Question

Chris went on a vacation for a week and asked his brother Paul to feed his old cat Charlie. But Paul is forgetful, and Chris is 70% sure Paul will forget to feed his cat. Without food, Charlie will die with probability 0.5. With food, he will die with probability 0.03. Chris came back from vacation and found Charlie alive. What is the probability that Paul forgot to feed Charlie (round off to third decimal place)?

Answer 1

Bayes' Theorem: P(A | B) = P(A & B)/P(B)

P(Paul forgot to feed Charlie) = 0.70

P(Charlie will die | Paul forgot to feed Charlie) = 0.50

P(Charlie is alive | Paul forgot to feed Charlie) = 1 - 0.50 = 0.50

P(Paul fed Charlie) = 1 - 0.70 = 0.30

P(Charlie will die | Paul fed Charlie) = 0.03

P(Charlie is alive | Paul fed Charlie) = 1 - 0.03 = 0.97

P(Paul forgot to feed Charlie | Charlie is alive) = P(Paul forgot to feed Charlie and Charlie is alive) / P(Charlie is alive)

= (0.70x0.50)/(0.70x0.50 + 0.30x0.97)

= 0.546