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

Answer:

There is a 54.6% probability that Paul forgot to feed the cat.

Step-by-step explanation:

We have these following probabilities:

A 70% probability that Paul forgets to feed the cat.

A 30% probability that Paul feeds the cat.

If not fed, a 50% probability that the cat dies.

If fed, a 3% probability that the cat dies.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that Paul forgot to feed the cat, given that he was alive.

So

P(B) is the probability of Paul forgetting to feed the car. So $P(B) = 0.7$

P(A/B) is the probability of the cat being alive, given that he was not fed. So $P(A/B) = 0.5$

P(A) is the probability of the cat being alive. If he is fed, there is a 97% probability of him being alive. There is a 30% probability that he is fed.

If he is not fed, there is a 50% probability of him being alive. There is a 70% probability that he is not fed. So:

$P(A) = 0.97*0.30 + 0.5*0.7 = 0.641$

The probability that Paul forgot to fed the cat, given that Chris found the cat alive is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.7}{0.641} = 0.546$

There is a 54.6% probability that Paul forgot to feed the cat.