Law & Order

Probability Level 3

As part of a prison break, prisoners dug two escape tunnels, A and B.

7 prisoners escaped through tunnel A, of whom 3 were recaptured.

9 prisoners escaped through tunnel B, of whom 4 were recaptured.

The warden sat down to speak with one of the recaptured prisoners. What is the probability the prisoner made his escape through tunnel A?

1/2 3/7 27/55 7/16

Denton Young by Denton Young , 47, USA

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2 solutions

Denton Young
Jun 2, 2016

The chances that a prisoner will be recaptured from tunnel A = 3/7.

The chances that a prisoner will be recaptured from tunnel B = 4/9.

Given that the prisoner being spoken to was recaptured, the chance he made his escape through tunnel A is therefore (3/7) / (3/7 + 4/9) = 27/55.

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Moderator note:

Great question. Conditional probability always turns out trickier than the first glance.

If you have 7 marbles, and 3 of them are white and 4 are black, isn't the probability of drawing a white 3/7?

Isn't it given that 7 people were captured, 3 of them are from group A and 4 are from group B, so shouldn't the probability be the same?

Yasin Tarabar - 5 years ago

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I also have the same question. Why is 3/7 wrong?

Atomsky Jahid - 5 years ago

I think there's a flaw in your question. You didn't say "chances" in the question, but you used it in the solution. 3 out of 7 were a fact in the problem, not a probability. As you stated the fact, a problem solver would be sure that there are 3 prisoners from route A. Hence, the answer would be 3/7. But, if you stated that the chances of recapturing from route A is 3/7, the scenario would have been different.

Atomsky Jahid - 5 years ago

Please clarify why it isnt 3/7..we know for sure that of the total 7 recaptured 3 are from A..so why isn't it simply 3/7??

Istiak Reza - 5 years ago
Tom Engelsman
May 11, 2021

Let A = A = the event the prisoner was inside Tunnel A, B = B = the event a prisoner was captured. Computing:

P ( A ∣ B ) = P ( A ∩ B ) P ( B ) = 3 / 16 7 / 16 = 3 7 . \large P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{3/16}{7/16} = \boxed{\frac{3}{7}}.

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