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1. A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least 1 woman ?

Solution:
Total number of persons = (3 + 2) = 5
n(S)=5C3=5C2     =5×42×1   = 10
Let E be the event of selecting 3 members having at least 1 women
Then, n(E) = n [(1 women and 2 men ) or (2 women and 1 man)]
= n (1 woman and 2 men) + n (2 women and 1 man)
=(2C1×3C2)+(2C2×3C1)=(2C1×3C1)+(1×3C1)=(2×3)+(1×3)=(6+3)=9
P(E)=n(E)n(S)=910

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