A step-by-step walkthrough of SOA Exam P sample question 213: a factory tests 100 light bulbs, each defective with chance 0.02 independently, and we want the chance that exactly two are defective given that two or fewer are. The count is binomial (100, 0.02) and the event sits inside the condition, so we cut the distribution to 0, 1 and 2 and rescale: P(0) = 0.1326, P(1) = 0.2707, P(2) = 0.2734, and 0.2734/0.6767 = 0.404. We check it with ratios of neighbouring chances (1 : 2.04 : 2.06, so 0.98^100 cancels) and show why forgetting the condition gives the tempting 0.273, plus where 0.133, 0.271 and 0.677 come from. Ideal practice for the binomial distribution and conditional probability on the actuarial Exam P.
0:00 Closing time at a light bulb factory
0:24 Question 213: a box that passed
0:34 The facts and the choices
1:09 The plan: cut, then rescale
1:32 Solution: 0.2734/0.6767 = 0.404
2:13 Check: ratios of neighbours
2:52 The trap: no condition
3:29 Takeaway
source