Probability and combinations

[venmic]

can some one give me a complete explantion to his


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

[Anurag@Gurome]

can some one give me a complete explantion to his


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

Total no. of possible outcomes = 8C3 = 8 * 7 = 56 ways
If all 3 magazines are fashion magazines, then it can occur in 4C3 = 4 ways.
If 2 are fashion magazines, and 1 is sports magazine, then this can occur in 4C2 * 4C1 = 6 * 4 = 24 ways
If 1 is fashion magazine and 2 are sports magazines, then this can occur in 4C1 * 4C2 = 4 * 6 = 24 ways
So, no. of ways so that at least one of the fashion magazines is selected = 4 + 24 + 24 = 52 ways

Hence, required probability = 52/56 = 13/14

The correct answer is E.

[Stuart@KaplanGMAT]

can some one give me a complete explantion to his


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

Hi,

on most complicated probability problems on the GMAT, the "one minus" approach is the quickest way to solve.

This approach is based on the equation:

Prob(what you want) + Prob(what you don't want) = 1

which rearranges to:

Prob(what you want) = 1 - Prob(what you don't want).

Applying that formula to this question, we see that we want at least 1 fashion magazine; so, what we don't want is 0 fashion magazines.

Prob(no fashion magazines) = Prob(3 sports magazines)

We can now solve using either combinations or straight probability.

Using combinations:

Prob = (# desired outcomes)/(total # of possibilities)
Prob(3 sports) = 4C3/8C3
Prob(3 sports) = (4/1)/(8*7*6/3*2)
Prob(3 sports) = 4/56 = 1/14

Prob(at least 1 fashion) = 1 - 1/14 = 13/14

Using straight probability:

Prob of multiple independent events all occurring = the product of the probabilities each event occurring

Prob(1st one sports)*Prob(2nd one sports)*Prob(3rd one sports)
= 4/8 * 3/7 * 2/6 = 4*3*2/8*7*6 = 4/8*7 = 4/56 = 1/14

Prob(at least 1 fashion) = 1 - 1/14 = 13/14

[manpsingh87]

can some one give me a complete explantion to his


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

total no. of magazines=8;
out of 8 magazines any 3 can be selected in 8C3 ways=56;
probability that atleast one of the selected magazine is of fashion= 1- probability that none of the selected magazine is of fashion.

probability that none of the selected magazine is of fashion.= probability that all the selected magazine is of sports= 4C3;

hence required probability=1-4/56=13/14 hence E

[nitestr]

Hi Stuart,

One quick question on this. By multiplying the Prob, is the order being counted as well?

Prob(1st one sports)*Prob(2nd one sports)*Prob(3rd one sports)
= 4/8 * 3/7 * 2/6 = 4*3*2/8*7*6 = 4/8*7 = 4/56 = 1/14


Thanks

ns