/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 113 A salesperson figures that the p... [FREE SOLUTION] | 91Ó°ÊÓ

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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

If an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2},\) and \(S_{3}\) can occur, with these probabilities: $$ P\left(S_{1}\right)=.2 \quad P\left(S_{2}\right)=.5 \quad P\left(S_{3}\right)=.3 $$ The probabilities of a fourth event \(A\) occurring, given that event \(S_{1}, S_{2},\) or \(S_{3}\) occurs, are $$ P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3 $$ If event \(A\) is observed, find \(P\left(S_{1} \mid A\right), P\left(S_{2} \mid A\right),\) and \(P\left(S_{3} \mid A\right)\).

Two cold tablets are accidentally placed in a box containing two aspirin tablets. The four tablets are identical in appearance. One tablet is selected at random from the box and is swallowed by the first patient. A tablet is then selected at random from the three remaining tablets and is swallowed by the second patient. Define the following events as specific collections of simple events: a. The sample space \(S\). b. The event \(A\) that the first patient obtained a cold tablet. c. The event \(B\) that exactly one of the two patients obtained a cold tablet. d. The event \(C\) that neither patient obtained a cold tablet.

Refer to the experiment conducted by Gregor Mendel in Exercise 4.64 Suppose you are interested in following two independent traits in snap peas-seed texture \((\mathrm{S}=\) smooth, \(\mathrm{s}=\) wrinkled ) and seed color \((\mathrm{Y}=\) yellow, \(\mathrm{y}=\) green \()-\) in a second-generation cross of heterozygous parents. Remember that the capital letter represents the dominant trait. Complete the table with the gene pairs for both traits. All possible pairings are equally likely. a. What proportion of the offspring from this cross will have smooth yellow peas? b. What proportion of the offspring will have smooth green peas? c. What proportion of the offspring will have wrinkled yellow peas? d. What proportion of the offspring will have wrinkled green peas? e. Given that an offspring has smooth yellow peas, what is the probability that this offspring carries one s allele? One s allele and one \(y\) allele?

A survey of people in a given region showed that \(20 \%\) were smokers. The probability of death due to lung cancer, given that a person smoked, was roughly 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is \(.006,\) what is the probability of death due to lung cancer given that a person is a smoker?

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