91Ó°ÊÓ

Since there are m different types of coupons, each time a new coupon is obtained, it is a type i coupon with probability \(p_i, i=1, \ldots, m\). We are looking for the probability that the nth coupon is a new type.

We can calculate the probability of the nth coupon being a new type by conditioning on the type of this coupon. So let's denote this event as A, which is the event that the nth coupon is a new type.

To calculate P(A), we can use the Law of Total Probability, which states that if we partition the event space, we can find the probability of an event happening by summing the conditional probabilities of that event when it is conditioned on a partition: \[P(A)=\sum_{i=1}^{m} P(A | B_i)P(B_i)\] Here, \(B_i\) denotes the event of obtaining a type i coupon as the nth coupon. We are given the probabilities \(p_i = P(B_i)\) for \(1 \le i \le m\). Now, we need to find the conditional probabilities \(P(A|B_i)\), which represent the probability that the nth coupon is a new type given that it is of type i. If the nth coupon is of type i, then the probability that it's a new type is the probability that no type i coupons have been collected in the first \((n-1)\) coupons. The probability of not collecting a type i coupon in a single draw is \((1-p_i)\). So, the probability of not collecting any type i coupon in the first \((n-1)\) draws is \((1-p_i)^{n-1}\). Therefore, \(P(A|B_i) = (1-p_i)^{n-1}\).

We can now use the Law of Total Probability to calculate the probability that the nth coupon is a new type: \[P(A)=\sum_{i=1}^{m} P(A | B_i)P(B_i) = \sum_{i=1}^{m} (1-p_i)^{n-1}p_i\] So the probability that the nth coupon is a new type is equal to the sum of conditional probabilities of the event that the nth coupon is a new type given that it's of type i, multiplied by the probability of obtaining a type i coupon: \[P(A) = \sum_{i=1}^{m} (1-p_i)^{n-1}p_i\]