| Course | GB513 Business Analytics |
|---|---|
| Unit | Unit 3 |
| Paper type | Probability problem set |
| Length | About 840 words, 6 pages double-spaced |
| Format | APA 7 student paper |
| School | Purdue University Global |
| Program | MBA |
| Updated | October 2026 |
Free sample paper for GB513 Unit 3
Six Short Questions at an Electronics Retailer: A Unit 3 Probability Problem Set
[Student Name]
Purdue University Global
GB513: Business Analytics
Unit 3 Probability Problem Set
[Instructor Name]
[Date]
The electronics retailer, its customers and every figure are composite and illustrative; the methods cited are real.
Problem 1: Reading a Contingency Table
Given. Voltline Electronics recorded 500 recent customers by age group and whether they bought an extended warranty. Of 160 customers aged 18 to 34, 32 bought one. Of 210 aged 35 to 54, 63 did. Of 130 aged 55 and over, 52 did. In total, 147 of 500 bought a warranty.
Rule. Joint probability is a cell divided by the grand total; marginal probability is a row or column total divided by the grand total; conditional probability is a cell divided by its row total (Anderson et al., 2020).
Calculation. The joint probability that a customer is 55 or over and bought a warranty is 52 divided by 500, or 0.1040. The marginal probability of buying a warranty is 147 divided by 500, or 0.2940. The marginal probability of being 55 or over is 130 divided by 500, or 0.2600. The conditional probability of buying a warranty given the customer is 55 or over is 52 divided by 130, or 0.4000.
Result. About 29 percent of all customers buy a warranty, but among customers 55 and over the rate is 40 percent.
Problem 2: Testing Independence
Given. The events are A, the customer is 55 or over, and W, the customer buys a warranty.
Rule. Two events are independent if the probability of one does not change when the other is known, that is, if P(W given A) equals P(W).
Calculation. P(W given A) is 0.4000, and P(W) is 0.2940. They are not equal.
Result. Age and warranty purchase are not independent: older customers are more likely to buy, which suggests sales staff could offer the warranty with more confidence to that group.
Problem 3: Defective Units in a Shipment
Given. About 4 percent of a particular tablet model arrive with a defect. A store receives a shipment of 20.
Rule. The number of defective units follows a binomial distribution, because the shipment has a set count of units, each defective or not independently, with one shared defect rate (Levine et al., 2021).
Calculation. The probability of no defects is BINOM.DIST(0, 20, 0.04, FALSE), or 0.4420. The probability of two or more defects is 1 minus BINOM.DIST(1, 20, 0.04, TRUE), or 0.1897.
Result. A store has about a 44 percent chance of a shipment with no defects and about a 19 percent chance of two or more.
Problem 4: Expected Defects and At Most One
Given. The same shipment of 20 at 4 percent.
Rule. The expected value of a binomial variable is n times p.
Calculation. Expected defects are 20 times 0.04, or 0.8. The probability of at most one defect is BINOM.DIST(1, 20, 0.04, TRUE), or 0.8103.
Result. On average a shipment holds less than one defective tablet, and about 81 percent of shipments hold one or none. A store that budgets for one return per shipment of this model will be right four times in five.
Problem 5: Walk-In Customers at the Service Counter
Given. The service counter averages 6.5 walk-in customers an hour.
Rule. Arrivals of independent customers at a steady average rate follow a Poisson distribution.
Calculation. The probability of ten or more walk-ins in an hour is 1 minus POISSON.DIST(9, 6.5, TRUE), or 0.1226. The probability of three or fewer is POISSON.DIST(3, 6.5, TRUE), or 0.1118.
Result. In about 12 percent of hours the counter will see ten or more customers, enough to justify a second staff member on call during peak periods.
Each answer closes with a plain statement a store manager could use, because a probability no one can use has not finished the problem.
Problem 6: Comparing Two Warranty Prices
Given. Voltline sells a three-year warranty for $89. About 15 percent of covered devices need a repair during the warranty, at an average cost of $310. The current take rate is 29.4 percent. Marketing estimates that at $129, the take rate would fall to 18 percent.
Rule. The expected value of a random variable is the sum of each outcome times its probability.
Calculation. Expected repair cost per warranty is 0.15 times $310, or $46.50. At $89, expected profit per warranty sold is $42.50, and per customer it is 0.294 times $42.50, or about $12.50. At $129, expected profit per warranty sold is $82.50, and per customer it is 0.18 times $82.50, or about $14.85.
Result. The higher price earns about $2.35 more per customer on average, even though fewer customers buy, so Voltline should test the $129 price in a few stores before changing it everywhere. James Evans notes that expected value supports decisions under uncertainty but depends on the accuracy of the probability estimates behind it (Evans, 2020); the 18 percent take rate is an estimate, which is why a test comes first.
References
Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., & Cochran, J. J. (2020). Statistics for business and economics (14th ed.). Cengage.
Evans, J. R. (2020). Business analytics: Methods, models, and decisions (3rd ed.). Pearson.
Levine, D. M., Stephan, D. F., & Szabat, K. A. (2021). Statistics for managers using Microsoft Excel (9th ed.). Pearson.
The Purdue Global GB513 Unit 3 instructions
GB513 Unit 3 asks you to solve a set of short probability problems in business settings. Work each problem in the same four parts: restate what is given, name the probability rule or distribution and why it fits, show the calculation with its Excel function and inputs, and write a result sentence a manager could use. For contingency tables, distinguish joint, marginal and conditional probabilities. Test independence by comparing a conditional probability with a marginal one. Choose binomial for a fixed number of trials and Poisson for counts over an interval. Use expected value to compare decisions, and note how much the answer depends on estimated probabilities. Round consistently and restate results as percentages where a manager would read them that way, citing statistics texts in APA 7.
How the GB 513 Unit 3 example is put together
Problem 1 reads Voltline Electronics' table of 500 customers, with a joint probability of 0.1040, a marginal warranty probability of 0.2940 and a conditional probability of 0.4000 for customers 55 and over, citing Anderson and colleagues. Problem 2 finds age and warranty purchase are not independent. Problem 3 uses BINOM.DIST for a shipment of 20 at 4 percent, giving 0.4420 for no defects and 0.1897 for two or more, citing Levine, Stephan and Szabat. Problem 4 gives expected defects of 0.8 and a 0.8103 chance of one or none. Problem 5 uses POISSON.DIST at 6.5 an hour for ten or more and three or fewer, and the highlighted sentence explains the result sentences. Problem 6 compares $89 and $129 warranties at expected repair cost of $46.50, citing Evans on estimate accuracy.
Where the points sit in the GB513 Unit 3 rubric
GB513 instructors look for the right rule or distribution, a visible calculation and a usable result for every problem. Restating the given information shows the problem was understood. Saying why each distribution suits its problem, for example the set number of units in a binomial shipment, is rewarded. Showing Excel functions with their inputs lets a grader check tails and cumulative flags. An independence test that compares a conditional with a marginal probability shows conceptual understanding. Expected value comparisons should note their dependence on estimated probabilities. Result sentences that a manager could act on carry significant weight. Three texts are cited in APA 7, one for probability rules, one for Excel distribution functions and Evans's analytics text for expected value under uncertainty.
GB513 Unit 3 help: mistakes that cost points
Many GB513 Unit 3 problem sets give numbers without saying which distribution was used or why. Others compute the wrong tail, such as the probability of exactly two instead of two or more. A common gap is the result sentence: a probability of 0.1226 means little until it is restated as about one hour in eight. Some writers treat estimated probabilities in an expected value problem as certain. Restate each problem, name and justify the rule, show the Excel function and inputs, check the tail and end with a plain sentence a manager could use.
Get a GB513 Unit 3 example written to your instructions
The scenarios your GB513 section set for Unit 3 are what the sample works, not the retailer used above. Send the problem list with its instructions and rubric; a custom set arrives in 24 to 48 hours, each answer carrying its function and its business sentence. There is no charge for the first one. The paper above is an original model document written by our desk, not a submitted student paper and not an official Purdue University Global document.
More GB513 papers and MBA samples
- GB513 Unit 1: Descriptive Statistics Exercise
- GB513 Unit 2: Data Summary Analysis
- GB513 Unit 4:
- GB513 Unit 5: Confidence Interval Analysis
- GB513 Unit 6: Hypothesis Test Memo
- GB513 Unit 7: Two-sample Comparison
- GB513 Unit 8: Simple Regression Analysis
- GB513 Unit 9: Multiple Regression Model
- GB513 Unit 10: Applied Forecast Analysis
- GB512 Unit 3: Routine Business Memo
- GB500 Unit 3: Marketing Mix Application
GB513 Unit 3 questions, answered
How do I know whether a problem is binomial or Poisson?
Look at what is being counted. If there is a fixed number of trials, such as twenty units inspected, and each is a success or failure, the problem is binomial. If the count happens over an interval of time or space with no fixed ceiling, such as arrivals per hour, it is Poisson. The sample puts that reasoning in one line before each calculation.
Does the problem set need the Excel functions written out?
In many GB513 sections, yes, or at least the inputs to them. Showing =BINOM.DIST(2,20,0.05,TRUE) lets a grader see which tail you computed and whether the cumulative flag was right. When the prompt calls for hand calculation instead, the example shows the formula with values substituted. Either way the method must be visible, not just the answer.
Where can I find a free GB513 Unit 3 sample paper?
This page carries a full GB513 Unit 3 probability problem set for an electronics retailer, with a contingency table, independence test, binomial and Poisson problems, an expected value comparison and APA 7 references.
How does the set test whether two events are independent?
It compares the probability that a customer 55 or over buys a warranty, 0.40, with the probability that any customer buys one, 0.294. Because they differ, age and warranty purchase are not independent.
Why test the $129 price before adopting it?
The expected value comparison depends on an estimated take rate of 18 percent at the higher price. If fewer customers buy than estimated, the advantage could disappear, so a test in a few stores checks the estimate first.