## anonymous 5 years ago When a thumbtack is flipped, the probability that it lands up is 0.6. The thumbtack is flipped five times. a) Find the probability that the thumbtack lands up exactly twice in the five flips. b) Find the probability that the thumbtack lands up at least twice in the five flips. c) Let X be the number of times the thumbtack lands up in the five flips. Find the mean u(x), and the variance Var(X).

This is a binomial probability problem. Let the probability of success be p=0.6, and therefore failure, q=0.4. Success is 'thumbtack lands up'. (a) The probability that it lands up exactly twice in five trials is$(5;2)(0.6)^2(0.4)^3=\frac{5!}{2!(5-2)!}(0.6)^2(0.4)^3=\frac{144}{625}$ (b) The probability that it lands at least twice is 1- (no. times landing once in five):$1-(5;1)(0.6)(0.4)^4=1-\frac{5!}{1!(5-1)!}(0.6)(0.4)^4=\frac{577}{625}$ (c) The mean in a binomial distribution is given by np, so here, $\mu=np=5 \times 0.6 = 3$and the variance by npq, so$var=npq=5 \times 0.6 \times 0.4=1.2$