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Chapter 4: Problem 106

A heavy-equipment salesman can contact either one or two customers per day with probabilities \(1 / 3\) and \(2 / 3,\) respectively. Each contact will result in either no sale or a \(\$ 50,000\) sale with probabilities \(9 / 10\) and \(1 / 10,\) respectively. What is the expected value of his daily sales?

Short Answer

Expert verified
Answer: The expected value of the heavy-equipment salesman's daily sales is $8,000.

Step by step solution

01

Identify the possible outcomes and their probabilities

There are four possible outcomes in this situation: 1. The salesman contacts one customer and makes a sale. 2. The salesman contacts one customer and doesn't make a sale. 3. The salesman contacts two customers and makes a sale with one or both of them. 4. The salesman contacts two customers and doesn't make a sale with either of them. Let's list the probabilities for each outcome: 1. Probability of contacting one customer - \(1/3\) Probability of making a sale - \(1/10\) 2. Probability of contacting one customer - \(1/3\) Probability of not making a sale - \(9/10\) 3. Probability of contacting two customers - \(2/3\) Probability of making a sale with one or both - We will need to calculate this. 4. Probability of contacting two customers - \(2/3\) Probability of not making a sale with either - \(9/10 \times 9/10 = 81/100\)
02

Calculate the probability of making a sale with at least one customer out of two

To calculate the probability of making a sale with at least one customer out of two, we can find the complementary probability, i.e., the probability of not making a sale with any of them. From outcome 4, we know that the probability of not making a sale with both customers is \(81/100\). So, the probability of making a sale with at least one customer out of two is: \(1 - (81/100) = 19/100\)
03

Calculate the expected value for each outcome

Now, we will calculate the expected value for each outcome by multiplying their respective probability with the sale amount. 1. Expected value for outcome 1: \((1/3) \times (1/10) \times \$50,000 = \$1,666.67\) 2. Since there is no sale in outcome 2, there is no expected value for this. 3. Expected value for outcome 3: \((2/3) \times (19/100) \times \$50,000 = \$6,333.33\) 4. Similarly, as there is no sale in outcome 4, there is no expected value for this.
04

Determine the expected value of daily sales

We will now add the expected values of all outcomes to find the expected daily sales: Expected value of daily sales = \((\$1,666.67) + (\$0) + (\$6,333.33) + (\$0)\) Expected value of daily sales = \( \$8,000\) Hence, the expected value of the heavy-equipment salesman's daily sales is \( \$8,000\).

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Key Concepts

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

Probability Outcomes
When dealing with random events, like the number of sales a salesman makes in a day, it's crucial to understand the various possible outcomes that can occur. A probability outcome is any possible result that might arise from a chance event, and each outcome has a specific chance or probability of happening.

For the heavy-equipment salesman's case, there are four distinct outcomes based on the combination of customers contacted and the sales made. The outcome is not just a sale or no sale; it must account for the number of customers as well. Understanding the probability outcomes is the foundation for further calculations, including the probability and expected value of daily sales.
Probability Calculation
The probability calculation involves determining the likelihood of each outcome. It鈥檚 the process of quantifying the chance of an event happening. In our exercise, the probabilities are a mix of independent events (the number of customers a salesman contacts) and dependent events (making a sale based on the number of contacts).

Consider the second outcome where the salesman contacts one customer and doesn't make a sale. Here, the combined probability is the product of the individual probabilities: \[\begin{equation} \frac{1}{3} (for contacting one customer) \times \frac{9}{10} (for not making a sale) \end{equation}\]
Calculating such probabilities for multiple scenarios, especially with more than two outcomes, may seem daunting. By breaking it down into individual events and understanding that the total can not exceed 1, this complex task becomes manageable.
Expected Value Calculation
Expected value calculation is a weighted average of all possible outcomes of a random variable, with the weights being their respective probabilities. It represents the long-term average of repeated instances of the same probabilistic event and is crucial in decision-making processes where outcomes are uncertain.

For example, we calculated the expected value of the salesman's sales for each outcome separately. Then, as seen in the exercise, we sum these individual expected values to obtain the overall expected daily sales:\[\begin{equation}Expected \ daily \ sales = \sum (Probability \ of \ each \ outcome \ x \ Sale \ amount)\end{equation}\]This approach gives a realistic expectation of the salesperson's performance over time, taking into account the variability of daily outcomes.

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Most popular questions from this chapter

A fire-detection device uses three temperature-sensitive cells acting independently of one another in such a manner that any one or more can activate the alarm. Each cell has a probability \(p=.8\) of activating the alarm when the temperature reaches \(100^{\circ} \mathrm{F}\) or higher. Let \(x\) equal the number of cells activating the alarm when the temperature reaches \(100^{\circ} \mathrm{F}\). a. Find the probability distribution of \(x\). b. Find the probability that the alarm will function when the temperature reaches \(100^{\circ} \mathrm{F}\). c. Find the expected value and the variance for the random variable \(x\).

A student prepares for an exam by studying a list of 10 problems. She can solve 6 of them. For the exam, the instructor selects 5 questions at random from the list of \(10 .\) What is the probability that the student can solve all 5 problems on the exam?

You own 4 pairs of jeans, 12 clean T-shirts, and 4 wearable pairs of sneakers. How many outfits (jeans, T-shirt, and sneakers) can you create?

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.

Use event relationships to fill in the blanks in the table below. $$ \begin{array}{|c|c|l|c|c|c|} \hline \boldsymbol{P}(\boldsymbol{A}) & \boldsymbol{P}(\boldsymbol{B}) & \text { Conditions for Events } \boldsymbol{A} \text { and } \boldsymbol{B} & \boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{B}) & \boldsymbol{P}(\boldsymbol{A} \cup \boldsymbol{B}) & \boldsymbol{P}(\boldsymbol{A} \mid \boldsymbol{B}) \\ \hline .3 & .4 & \text { Mutually exclusive } & & & \\ \hline .3 & .4 & \text { Independent } & & & \\ \hline .1 & .5 & & & & .1 \\ \hline .2 & .5 & & 0 & & \\ \hline \end{array} $$
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