91Ó°ÊÓ

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?

Step by step solution

01

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.

02

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.

03

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%.

04

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%.

05

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%.

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Key Concepts

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

Sample Space

A sample space in probability is the set of all possible outcomes of a random experiment. In the context of the given exercise, the task involves choosing two problems out of a total of five. These individual problems are labeled as A, B, C, D, and E. Each possible pair constitutes an element of the sample space. Therefore, the sample space here can be represented by these combinations: (A, B), (A, C), (A, D), (A, E), (B, C), (B, D), (B, E), (C, D), (C, E), and (D, E). Thus, there are 10 such pairs, forming our sample space. Understanding the sample space is crucial since it forms the foundation for calculating probabilities of different events occurring within this space.

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of objects. In this exercise, we use combinatorial methods to determine how many ways two homework problems can be selected from a total of five. This is a classic combinatorial problem where order does not matter, thus we use the formula for combinations: \(C(n, k) = \frac{n!}{k!(n-k)!}\). Here, \(n = 5\) and \(k = 2\), so \(C(5, 2) = \frac{5!}{2!(5-2)!} = 10\). This formula shows us that there are 10 unique pairs of problems that can be chosen. Combinatorics provides a systematic way of counting these possibilities, which is essential for accurate probability calculations.

Equally Likely Outcomes

In a probability experiment, outcomes are considered equally likely if each one has the same probability of occurring. In our exercise about homework, the problems are selected at random which implies no bias or preference; therefore, each pair of problems is an equally likely outcome. This means each of the 10 pairs has the same chance of being selected, with a probability of \(\frac{1}{10}\). This assumption is crucial for making fair predictions and ensuring the correctness of further probability calculations. Recognizing when outcomes are equally likely helps in applying probability in a uniform and balanced manner across all scenarios involved in the experiment.