A-Level Probability: worked solution

Question

With three receipts each showing one of 10 equally likely characters A–J, given that exactly one is A, find the probability the remaining two are B and C.

Each of the three receipts independently shows one of the 10 characters A to J, so each particular character has probability 0.1 and each other outcome has probability 0.9. This is a conditional probability: P(the other two are B and C | exactly one A) = P(both events) / P(exactly one A). Step 1. Find P(exactly one A), the event being conditioned on. Choose which one of the 3 receipts carries the A: 3 ways. That receipt shows A with probability 0.1; each of the other two is not A, with probability 0.9 each. P(exactly one A) = 3 × 0.1 × 0.9² 0.9² = 0.81 P(exactly one A) = 3 × 0.1 × 0.81 = 0.243 Step 2. Find the probability of both events together, i.e. the three receipts show A, B and C in some order. Note this event automatically contains "exactly one A", so no extra condition is needed. The three distinct letters A, B, C can be placed on the three receipts in 3! = 6 orders. Each particular order has probability 0.1 × 0.1 × 0.1 = 0.001. P = 6 × 0.001 = 0.006 Step 3. Divide. P = 0.006 / 0.243 Multiply top and bottom by 1000: = 6/243 Divide top and bottom by 3: = 2/81 = 0.0247 (3 s.f.)