Statistics And Probability Question 213
Question: A committee has to be made of 5 members from 6 men and 4 women. The probability that at least one woman is present in committee, is
Options:
A) $ \frac{1}{42} $
B) $ \frac{41}{42} $
C) $ \frac{2}{63} $
D) $ \frac{1}{7} $
Show Answer
Answer:
Correct Answer: B
Solution:
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Total number of ways $ ={}^{4}C_1\times {}^{6}C_4+{}^{4}C_2\times {}^{6}C_3+{}^{4}C_3\times {}^{6}C_2+{}^{4}C_4\times {}^{6}C_1+{}^{6}C_5 $ $ =60+120+60+6+6=252 $ No. of ways in which at least one woman exist are $ ={}^{4}C_1\times {}^{6}C_4+{}^{4}C_2\times {}^{6}C_3+{}^{4}C_3\times {}^{6}C_2+{}^{4}C_4\times {}^{6}C_1=246 $ Hence required probability $ =\frac{246}{252}=\frac{41}{42} $ .
BETA
