1. **Problem statement:** We have 100 letters and 100 envelopes, and letters are randomly inserted into envelopes. We want to find the probabilities for different numbers of letters being correctly placed. 2. **Key concept:** This is a problem of derangements and permutations. The total number of ways to place 100 letters into 100 envelopes is $100!$. 3. **a) Probability no letter is correct (derangement):** The number of derangements of $n$ items, denoted $!n$, is approximately $$!n = n! \sum_{k=0}^n \frac{(-1)^k}{k!}.$$ For $n=100$, the probability is $$P(0) = \frac{!100}{100!} = \sum_{k=0}^{100} \frac{(-1)^k}{k!} \approx e^{-1} \approx 0.3679.$$ 4. **b) Probability exactly one letter is correct:** Number of ways to fix exactly one letter correctly is zero because if exactly one letter is fixed, the rest must be deranged, but derangements of $n-1$ with one fixed point is zero. So, $$P(1) = 0.$$ 5. **c) Probability exactly 98 letters are correct:** If 98 letters are correct, then 2 letters remain. The remaining 2 letters must be placed incorrectly. But for 2 letters, the number of derangements is 1 (swap the two). So, Number of ways = $\binom{100}{98} \times !2 = \binom{100}{2} \times 1 = 4950$. Probability: $$P(98) = \frac{4950}{100!}$$ which is negligible (practically zero). 6. **d) Probability exactly 99 letters are correct:** If 99 letters are correct, the last letter must also be correct (only one envelope left), so this is impossible. $$P(99) = 0.$$ 7. **e) Probability all letters are correct:** Only one way to have all correct: $$P(100) = \frac{1}{100!}$$ which is extremely small. **Final answers:** $$P(0) \approx 0.3679, \quad P(1) = 0, \quad P(98) \approx 0, \quad P(99) = 0, \quad P(100) = \frac{1}{100!}.$$