Question: A committee of 5 persons is to be constituted from a group of 6 gents and 8 ladies. If the selection is made randomly, find the probability that there are 3 ladies and 2 gents in the committee
Options:
A) $ \frac{60}{149} $
B) $ \frac{60}{143} $
C) $ \frac{47}{140} $
D) $ \frac{42}{139} $
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Answer:
Correct Answer: B
Solution:
- 5 person out of 14 can be selected in $ {}^{14}C _5 $ ways.
3 ladies out of 8 can be selected in $ {}^{8}C _3 $ ways.
2 gents out of 6 can be selected in $ {}^{6}C _2 $ ways.
$ \therefore $ Required probability $ =\frac{{}^{8}C _3\times {}^{6}C _2}{{}^{14}C _5} $
$ =\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{6\times 5}{2\times 1}\times \frac{5\times 4\times 3\times 2\times 1}{14\times 13\times 12\times 11\times 10} $
$ =\frac{60}{143} $