23% of the time the older computers in the computer lab go bad. 6 was randomly chosen to be fixed:?

Question

(a) What is the probability that more than 4 of the 6 will be defective?
(b) How many of the 6 computers do you expect to be bad at the time of testing?
(c) Suppsoe there are 50 computers in the lab and the probability that a computer is virus free is 4%. Using 2 methods, calculate the probability of 6 students using the lab and escaping the virus plague. That is, what is the probability of exactly 6 diskettes remaining virus free if all 50 PC's were used (by students with diskettes) in a lab tutorial?

Answer 1

(a) ((6!/(5!1!)) * (.23^5)*(.77^1)) + ((6!(6!1!)) * (.23^6)*(.77^0)) = .31216% approximately.

(b) 6*.23 = 1.38

(c) ((50!/(44!6!))*(.04^6) = 6.5% approximately.