Question
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 ?
Answer: Option D
Total number of persons = (3 + 2) = 5
$$\therefore n(S) = {}^5\mathop C\nolimits_3 = {}^5\mathop C\nolimits_2 $$ $$ = \frac{{5 \times 4}}{{2 \times 1}}$$ = 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)
$$\eqalign{
& = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_2 } \right) + \left( {{}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right) \cr
& = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_1 } \right) + \left( {1 \times {}^3\mathop C\nolimits_1 } \right) \cr
& = \left( {2 \times 3} \right) + \left( {1 \times 3} \right) \cr
& = \left( {6 + 3} \right) \cr
& = 9 \cr} $$
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{10}}$$
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Total number of persons = (3 + 2) = 5
$$\therefore n(S) = {}^5\mathop C\nolimits_3 = {}^5\mathop C\nolimits_2 $$ $$ = \frac{{5 \times 4}}{{2 \times 1}}$$ = 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)
$$\eqalign{
& = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_2 } \right) + \left( {{}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right) \cr
& = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_1 } \right) + \left( {1 \times {}^3\mathop C\nolimits_1 } \right) \cr
& = \left( {2 \times 3} \right) + \left( {1 \times 3} \right) \cr
& = \left( {6 + 3} \right) \cr
& = 9 \cr} $$
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{10}}$$
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