Jcpenny wrote:
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
We can use the formula:
P(at least one fashion magazine selected) = 1 - P(no fashion magazines are selected)
The number of ways to select 3 non-fashion magazines is 4C3 = 4!/[3!(4-3)!] = 4!/3! = 4 ways.
The number of ways to select 3 magazines from 8 is:
8C3 = 8!/[3!(8 - 3)!] = 8!/[3!5!] = (8 x 7 x 6)/( 3 x 2) = = 56 ways
So P(at least one fashion magazine selected) = 1 - 4/56 = 52/56 = 13/14.
Alternate Solution:
We can use the formula:
P(at least one fashion magazine selected) = 1 - P(no fashion magazines are selected)
The probability that no fashion magazines are selected is the same as saying that only sports magazines were selected, which is: 4/8 x 3/7 x 2/6 = 24/336 = 4/56 = 1/14.
Thus, P(at least one fashion magazine selected) = 1 - 1/14 = 14/14 - 1/14 = 13/14.
Answer: E
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