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
