A factory tests 100 light bulbs for defects. The probability that a bulb is defective is 0.02. The occurrences of defects among the light bulbs are mutually independent events.

Calculate the probability that exactly two are defective given that the number of defective bulbs is two or fewer.

• 0.133
• 0.271
• 0.273
• 0.404
• 0.677

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

Solution: D

The number of defects has a binomial distribution with n = 100 and p = 0.02.

[[math]] \begin{align*} \operatorname{P}[X = 2 | X \leq 2] &= \frac{\operatorname{P}[X = 2]}{\operatorname{P}[ X \leq 2]}\\ & = \frac{\binom{100}{2}(0.02)^2(0.98)^{98}}{\binom{100}{0}(0.02)^0(0.98)^{98} + \binom{100}{1}(0.02)^1(0.98)^{99} + \binom{100}{2}(0.02)^2(0.98)^{98}} \\ &= \frac{0.27341}{0.13262 + 0.27065 + 0.27341}\\ & = 0.404. \end{align*} [[/math]]

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