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A salesperson figures that the probability of her consummating a sale during the first contact with a client is .4 but improves to .55 on the second contact if the client did not buy during the first contact. Suppose this salesperson makes one and only one callback to any client. If she contacts a client, calculate the probabilities for these events: a. The client will buy. b. The client will not buy.

Short Answer

Expert verified
Answer: The probability of a client buying a product is 0.73, and the probability of a client not buying a product is 0.27.

Step by step solution

01

Understand the given probabilities

We are given the following probabilities: P(Sale during first contact) = 0.4 And if no sale during the first contact, P(Sale during second contact) = 0.55
02

Calculate the probability of a sale happening at any point

To find the probability of a client buying the product at any point, we need to calculate the probability of a sale happening either in the first contact or in the second contact, given that there was no sale in the first contact. We can write this as: P(Client will buy) = P(Sale during first contact) + P(No Sale during first contact) × P(Sale during second contact) Using the given probabilities, we have: P(Client will buy) = 0.4 + (1 - 0.4) × 0.55 P(Client will buy) = 0.4 + 0.6 × 0.55 P(Client will buy) = 0.4 + 0.33 P(Client will buy) = 0.73
03

Calculate the probability of a client not buying at any point

To find the probability of a client not buying the product at any point, the only scenario will be if the client doesn't buy during both contacts. We can write this as: P(Client will not buy) = P(No Sale during first contact) × P(No Sale during second contact) Using the probabilities, we have: P(Client will not buy) = (1 - 0.4) × (1 - 0.55) P(Client will not buy) = 0.6 × 0.45 P(Client will not buy) = 0.27
04

Present the final probabilities

After calculating the probabilities for both scenarios, we have found: a. The probability of the client buying a product is 0.73. b. The probability of the client not buying a product is 0.27.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Sales
In the context of sales and marketing, the probability of sales is a crucial concept referring to the likelihood that a potential client will purchase a product or service. Understanding this probability helps businesses forecast revenue, allocate resources effectively, and design better sales strategies.

In the case of our exercise, the salesperson seeks to improve her chances by employing a two-contact strategy. The first part of the strategy is a direct measure of success on initial contact. Here, we were given a probability of 0.4, or 40%, which represents a fairly optimistic scenario. The second part hinges on conditional probability. If the first contact doesn't result in a sale, the probability of a successful sale on the second contact is 0.55, or 55%.

To better serve students, we should emphasize that each contact point with a client offers a unique opportunity to secure a sale, and that these individual probabilities contribute cumulatively to the overall chance of closing a sale.

Improving Sales Probabilities

There are strategies to increase the probability of sales, such as enhancing product presentation, personalizing customer interactions, and performing targeted follow-ups. These factors can directly influence the probabilities we're working with and are worth mentioning as they relate to real-world applications of the theoretical exercises.
Conditional Probability
The concept of conditional probability is fundamental in predicting outcomes in various scenarios given that a particular condition has been met. It essentially answers the question: 'What is the probability of event A occurring given that event B has already happened?'

In our exercise, the salesperson's second contact is a perfect example of conditional probability. The probability of a sale during the second contact (0.55) is contingent upon the event that there was no sale during the first contact. It's essential here to note that conditional probability is always dependent on the outcome of another event.

Real-Life Applications

Conditional probability is not just a theoretical construct; it has practical applications in fields like finance (assessing market trends given an event), medicine (determining treatment effectiveness given a diagnosis), and even daily decision making (carry an umbrella given a weather forecast). For students to fully grasp the concept, we should connect these theoretical underpinnings to tangible real-world situations.
Probability Theory
Probability theory is essentially the bedrock upon which the previous concepts are built. It's a branch of mathematics concerned with the analysis of random phenomena and the calculation of likelihoods of different outcomes. This theory provides the tools and frameworks used to model uncertainty and make informed predictions.

Through the perspective of probability theory, the exercise showcases the computation of the total probability of an event by considering all the possible ways the event can happen. Here we calculated the 'total probability' of a sale through multiple avenues – initially during the first contact and subsequently via a follow-up.

Grasping the Fundamentals

Key to understanding probability theory is becoming comfortable with its various rules and theorems, such as the addition rule and Bayes' theorem. Clear visual aids, relatable scenarios, and interactive models can help break down complex topics and facilitate student learning. Placing emphasis on these facets within the educational content will not only make it more digestible but also enhance its practical value.

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Most popular questions from this chapter

A survey classified a large number of adults according to whether they were judged to need eyeglasses to correct their reading vision and whether they used eyeglasses when reading. The proportions falling into the four categories are shown in the table. (Note that a small proportion, .02, of adults used eyeglasses when in fact they were judged not to need them.) $$\begin{array}{lcc} & \begin{array}{l}\text { Used Eyeglasses } \\\\\text { for Reading }\end{array} \\\\\hline \text { Judged to Need } & & \\ \text { Eyeglasses } & \text { Yes } & \text { No } \\\\\hline \text { Yes } & .44 & .14 \\\\\text { No } & .02 & .40\end{array}$$ If a single adult is selected from this large group, find the probability of each event: a. The adult is judged to need eyeglasses. b. The adult needs eyeglasses for reading but does not use them. c. The adult uses eyeglasses for reading whether he or she needs them or not.

Let \(x\) equal the number observed on the throw of a single balanced die. a. Find and graph the probability distribution for \(x\). b. What is the average or expected value of \(x ?\) c. What is the standard deviation of \(x\) ? d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the graph in part a. What proportion of all the measurements would fall into this range?

A jar contains four coins: a nickel, a dime, a quarter, and a half-dollar. Three coins are randomly selected from the jar. a. List the simple events in \(S\). b. What is the probability that the selection will contain the half-dollar? c. What is the probability that the total amount drawn will equal \(60 \phi\) or more?

A piece of electronic equipment contains six computer chips, two of which are defective. Three chips are selected at random, removed from the piece of equipment, and inspected. Let \(x\) equal the number of defectives observed, where \(x=0,1,\) or 2 . Find the probability distribution for \(x\). Express the results graphically as a probability histogram.

An experiment involves tossing a single die. These are some events: A: Observe a 2 \(B:\) Observe an even number \(C:\) Observe a number greater than 2 \(D:\) Observe both \(A\) and \(B\) \(E:\) Observe \(A\) or \(B\) or both \(F:\) Observe both \(A\) and \(C\) a. List the simple events in the sample space. b. List the simple events in each of the events \(A\) through \(F\) c. What probabilities should you assign to the simple events? d. Calculate the probabilities of the six events \(A\) through \(F\) by adding the appropriate simple-event probabilities.

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