Problem 67 - Algebra and Trigonometry Real Mathematics, Real People

Question

A sales representative makes sales on approximately one-fifth of all calls. On a given day, the representative contacts six potential clients. What is the probability that a sale will be made with (a) all six contacts, (b) none of the contacts, and (c) at least one contact?

Step-by-Step Solution

Answer

The probability of making a sale to all six clients is a very small but measurable value, the probability of making no sales is larger, and the probability of making at least one sale is significantly higher, but less than 1.

Step 1: Identify Parameters

Firstly, identify the parameters to use in calculation. The number of contacts `n` equals 6, the success probability `p` is 1/5 or 0.2.

Step 2: Calculate Probability for all six Contacts

Calculate the probability `P` for all six potential clients to make a deal, using the formula giving `k` = 6. Specifically, \(P(k = 6; n = 6,p =0.2) = C(6, 6) * (0.2^6) * (0.8^0)\). Calculate this value.

Step 3: Calculate Probability for No Contact

Calculate the probability `P` for none of the potential clients to make a deal. Use the formula, but this time with k = 0 using \(P(k = 0; n = 6,p =0.2) = C(6, 0) * (0.2^0) * (0.8^6)\). Calculate this value.

Step 4: Calculate Probability for at Least One Contact

To calculate the likelyhood of making at least one sale, calculate the compliment of making no sale using \(P(at least 1) = 1 - P(k = 0)\). You can use the result from Step 3 that \(P(k=0)\).

Key Concepts

Binomial Probability Distribution

Understanding the binomial probability distribution is crucial for analyzing scenarios where there are only two possible outcomes, typically referred to as 'success' and 'failure'. In our example involving a sales representative, success would be making a sale, while failure is not making a sale on a call. The distribution is defined by two parameters: the number of trials (n), and the probability of success in a single trial (p).

In the given problem, the number of trials is the six potential clients contacted (n=6), and the probability of success for making a sale on a call is 1/5, or 0.2 (p=0.2). Binomial probability distribution can help us calculate the probability of having exactly k successes (sales) in n trials (contacts). The general formula for the binomial probability is: \[P(k; n, p) = C(n, k) \times (p^k) \times ((1-p)^{n-k})\] where \(C(n, k)\) represents the number of combinations of n items taken k at a time, which is a part of combinatorics.

For instance, to find the probability of all six clients making a purchase, we set \(k=6\), and apply the formula to get: \[P(k = 6; n = 6, p = 0.2) = C(6, 6) \times (0.2^6) \times (0.8^0)\], which simplifies to a calculation involving combinations and powers of the success and failure probabilities.

Probability Theory

Probability theory is a branch of mathematics that deals with calculating the likelihood of events. It provides a formal framework to quantify uncertainty and is the foundation for statistical analysis and decision making under uncertainty.

In our case with the sales representative, the events are whether a sale is made or not made during each call. Probability theory helps us to determine the likelihood of these events. For example, when calculating the probability of making no sale, we are actually asking, 'What is the chance that every call ends without a sale?'. Given the success probability (p=0.2) and number of trials (n=6), we use the binomial formula to find out the likelihood of zero sales, indicated by \(k=0\).

The beauty of probability theory lies in its ability to measure the occurrence of an event even when there is randomness involved. Importantly, the theory assumes that each call is independent of the others, which is a reasonable assumption for many sales situations. It allows the use of the binomial model without concern for the order in which sales occur.

Combinatorics

Combinatorics is the study of counting, arrangement, and combination of objects. It plays a significant role in probability, especially when we are dealing with discrete outcomes and events. The concept of combinations, often noted as \(C(n, k)\) or \(_nC_k\), refers to finding the number of ways to choose k items from a set of n distinct items without regard to the order of selection.

Applying combinatory analysis allows us to compute the term \(C(n, k)\) in the binomial probability formula. For example, \(C(6, 0)\) is the number of ways to choose zero clients to make a sale from six potential clients, which is always 1, since there is only one way to not choose any clients at all.

The combination formula is: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \(n!\) represents n factorial, the product of all positive integers up to n. This aspect of combinatorics simplifies the complex problem of calculating probabilities in scenarios with multiple trials and varied results, such as multiple sales calls. By grasping these concepts, students can tackle a wide range of probability problems in different contexts and appreciate the incredible utility of combinatorics in the mathematical toolkit.