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 ith 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 is \(\frac{2}{3}\).

Step-by-step solution

Calculate the total number of arrangements of the letters

The total number of arrangements of 3 letters could be calculated as the factorial of 3, i.e., \(3! = 3 \times 2 \times 1 = 6\).

Determine the number of derangements

For three letters, there are actually only two derangements: namely ABC to BCA and ABC to CAB. (Please note that a derangement is an arrangement in which none of the letters appear in the original spot.) So the number of derangements is 2.

Calculate the probability of no letters in the correct envelope

This is simply the number of derangements divided by the number of total arrangements, i.e., the probability \(P(D)\) that no letter is correctly placed is given by \(P(D) = \frac{Number of derangements}{Total number of arrangements} = \frac{2}{6} = \frac{1}{3 }\).

Compute the probability of at least one letter in the correct envelope

The probability \(P(C)\) that at least one letter is in the correct envelope is given by \(P(C) = 1 - P(D) = 1 - \frac{1}{3} = \frac{2}{3}\).