Q.Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

The key idea is that this is a derangement problem — we want the probability that not all letters are misplaced. The total number of ways to insert 3 letters into 3 envelopes is $3! = 6$.

Step 1: Count the number of ways no letter is in its proper envelope (a derangement). For $n=3$, the number of derangements is

$!3 = 3! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left(1 - 1 + \frac12 - \frac16\right) = 6 \times \frac{1}{3} = 2.$ …

The key idea here is complementary counting. Instead of directly counting all arrangements where at least one letter is correct, it’s much easier to count the arrangements where no letter is correct — that is, every letter goes into a wrong envelope. Then subtract that from the total number of arrangements.

Why does this work? Because “at least one” includes many overlapping cases (exactly one correct, exactly two correct, all three correct). Counting those directly requires the inclusion-exclusion principle, which is doable but more work. The complement — “none correct” — is a single, clean case.

Let’s walk through it step by step.

1. Total number of ways to insert the letters

We have 3 distinct letters and 3 distinct envelopes. Each envelope gets exactly one letter. This is simply the number of permutations of 3 items:

2. What does “no letter in its proper envelope” mean?

This is called a derangement of 3 items. A derangement is a permutation where no element appears in its original position. For 3 items, we can list all permutations and see which ones are derangements.

Label the letters $L_1, L_2, L_3$ and the corresponding envelopes $E_1, E_2, E_3$. A proper placement means $L_i$ goes into $E_i$.

The 6 permutations (showing which envelope each letter goes to, in order $L_1, L_2, L_3$) are:

1. $(E_1, E_2, E_3)$ — all correct
2. $(E_1, E_3, E_2)$ — $L_1$ correct, others swapped
3. $(E_2, E_1, E_3)$ — $L_3$ correct, others swapped
4. $(E_2, E_3, E_1)$ — none correct
5. $(E_3, E_1, E_2)$ — none correct
6. $(E_3, E_2, E_1)$ — $L_2$ correct, others swapped

So the derangements (no letter in its own envelope) are only permutations 4 and 5. That’s 2 derangements. …