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Conditional Probability and Bayes’ Theorem June 6, 2007

Posted by Peter in Exam 1/P.
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Here’s a question that seems to make the rounds in the Exam 1/P circles every so often, because it’s an excellent example of the use of Bayes’ Theorem to compute conditional probability.

A writes to B and does not receive an answer. Assuming that one letter in n is lost in the mail, find the chance that B received the letter. It is to be assumed that B would have answered the letter if he had received it.

Solution: What makes this question tricky is (1) no numerical probabilities are supplied, and (2) it is not clear how to specify the appropriate events. A typical candidate shouldn’t have issues with (1), but resolving (2) takes some careful consideration. We want to find the probability that B received A’s letter, given that A did not receive a response from B. The lack of response could have happened in two ways: either A’s letter did not reach B, or it did but B’s response didn’t reach A. This naturally suggests that the event to condition on is the initial delivery of A’s letter to B. So let us denote the following events:

  • D = A’s letter is delivered to B
  • R = A receives B’s response

along with their complementary events, D’ and R’. Then the desired probability is

\displaystyle \Pr[D|R']  = \frac{\Pr[R'|D]\Pr[D]}{\Pr[R']} = \frac{\Pr[R'|D]\Pr[D]}{\Pr[R'|D]\Pr[D] + \Pr[R'|D']\Pr[D']},

where the leftmost equality follows from Bayes’ Theorem, and the rightmost equality follows from the law of total probability. It’s easy to see that the given information corresponds to

\Pr[D] = \frac{n-1}{n}, \quad \Pr[D'] = \frac{1}{n},

since this is simply the probability of a successful delivery. Similarly,

\Pr[R'|D] = \frac{1}{n}, \quad \Pr[R'|D'] = 1,

since the probability of a lost response is 1/n given that A’s letter was successfully delivered to B, and 1 if A’s letter to B was lost (and thus B never writes a response). It is now a straightforward exercise in algebra to substitute and simplify:

\displaystyle \Pr[D|R'] = \frac{\frac{1}{n}\left(\frac{n-1}{n}\right)}{\frac{1}{n}\left(\frac{n-1}{n}\right) + 1\left(\frac{1}{n}\right)} =\frac{n-1}{2n-1}.

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