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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 one woman ?
- A$\frac{1}{10}$
- B$\frac{9}{20}$
- C$\frac{9}{10}$
- D$\frac{1}{20}$
Answer not set
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0
CCommunity
Solution
The number of ways = Two men and one woman + One men and two women
$= \frac{^3C_2 \times ^2C_1}{^5C_3} + \frac{^3C_1 \times ^2C_2}{^5C_3}$
$= \frac{6+3}{20}$
$= 9/20$
Hence, the correct option is C.
The number of ways = Two men and one woman + One men and two women
$= \frac{^3C_2 \times ^2C_1}{^5C_3} + \frac{^3C_1 \times ^2C_2}{^5C_3}$
$= \frac{6+3}{20}$
$= 9/20$
Hence, the correct option is C.
0
CCommunity
correct option is C.