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.

• 0.064
• 0.138
• 0.148
• 0.23
• 0.246

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

The intersection of the two events (third malfunction on the fifth day and not three malfunctions on first three days) is the same as the first of those events. So the numerator of the conditional probability is the negative binomial probability of the third success (malfunction) on the fifth day, which is

[[math]] \binom{4}{2} (0.4)^2(0.6)^2(0.4) = 0.13824. [[/math]]

The denominator is the probability of not having three malfunctions in three days, which is

[[math]] 1-(0.4)^3 = 0.936. [[/math]]

The conditional probability is 0.13824/0.936 = 0.1477.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.