Question

A secretary types three letters and the three corresponding envelopes. In a hurry, he places at random one letter in each envelope. What is the probability that at least one letter is in the correct envelope? Hint: Let \(C_{i}\) be the event that the \(i\) th letter is in the correct envelope. Expand \(P\left(C_{1} \cup C_{2} \cup C_{3}\right)\) to determine the probability.

Short answer

The probability that at least one letter is in the right envelope when randomly placed is \(4/6\) or approximately 0.67, or 67%.

Step-by-step solution

Analyze the Possible Outcomes

There are three letters and three envelopes, so there are \(3!\) or 6 possible ways the letters could be arranged within the envelopes. If we label the letters and envelopes as 1, 2, and 3, the six possible arrangements are: 123, 132, 213, 231, 312, 321.

Identify the Favorable Outcomes

Now, we want at least one letter in its correct envelope. To use the principle of inclusion-exclusion, first count the ways in which exactly one envelope contains the correct letter. There are three ways this could happen (one for each letter). Next, count the number of ways in which exactly two envelopes contain the correct letters. Since if two letters are in their correct envelopes, the third must be as well, there are zero ways for this to happen. Finally, there is one way in which all three envelopes contain the correct letters.

Apply the Principle of Inclusion-Exclusion

The principle tells us \(P\left(C_{1}\cup C_{2} \cup C_{3}\right)\) equals the sum of the probabilities of each \(C_{i}\), minus the sum of the probabilities of each pair of \(C_{i}\) intersecting, plus the probability of all the \(C_{i}\) intersecting. When applying these calculations, we get \((3 - 0 + 1) / 6 = 4/6.\)