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There are 8 magazines lying on a table; 4 are fashion magazines and th
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26 Jan 2022, 07:28
26 Jan 2022, 07:28
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66% (01:31) correct
33% (01:35) wrong based on 6 sessions
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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
A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14
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Re: There are 8 magazines lying on a table; 4 are fashion magazines and th
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26 Jan 2022, 08:05
26 Jan 2022, 08:05
Carcass 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
A. 1/2
B. 2/3
C. 32/35
D. 11/12
E. 13/14
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 fashion magazine) = 1 - P(not getting at least 1 fashion magazine)
What does it mean to not get at least 1 fashion magazine? It means getting zero fashion magazines.
In other words, all 3 selected magazines are sports magazines.
So, we can write: P(getting at least 1 fashion magazine) = 1 - P(all 3 selections are sports magazines)
P(all 3 selections are sports magazines)
P(all 3 selections are sports magazines) = P(1st is a sports magazine AND 2nd is a sports magazine AND 3rd is a sports magazine)
= P(1st is a sports magazine) x P(2nd is a sports magazine) x P(3rd is a sports magazine)
= 4/8 x 3/7 x 2/6
= 1/2 x 3/7 x 1/3
= 3/42
= 1/14
So.....
P(getting at least 1 fashion magazine) = 1 - P(all 3 selections are sports magazines)
= 1 - 1/14
= 13/14
Answer: E
gmatclubot
Re: There are 8 magazines lying on a table; 4 are fashion magazines and th [#permalink]
26 Jan 2022, 08:05
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