venmic wrote: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
