Problem 11 - A First Course in Probability

Question

A television store owner figures that 45 percent of the customers entering his store will purchase an ordinary television set, 15 percent will purchase a plasma television set, and 40 percent will just be browsing. If 5 customers enter his store on a given day, what is the probability that he will sell exactly 2 ordinary sets and 1 plasma set on that day?

Step-by-Step Solution

Answer

The probability that the store owner will sell exactly 2 ordinary sets and 1 plasma set on that day is approximately 14.58%.

Step 1: Identify the probability of each type of purchase

We need to first convert the given percentages for each type of purchase into probabilities by dividing by 100. We have the following probabilities: - \(P(\text{Ordinary}) = 0.45\) - \(P(\text{Plasma}) = 0.15\) - \(P(\text{Browsing}) = 0.40\)

Step 2: Set up the desired combination of purchases

We are trying to find the probability of having exactly 2 ordinary purchases and 1 plasma purchase. So, we have: - Ordinary: 2 purchases - Plasma: 1 purchase - Browsing: 2 customers (since there are 5 customers in total)

Step 3: Determine the probability for each type of purchase

Now we will determine the probability for each type of purchase using the respective probabilities: - Ordinary: \(P(\text{Ordinary})^2 = (0.45)^2 = 0.2025\) - Plasma: \(P(\text{Plasma})^1 = (0.15)^1 = 0.15\) - Browsing: \(P(\text{Browsing})^2 = (0.40)^2 = 0.16\)

Step 4: Determine the number of combinations of these purchases

We have 5 customers, so we need to find the number of ways to choose 2 customers out of 5 for browsing and 1 customer out of the remaining 3 for the plasma purchase. This can be achieved using the combination formula: - Browsing: \(C(5,2) = \frac{5!}{2!(5-2)!} = 10\) - Plasma: \(C(3,1) = \frac{3!}{1!(3-1)!} = 3\) Now, we need to find the total number of ways to achieve this specific combination of purchases. In this case, since the browsing and plasma purchases are separate and independent, we can simply multiply the number of combinations together to find the total number of combinations: - Total combinations: \(10 \times 3 = 30\)

Step 5: Calculate the probability of the desired outcome

Now we will multiply the probabilities for each type of purchase together and then multiply by the total number of combinations to find the probability of the desired outcome: - Probability: \(0.2025 \times 0.15 \times 0.16 \times 30 = 0.1458\) Therefore, the probability that the store owner will sell exactly 2 ordinary sets and 1 plasma set on that day is approximately 14.58%.

Key Concepts

Combination Formula

Understanding how to calculate the number of combinations is crucial for solving many probability problems. The combination formula helps to determine the number of ways in which we can choose a subset of items from a larger set where the order of selection does not matter. The formula is commonly written as:\[\begin{equation}C(n, k) = \frac{n!}{k! \times (n-k)!}\end{equation}\]Where:

• n represents the total number of items,
• k is the number of items to choose,
• n! denotes the factorial of n which is the product of all positive integers up to n.

For example, if we have 5 customers and we want to select any 2 to purchase a specific product, we use the combination formula to calculate the number of possible pairs that can be made from these 5 customers without considering the sequence in which they are chosen. This approach is fundamental when dealing with scenarios where outcomes with the same elements are considered identical, regardless of their order. It's not just about how many different items exist but about how many ways we can group those items.

Event Independence

In probability theory, event independence plays a pivotal role, particularly in the context of multiple events. Two events are considered independent if the occurrence of one does not affect the probability of the occurrence of the other.When events are independent, the probability of both events occurring is the product of their separate probabilities. The mathematical expression for this is:\[\begin{equation}P(A \text{ and } B) = P(A) \times P(B)\end{equation}\]This concept is instrumental in our problem with the television store owner. The events of selling ordinary TVs and plasma TVs are independent of each other, and the probability of selling a certain combination of these can be calculated by multiplying the independent probabilities of each event. This approach vastly simplifies complex problems involving multiple independent choices or events, allowing us to handle them in a more manageable way.

Binomial Probability

The binomial probability is a type of distribution that arises when we perform a series of independent 'Bernoulli trials'. A Bernoulli trial is a random experiment where there are only two possible outcomes, often termed success and failure. In the context of our exercise, selling or not selling a television to a customer can be viewed as a Bernoulli trial.The binomial probability formula is given by:\[\begin{equation}P(X=k) = C(n, k) \times (p)^k \times (1-p)^{n-k}\end{equation}\]

• X is the random variable representing the number of successes,
• k is the desired number of successes,
• p is the probability of success on any given trial,
• n is the number of trials.

This formula calculates the probability of getting exactly k successes in n independent trials. It combines the concepts of combination (to find all possible ways of success) and event independence (since each trial is independent). In our television store scenario, the binomial probability concept helps the owner to predict sales patterns, thus providing valuable business insights. This model is extremely useful in a wide range of fields, including business, biology, and social sciences, for analyzing situations with two possible outcomes.