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by kadishmj » Wed Sep 05, 2007 11:57 am
There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected?

A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14



.
.
.
.
.
OA is E. 13/14
I would appreciate it if someone could explain the method for getting the solution. Thanks!
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by ri2007 » Wed Sep 05, 2007 12:43 pm
First calculate the probability that not even one fashion magazine is selected =

(4/8) * (3/7) * (2/6) =1/14

The probability of getting at least one fashion magazine is

1 - probability that not even one fashion magazine is selected = 13/14
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by kadishmj » Wed Sep 05, 2007 1:15 pm
ri2007 wrote:First calculate the probability that not even one fashion magazine is selected =

(4/8) * (3/7) * (2/6) =1/14

The probability of getting at least one fashion magazine is

1 - probability that not even one fashion magazine is selected = 13/14
Thanks for the response ri2007, but if I'm calculating the probability that not even one fashion magazine is selected, I would still have to get 4 magazines, so the calculation should be :

(4/8) * (3/7) * (2/6) * (1/5) = 1/70

right?
Last edited by kadishmj on Wed Sep 05, 2007 1:19 pm, edited 1 time in total.
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by ri2007 » Wed Sep 05, 2007 1:18 pm
no because you have to calculate the probability that not even one fashion magazine is selected in the 3 chances that you pick. You can check out the Princeton Review Crack the GMAT they have a almost identical question with a great explaination.
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by agps » Wed Sep 05, 2007 3:48 pm
the calculation 4/8*3/7*2/6 is the probability that the 1st, 2nd and 3rd choices are sports magazines. = 24/336 = 1/14

now you want all other options except this one. so 1-1/14 is the answer or 13/14
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by vinaysingh » Tue Sep 11, 2007 11:15 pm
Answer should be 13/14
The solution goes like this:-
Total events 8C3 = 56

For at least one Fashion magazine, the cases can be as follows

1. F S S
2. F F S
3. F F F

For 1. the total events will be 4C1 x 4C2 (We are selecting 1 F Mag from a group of 4 and 2 S Mag from a group of 4) = 4 x 6 = 24
For 2 . the total events will be 4C2 x 4C1 = 24
For 3 the total events will be 4C3 = 4

Total favorable events will be 24 + 24 + 4 = 52
Prob = 52/56 = 13/14 (The ans)
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