Question

A professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E} .\) One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?

Short answer

The sample space for picking 2 out of 5 problems consists of 10 possible pairs and all outcomes are equally likely. The probability of being able to turn in both problems, if you solved the first or the last three problems, is still 30% as there are 3 successful outcomes. If 4 problems were completed, the probability increases to 40% as there are 4 successful outcomes.

Step-by-step solution

Identifying the Sample Space

To find the sample space for selecting two problems out of five, consider each problem as an item, namely: A, B, C, D, E. Use the combination principle to find the number of ways to choose 2 problems out of 5. The formula 'n choose k' where n = total items and k = items chosen at a time is \(C(n, k) = n! / [k!(n - k)!]\). Substituting n = 5 and k = 2 yields: \(C(5, 2) = 5! / [2!(5 - 2)!] = 10\). Therefore, the sample space includes 10 possible pairs of problems.

Determining of Equally Likely Outcomes

As no preference is given for the selection of any problems, all outcomes are equally likely. The chance of each pair being picked is equal due to the randomness of the selection.

Calculating Probability of Being Able to Deliver both Problems

To find the probability, identify the successful outcomes first. Since you've only completed problems A, B, and C, the possible successful pairs could be (A, B), (B, C), and (A, C). Thus, there are 3 successful outcomes. Probability is calculated as 'Number of successful outcomes' divided by 'Total outcomes'. Substituting gives Probability = 3/10 = 0.3 or 30%.

Verifying if the Probability Changes If Completed Problems Change

If you completed the last three problems instead of the first three, the successful outcomes would change, but the number would still remain 3. Therefore, the probability would not change and still remain at 30%.

Checking the Change in Probability If More Problems Were Completed

Now, if you completed 4 out of the 5 problems, you will be able to deliver any randomly selected pair, unless the pair includes the one problem you didn't solve. This would result in 4 successful outcomes and hence, Probability = 4/10 = 0.4 or 40%.