The setup is simple enough but the conditioning is what gets you.
Recognize this as a conditional probability problem and apply Bayes' theorem. Define events clearly: A1, A2, A3 for picking each cat, and L for losing. Compute P(A1|L) = P(L|A1)P(A1) / P(L), where P(L) = sum P(L|Ai)P(Ai).
Pro tip: Always state your assumptions and walk through the calculation step by step; interviewers value clear reasoning over just the final answer. Double-check arithmetic and consider if the result makes intuitive sense.
Let A1, A2, A3 be the events of picking the most, middle, and least athletic cat, respectively. Each has prior probability 1/3. Let L be the event that your cat loses. Then P(L|A1)=1/2, P(L|A2)=2/3, P(L|A3)=5/6.
Use the law of total probability: P(L) = P(L|A1)P(A1) + P(L|A2)P(A2) + P(L|A3)P(A3) = (1/2)(1/3) + (2/3)(1/3) + (5/6)(1/3) = 1/6 + 2/9 + 5/18 = 3/18 + 4/18 + 5/18 = 12/18 = 2/3.
Compute the posterior probability that your cat was the most athletic given that it lost: P(A1|L) = P(L|A1)P(A1) / P(L) = (1/2 * 1/3) / (2/3) = (1/6) / (2/3) = 1/4.
Check that the result is reasonable: since the most athletic cat is least likely to lose, the probability it was chosen given a loss should be lower than the prior 1/3. Indeed, 1/4 < 1/3. Also, ensure the sum of posterior probabilities for all cats given loss equals 1: P(A1|L)=1/4, P(A2|L)= (2/3*1/3)/(2/3)=1/3, P(A3|L)= (5/6*1/3)/(2/3)=5/12; sum = 1/4+1/3+5/12 = 3/12+4/12+5/12=1.
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