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jyoti kaurav

· started a discussion

· 1 Months ago

wrong combination (Question:164930.0)

5C4 = 5! / 4!*1!
9C4 = 9! / 4!*5!

Question:
A committee of four members is to be formed out of 5 men and 4 women. What is the probability that the committee consists of at least one woman?
Options:
A) \(\cfrac{5}{126}\)
B) \(\cfrac{121}{126}\)
C) \(\cfrac{18}{129}\)
D) \(\cfrac{111}{129}\)
E) None of these
Solution:
Ans: (b) The probability that the committee has no woman = 5C4 / 9C4

\(=\cfrac{5!/1!}{9!/4!}\)

\(=\cfrac{5 ×4 ×3 ×2 ×1}{9×8×7×6}\)       \(=\cfrac{5}{126}\)  


Therefore, the probability that the committee has at least one woman 

= 1 – \(\cfrac{5}{126}\) = \(=\cfrac{121}{126}\)

jyoti kaurav

· commented

· 1 Months ago

so prob. for no women is
5!*5! / 9!
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