91Ó°ÊÓ

To solve this problem, we need to find the probability of having exactly one type i coupon in the final collection. Let's condition on T: - If T=0, it means that the collection starts with type i. Then we need to collect the remaining n-1 types without getting another type i. This is similar to distributing n-1 distinct items (types of coupons) in n-1 distinct boxes (remaining coupon types), which can be expressed using the Stirling numbers of the second kind. The probability of this happening can be expressed as \[\frac{S(n - 1, n - 1)}{n^{n-1}}.\] - If T=k (1 ≤ k ≤ n-1), it means that there are k types collected before the first type i appears. And we need to collect the remaining n-k-1 types without getting another type i coupon. This can be expressed by distributing n-k-1 distinct items in n-k distinct boxes. The probability of this happening can be written as \[\frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\] Now, we can compute the overall probability by summing these probabilities for T = 0 to T = n-1: \[P(\text{exactly one type i}) = \sum_{k = 0}^{n - 1} \frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\]

Now, we want to find the expected number of types that appear exactly once in the final collection. To do this, we'll use the linearity of expectation, which states that the expected value of the sum of random variables is the sum of their expected values. So, for each type of coupon, we can compute the probability of having only one of that type in the final collection, and then sum these probabilities: \[E(\text{number of types that appear exactly once}) = \sum_{i = 1}^{n} P(\text{exactly one type i}).\] Using the result from part (a), we have: \[E(\text{number of types that appear exactly once}) = n \sum_{k = 0}^{n - 1} \frac{S(n - k - 1, n - k)}{n^{n-k-1}}.\]