tlntd42
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
I know its combinations but i cant seem to get the answer
You might want to post this question in the math section. But here's how I'd solve it -
In this case I think it's easier to solve for the probability that
none of the magazines selected will be fashion, then subtract that number from 1. If there are 8 magazines, 4 of each, the probability that the first magazine selected is sports is 4/8, or 1/2. That leaves 7 magazines to choose from, 3 of which are sports. So the probability that the second one chosen is sports is 3/7. That leaves 6 with 2 sports, so the third probability is 2/6, or 1/3. Multiply those together: 1/2x3/7x1/3, for a total probability of getting all sports magazines = 1/14. So the opposite probability, i.e. that at least one of the magazines is fashion, = 1 - 1/14, or 13/14.
Clear as mud?