Each member of a nine person committee attends any given meeting of the committee with a probability of 1/2. what is the probability that at least two-thirds of the membership of the committee will be present at its next meeting?

To solve the question we proceed as follows:
2/3*(9)=6
Thus the probability that at least 6 members will attend will be given by the sum of the probabilities that 6,7,8,9 members will attend the meeting.
Probability that n members will attend is given by:
9Cn=9!/((9-n)!*n!)
thus
P(n=6)=9!(3!6!)/512=84/512
p(n=7)=9!/(2!*7!)/512=36/512
p(n=8)=9!/(1!*8!)/512=9/512
p(n=9)=9!/(0!*9!)/512=1/512
thus the probability that at least 6 members will attend is:
84/512+36/512+9/512+1/512
=130/512
=0.253

Hey, can you post the problem?