A three-person committee is chosen at random from a group of 5 women and 4 men. Find the probability that the committee contains at least one man.
Answer: The complement of the event that the committee contains at least one man is the event that there is no man in the committee.
Therefore, P(at least 1 man) = 1- P( no man)
[tex]=1- \frac{\binom{5}{3}}{\binom{9}{3}}[/tex]
[tex]=1-\frac{10}{84}[/tex]
[tex]\because \binom{5}{3}=\frac{5!}{(5-3)!3!} =10, \binom{9}{3}=\frac{9!}{(9-3)!3!} =84[/tex]
[tex]=1-0.1190[/tex]
[tex]=0.881[/tex]
Therefore, the probability that the committee contains at least one man is 0.881
