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.
2/3
Explanation
The problem involves permutations of three items. Total ways to place three letters into three addressed envelopes is 3! = 6. The event 'at least one letter in its proper envelope' can be handled by complementary counting: subtract the derangements (arrangements with no matches) from the total. For three items, there are exactly two derangements. Hence, favorable outcomes = 6 - 2 = 4, giving probability 4/6 = 2/3.
Solution Steps
Step 1: Total outcomes – Since each envelope must contain exactly one letter, the number of ways to insert the three letters into three envelopes is the number of permutations of three items: 3! = 6.
Step 2: Favorable outcomes – Count arrangements where at least one letter is in its correct envelope. Use complement: total minus arrangements where no letter is correct (derangements). For three items, derangements = 2 (the cycles (2,3,1) and (3,1,2)). So favorable = 6 - 2 = 4.
Step 3: Probability – Probability = favorable / total = 4/6 = 2/3.