Question

Suppose each new coupon collected is, independent of the past, a type \(i\) coupon with probability \(p_{i} .\) A total of \(n\) coupons is to be collected. Let \(A_{i}\) be the event that there is at least one type \(i\) in this set. For \(i \neq j\), compute \(P\left(A_{i} A_{j}\right)\) by (a) conditioning on \(N_{i}\), the number of type \(i\) coupons in the set of \(n\) coupons; (b) conditioning on \(F_{i}\), the first time a type \(i\) coupon is collected; (c) using the identity \(P\left(A_{i} \cup A_{j}\right)=P\left(A_{i}\right)+P\left(A_{j}\right)-P\left(A_{i} A_{j}\right)\).

Short answer

To compute the probability \(P\left(A_{i} A_{j}\right)\), where \(A_i\) represents the event of having at least one type \(i\) coupon, and \(A_j\) represents the event of having at least one type \(j\) coupon with \(i \neq j\), we can use the three following methods: (a) \(P\left(A_{i} A_{j}\right) = \sum_{k=1}^{n} \left[1-(1-p_j)^{n-k}\right] \binom{n}{k} p_i^k (1-p_i)^{n-k}\) (b) \(P\left(A_{i} A_{j}\right) = \sum_{f_i=1}^{n} \left[1-(1-p_j)^{n-f_i}\right] p_i (1-p_i)^{f_i-1}\) (c) \(P\left(A_{i} A_{j}\right) = (1-(1-p_i)^n) + (1-(1-p_j)^n) - \left[1 - (1 - p_i - p_j + p_ip_j)^{n}\right]\)

Step-by-step solution

(Compute Probability of Events Using Method A)

To compute the probability \(P\left(A_{i} A_{j}\right)\) by conditioning on \(N_{i}\), the number of type \(i\) coupon in the set of \(n\) coupons, we will first find the probability of having \(k\) type \(i\) coupons and then sum them to get the final result. Consider the case where there are \(k\) type \(i\) coupons: \(P\left(A_{j} | N_i=k\right) = 1-P\left(A_{j}^c | N_i=k\right)\) Here, \(A_{j}^c\) is the complementary event of \(A_j\), which means that no type \(j\) coupon is present. Since we have \(n-k\) remaining slots to fill and the probability of not having a type \(j\) coupon in one slot is \((1-p_j)\), the probability of \(A_{j}^c\) given \(N_i=k\) is: \(P\left(A_{j}^c | N_i=k\right) = (1-p_j)^{n-k}\) Thus, \(P\left(A_{j} | N_i=k\right) = 1-(1-p_j)^{n-k}\) Now, we will find the probability of having \(k\) type \(i\) coupons in the set \(P\left(N_i=k\right)\). This can be calculated using the binomial distribution: \(P\left(N_i=k\right) = \binom{n}{k} p_i^k (1-p_i)^{n-k}\) Now, to compute \(P\left(A_{i} A_{j}\right)\), we can sum over all possible values of \(k\): \(P\left(A_{i} A_{j}\right) = \sum_{k=1}^{n} P\left(A_{j} | N_i=k\right)P\left(N_i=k\right) = \sum_{k=1}^{n} \left[1-(1-p_j)^{n-k}\right] \binom{n}{k} p_i^k (1-p_i)^{n-k}\)

(Compute Probability of Events Using Method B)

To compute the probability \(P\left(A_{i} A_{j}\right)\) by conditioning on \(F_{i}\), the first time a type \(i\) coupon is collected, we will again sum over the possible values of \(f_i = 1,...,n\). \(P\left(A_{i} A_{j}\right) = \sum_{f_i=1}^{n} P\left(A_{j} | F_i=f_i\right)P\left(F_i=f_i\right)\) Similar to method A, we have: \(P\left(A_{j} | F_i=f_i\right) = 1-(1-p_j)^{n-f_i}\) To find \(P\left(F_i=f_i\right)\), we need to consider that a type \(i\) coupon was collected at position \(f_i\) and no type \(i\) coupons were collected in the first \(f_i-1\) positions: \(P\left(F_i=f_i\right) = p_i (1-p_i)^{f_i-1}\) Now, we can compute \(P\left(A_{i} A_{j}\right)\): \(P\left(A_{i} A_{j}\right)=\sum_{f_i=1}^{n} \left[1-(1-p_j)^{n-f_i}\right] p_i (1-p_i)^{f_i-1}\)

(Compute Probability of Events Using Method C)

To compute the probability \(P\left(A_{i} A_{j}\right)\) using the identity: \(P\left(A_{i} \cup A_{j}\right) = P\left(A_{i}\right) + P\left(A_{j}\right) - P\left(A_{i} A_{j}\right)\), we first need to find the probabilities \(P\left(A_{i}\right)\) and \(P\left(A_{j}\right)\). \(P\left(A_{i}\right) = 1 - P\left(A_{i}^c\right) = 1 - (1-p_i)^{n}\) \(P\left(A_{j}\right) = 1 - P\left(A_{j}^c\right) = 1 - (1-p_j)^{n}\) The probability of the union of these two events can be computed as: \(P\left(A_{i} \cup A_{j}\right) = 1 - P\left(A_{i}^c \cap A_{j}^c\right) = 1 - (1 - p_i - p_j + p_ip_j)^{n}\) Now we can find \(P\left(A_{i} A_{j}\right)\) using the identity: \(P\left(A_{i} A_{j}\right) = P\left(A_{i}\right) + P\left(A_{j}\right) - P\left(A_{i} \cup A_{j}\right) = (1-(1-p_i)^n) + (1-(1-p_j)^n) - \left[1 - (1 - p_i - p_j + p_ip_j)^{n}\right]\)

Key concepts

Binomial Distribution

When solving coupon probability problems, understanding the binomial distribution is a crucial step. This distribution applies to scenarios where there are exactly two outcomes for each trial, often termed as 'success' and 'failure'. In the context of collecting coupons, each attempt to collect a new coupon can be seen as a trial, and obtaining a specific type of coupon (say type i) can be considered a 'success'.

The binomial distribution allows us to compute the probability of having a certain number of successes (type i coupons) in a fixed number of trials (n coupons collected). The formula to calculate this is:

Conditional Probability

The concept of conditional probability is exemplified in steps 1 and 2 of our exercise. It deals with the likelihood of an event occurring, given that another event has already taken place. For example, when we want to find the probability of having at least one type j coupon given that k type i coupons are already present, we are using conditional probability.

To clarify this with our coupon collection, suppose we already have k type i coupons. The chance of now getting at least one type j coupon with the remaining n-k draws illustrates conditional probability, and this is calculated in step 1 of the provided solution as:

Probability Theory

At the heart of these computations is probability theory, the branch of mathematics that deals with the analysis of random phenomena. The fundamental rule used in step 3 of our solution is rooted in probability theory. It stipulates that the probability of at least one of two events happening, A_i or A_j, is the sum of the probabilities of each event occurring separately, minus the probability of both happening at the same time.

This fundamental rule is known as the Inclusion-Exclusion Principle, and it allows us to calculate the probability of a union of events as: