On any given day, a certain machine has either no malfunctions or exactly one malfunction. the probability of malfunction on any given day is 0.40. machine malfunctions on different days are mutually independent. calculate the probability that the machine has its third malfunction on the fifth day, given that the machine has not had three malfunctions in the first three days.

Let A = "3rd malfunction occurs on 5th day", and B = "not 3 malfunctions in first 3 days". Then we want p(A|B) = p(A&B)/p(B). In this problem, A is a subset of B, so p(A&B) = p(A).

For p(A), we have ...
There are 6 ways to have 2 malfunctions in the first 4 days. The probability of that is 6*(.4^2)(0.6^2) = 0.3456. Then the probability of the third malfunction on the 5th day is 0.4 times that,
.. p(3rd malfunction on 5th day) = 0.13824

For p(B), we have ...
There is only one way to have 3 malfunctions in the first 3 days. The probability of that is 0.4^3 = 0.064. Then p(B) = 1 - p(3 malfunctions in 3 days) = 0.936.

p(A|B) = p(A)/p(B) = 0.13824/0.936 ≈ 0.14769