Problem 56 - A First Course in Probability

Question

Suppose that you continually collect coupons and that there are \(m\) different types. Suppose also that each time a new coupon is obtained, it is a type \(i\) coupon with probability \(p_{i}, i=1, \ldots, m .\) Suppose that you have just collected your \(n\)th coupon. What is the probability that it is a new type? Hint: Condition on the type of this coupon.

Step-by-Step Solution

Answer

The probability of obtaining a new type of coupon as the nth coupon is given by: P(A) = \(\sum_{i=1}^m [ p_i^2 * (1-p_i)^{n-1} ]\)

Step 1: Identify the events

Let A be the event that the nth coupon is a new type, and B_i be the event that the nth coupon is of type i. We need to find the probability of event A.

Step 2: Calculate conditional probabilities

We will calculate the probability of event A given each event B_i, i.e. P(A|B_i). If the nth coupon is of type i, then we know that among the first n-1 coupons, there were no coupons of type i. The probability of not having coupon i would be (1-p_i). So, the probability of not having the type i coupon in each of the first n-1 draws is (1-p_i)^(n-1), and the probability of obtaining it on the nth draw is p_i. So, to find P(A|B_i), you need to multiply these probabilities: P(A|B_i) = p_i * (1-p_i)^(n-1)

Step 3: Calculate the probability of each event B_i

Now, we need to calculate the probability of B_i, which is the probability that the nth coupon is of type i. This is given directly by the problem: P(B_i) = p_i

Step 4: Calculate joint probabilities

Now, let's calculate the joint probabilities P(A∩B_i), which is the probability that both events A and B_i occur simultaneously. Since these events are dependent, we can obtain these probabilities by multiplying the conditional probabilities, P(A|B_i), and the probabilities of B_i: P(A∩B_i) = P(A|B_i) * P(B_i) = p_i * (1-p_i)^(n-1) * p_i = p_i^2(1-p_i)^(n-1)

Step 5: Calculate the probability of event A

To find the probability of event A (the nth coupon being a new type), we need to sum up the joint probabilities for all values of i, from 1 to m: P(A) = Σ [P(A∩B_i)] = Σ [ p_i^2 * (1-p_i)^(n-1) ] So, the probability of obtaining a new type of coupon as the nth coupon is given by: P(A) = Σ_{i=1}^m [ p_i^2 * (1-p_i)^(n-1) ]

Key Concepts

Probability

Probability is a fundamental concept in statistics and mathematics focused on measuring the chance or likelihood that a given event will occur.
In the context of the Coupon Collector's Problem, probability plays a crucial role. Here, the concept is used to determine how likely it is to draw a new type of coupon when continuously collecting different types of coupons.
When you have a set of possibilities, such as different types of coupons, each possibility can be assigned a probability. These probabilities help us understand and make informed predictions about events, such as drawing a specific type of coupon.
The probability of an event is always a number between 0 and 1. A probability of 0 means the event will not occur, while a probability of 1 means the event will certainly happen.
In the exercise, each type of coupon appears with a certain fixed probability, denoted as \( p_i \) for type \( i \). A deeper understanding of probabilities helps students predict and calculate outcomes in various scenarios, enhancing their decision-making skills.

Conditional Probability

Conditional probability is the likelihood of an event occurring, given that another event has already occurred. It focuses on the relationship between two or more events and provides a way to update the probability of an event based on new information.
In the Coupon Collector's Problem, we use conditional probability to explore the probability that the newly collected \( n \)-th coupon is a type not previously obtained. This involves considering the condition that many other coupons have already been collected.
The mathematical notation of conditional probability is \( P(A|B) \), which translates to the probability of event A occurring given that B has occurred. In the exercise, \( A \) is the event that the \( n \)-th coupon is a new type, and \( B_i \) is the event that it's of type \( i \).
To find \( P(A|B_i) \), we consider the probability that none of the first \( n-1 \) coupons are of type \( i \), which is \( (1-p_i)^{n-1} \), and that the \( n \)-th coupon is of type \( i \), which is \( p_i \). Successfully calculating conditional probabilities helps break down complex problems into simpler components and can greatly enhance problem-solving strategies.

Joint Probability

Joint probability refers to the likelihood of two events happening at the same time. It provides insight into the combined occurrence of two events and is essential for understanding the interaction between different events.
In our exercise, we're interested in the joint probability that both events \( A \) and \( B_i \) take place simultaneously, where \( A \) is the event of picking a new coupon type and \( B_i \) is the event of that coupon being of type \( i \).

• It's calculated as \( P(A \cap B_i) \) using the formula \( P(A|B_i) \) \( \times \) \( P(B_i) \).
• In this case, \( P(A \cap B_i) = p_i^2(1-p_i)^{n-1} \), which leverages both the conditional probability and the probability of event \( B_i \).

Joint probabilities are essential for analyzing how multiple factors can interact and influence an outcome, thus fostering a comprehensive understanding of real-world events.

Event Independence

The concept of event independence is crucial in probability theory. Two events are considered independent if the occurrence of one does not affect the probability of the other occurring. This can simplify calculations since independent events do not impact each other.
However, when dealing with the Coupon Collector's Problem, we assume that events are not independent. The probability of obtaining a new type of coupon (event \( A \)) is dependent on the event of drawing a specific coupon of type \( i \) (event \( B_i \)).
Understanding independence is vital because it helps clarify scenarios where the outcome of one event is completely separate from others, versus dependent scenarios where knowledge of one event influences the likelihood of another.

• If events were independent in this problem, then the probability of each event would simply multiply together without considering the condition or previous draws.
• Since this is not the case, each step in the calculation accounts for previous occurrences, which is reflective of many real-life situations where one result affects others.

Grasping the differences between independent and dependent events aids in accurately defining and calculating probabilities in complex analyses.