A-Level Discrete random variables: worked solution

Question

Given she is late with probability 0.215 when leaving at 6:48 am, and she attends five days weekly always leaving at that time, find the probability that her first fortnight contains exactly two late arrivals while the third late arrival occurs during week 6.

Step 1: turn the timeline into blocks of days. She attends 5 days a week, and each day is an independent trial with P(late) = 0.215, so P(not late) = 1 - 0.215 = 0.785. The first fortnight is weeks 1 and 2, which is 2 × 5 = 10 days. 'Exactly two late arrivals in the first fortnight' and 'the third late arrival occurs during week 6' together force three separate requirements: (a) exactly 2 late days in the first 10 days; (b) no late day at all in weeks 3, 4 and 5, that is 3 × 5 = 15 days, because a late day there would be the third one and it would not fall in week 6; (c) at least one late day during week 6, which is 5 days, so that the third late arrival does happen in that week. The three blocks of days are disjoint and the days are independent, so the required probability is the product of the three probabilities. Step 2: block (a), the first fortnight. Let R be the number of late days in 10 independent days, each with probability 0.215. R ~ B(10, 0.215) P(R = 2) = C(10, 2) × 0.215² × 0.785⁸ = 45 × 0.046225 × 0.144197... = 0.299948... Step 3: block (b), weeks 3 to 5. Let S be the number of late days in those 15 days. S ~ B(15, 0.215) P(S = 0) = C(15, 0) × 0.215⁰ × 0.785¹⁵ = 0.785¹⁵ = 0.0264877... Step 4: block (c), week 6. We need at least one late day in 5 days. Use the complement, since 'at least one' is the opposite of 'none': P(at least one late in 5 days) = 1 - P(none late in 5 days) = 1 - 0.785⁵ = 1 - 0.298091... = 0.701909... (Equivalently, summing the chance that the first late day of that week is on day 1, 2, 3, 4 or 5 gives the geometric sum 0.215(1 - 0.785⁵)/(1 - 0.785), and since 1 - 0.785 = 0.215 this simplifies to the same 1 - 0.785⁵.) Step 5: multiply the three independent blocks. P = P(R = 2) × P(S = 0) × [1 - 0.785⁵] = 0.299948... × 0.0264877... × 0.701909... = 0.0055766... = 0.00558 (3 s.f.)