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There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines...

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There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines...

by BTGmoderatorLU » Fri Jul 24, 2020 12:48 pm

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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 are 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

The OA is E
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Re: There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines...

by dashingnightmare » Fri Jul 24, 2020 10:27 pm
Total number of ways of selecting 3 magazines out of 8=8c3
no of ways of selecting 3 magazines so that none of them is fashion magazine=4c3
prob=4c3/8c3=1/14
atlease 1 fashion magazine=1-1/14=13/14
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Re: There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines...

by hahahehe » Sat Jul 25, 2020 4:25 am
We have 4 Sports magazines, and 4 Fashion magazines
We have to select 3 magazines
Total ways of selecting 3 magazines out of these 8 -> 8C3 (8 choose 3) = 8!/(3! * 5!) = 56
Now we need probability that "at least one" of the fashion magazines will be selected. This scenario would include: selection of exactly 1 fashion magazine, exactly 2 fashion magazines, or exactly 3 fashion magazines. Either we can calculate these individual probabilities and add them up, or we can find out the probability that no fashion magazine is selected, and subtract it from 1, since
1 = Probability(select 0 fashion magazines) + probability(select 1 fashion magazine) + probability(select 2 fashion magazine) + probability(Select 3 fashion magazine)

Now P(select no fashion magazine) = ways of selecting only sports magazines / total ways = 4C3 / 56
= 4/56 = 1/14

So our desired probability = 1 - 1/14 = 13/14
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