A-Level Probability: worked solution

Question

For the same free-throw game (Ben chosen with probability 0.7, single-attempt success 0.1 for Ben and 0.07 for John, at most three throws), suppose the team lost. Find the probability that John had been the selected shooter.

This asks for P(John was chosen | the team lost), which is a reversed conditional probability, so use P(John | lost) = P(John and lost) / P(lost) and build the denominator from the two possible shooters. Step 1: probability of losing under each shooter. The team loses only if the shooter misses all three attempts, and the attempts are independent. Ben misses one attempt with probability 1 - 0.1 = 0.9, so P(lost | Ben) = 0.9³ = 0.729 John misses one attempt with probability 1 - 0.07 = 0.93, so P(lost | John) = 0.93³ = 0.804357 Step 2: the numerator, using P(John) = 1 - 0.7 = 0.3. P(John and lost) = P(John) x P(lost | John) = 0.3 x 0.804357 = 0.2413071 Step 3: the denominator, by the law of total probability over the two shooters. P(lost) = P(Ben) x P(lost | Ben) + P(John) x P(lost | John) = 0.7 x 0.729 + 0.3 x 0.804357 = 0.5103 + 0.2413071 = 0.7516071 This is the complement of the winning probability found earlier, 1 - 0.2483929 = 0.7516071, which is a useful check. Step 4: divide. P(John | lost) = 0.2413071 / 0.7516071 = 0.32105... = 0.321 (3 s.f.)