Probablility help? Please explain :o)?

Among employed women, 20% have never been married. Select 9 employed women at random.


(a) The number in your sample who have never been married has a binomial distribution. What are n and p?


n =


p =





(b) What is the probability that exactly 2 of the 9 women in your sample have never been married?








(c) What is the probability that 2 or fewer women have never been married?

Probablility help? Please explain :o)?
a) n = 9, p = .20





b) P(X=k) = C(n,k) p^k (1-p)^(n-k), where n = number of trials, p = probability of success, k = number of successes, and C(n,k) is the combination of n objects taken k at a time (order does not matter).


C(n,k) = n! / [k!(n-k)!]





P(X=2) = C(9,2) .20^2 .80^7


= 36(.04)(.2097152)


= 0.301989888





c) Two or fewer means 2 or 1 or 0.


Find P(X=2) + P(X=1) + P(X=0).





P(X=2) was found in part b, so all you have to do is find P(X=1) and P(X=0), then add all three together.