Α 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.15/36 B.3/28 C.1/28 D.1/56 E.1/168

The official answer is B - I cannot understand the logic behind it though. What I was doing was the below - unfortunately is incorrect!

8!/3!*5!=56
3!/2!*1!=3
and then favourite outcomes / total number of outcomes => 3/56
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by killer1387 » Mon Mar 26, 2012 3:36 am
Α 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.15/36 B.3/28 C.1/28 D.1/56 E.1/168

-->
let most expensive one's be X and Y (X,Y+ other 6 computers)

now we have to make a group of three with X and Y in it. hence we have one place vacant, which can be filled in 6C1 ways.
total number of ways for making grps of 3= 8c3

hence required probability= 6c1/8c3=3/28

hence B
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by GMATGuruNY » Mon Mar 26, 2012 3:47 am
anastasios1984 wrote:Α 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.15/36 B.3/28 C.1/28 D.1/56 E.1/168
P = (good combinations)/(total possible combinations).

Total possible combinations:
The number of combinations of 3 that can be formed from 8 choices = 8C3 = 56.

Good outcomes:
A good combination includes the 2 most expensive computers.
Thus, there is ONLY ONE POSITION that needs to be filled here: the THIRD computer in the combination.
Since there are 6 remaining computers, the number of options for the third computer = 6.

(Good combinations)/(total possible combinations) = 6/56 = 3/28.

The correct answer is B.
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by anastasios1984 » Mon Mar 26, 2012 3:43 pm
Thank you both - very sharp explanations and prompt replies! Love this network - makes my life easier!
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