Problem 33 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In a television game show, the winner is asked to select three prizes from five different prizes, \(A, B\), \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E} .\) a. Describe a sample space of possible outcomes (order is not important). b. How many points are there in the sample space corresponding to a selection that includes A? c. How many points are there in the sample space corresponding to a selection that includes \(\mathrm{A}\) and \(\mathrm{B}\) ? d. How many points are there in the sample space corresponding to a selection that includes either \(\mathrm{A}\) or \(\mathrm{B}\) ?

Step-by-Step Solution

Answer

In summary: a. The sample space of possible outcomes has \(10\) points or combinations. b. There are \(6\) points in the sample space corresponding to a selection that includes prize \(A\). c. There are \(3\) points in the sample space corresponding to a selection that includes both prizes \(A\) and \(B\). d. There are \(9\) points in the sample space corresponding to a selection that includes either prize \(A\) or prize \(B\).

Step 1: Sample Space of Possible Outcomes

To find the sample space, we need to calculate the number of ways to choose three prizes from the five given prizes. This is a combination problem and can be calculated using the formula: \[\binom{n}{k} = \frac{n!}{k! (n-k)!}\] In this case, we have \(n = 5\) prizes and \(k = 3\), and the formula becomes: \[\binom{5}{3} = \frac{5!}{3! 2!} = \frac{120}{(6)(2)} = 10\] So, there are \(10\) possible outcomes in the sample space.

Step 2: Sample Space Including Prize A

Now, let's find the number of points in the sample space that includes prize \(A\). Since \(A\) is selected, we have to choose the other two prizes from the remaining four \((B, C, D, E)\). The formula becomes: \[\binom{4}{2} = \frac{4!}{2! 2!} = \frac{24}{(2)(2)} = 6\] So, there are \(6\) points in the sample space corresponding to a selection that includes prize \(A\).

Step 3: Sample Space Including Prizes A and B

Next, let's find the number of points in the sample space that includes both prizes \(A\) and \(B\). Since both \(A\) and \(B\) are selected, we have to choose the third prize from the remaining three \((C, D, E)\). The formula becomes: \[\binom{3}{1} = \frac{3!}{1! (3-1)!} = \frac{6}{(1)(2)} = 3\] So, there are \(3\) points in the sample space corresponding to a selection that includes both prizes \(A\) and \(B\).

Step 4: Sample Space Including Either Prize A or Prize B

Finally, let's find the number of points in the sample space that include either prize \(A\) or prize \(B\). To find this, we can simply add the number of outcomes including prize \(A\) and the number of outcomes including prize \(B\), then subtract the number of outcomes including both \(A\) and \(B\) to avoid double-counting: \[6 + 6 - 3 = 9\] So, there are \(9\) points in the sample space corresponding to a selection that includes either prize \(A\) or prize \(B\).

Key Concepts

Sample Space

In the realm of combinatorics, understanding the concept of a "sample space" is pivotal. The sample space is essentially the collection of all possible outcomes for a particular experiment or selection. When dealing with a simple selection of prizes on a game show, as described in the problem, the sample space involves all different ways you can choose items from a given set.
In our case, the challenge is to choose 3 prizes from a total of 5. Since the order of selection doesn't matter, we use combinations to determine the possible outcomes, ensuring that every possible selection is accounted for without repetition.

Combination Formula

The combination formula is crucial when you want to find out how many ways a certain number of selections can be made from a larger set. This is given by the formula: \[\binom{n}{k} = \frac{n!}{k! (n-k)!}\]
Here, \(n\) is the total number of items, and \(k\) is the number of items to choose. The exclamation mark \(!\) denotes a factorial, which is the product of all positive integers up to that number. For instance, \(3! = 3 \times 2 \times 1 = 6\).
Using this formula helps quickly find the number of combinations, such as selecting 3 prizes from 5, which results in 10 different outcomes without the need to enumerate each possibility individually.

Permutation and Combination Concepts

In combinatorics, permutations and combinations are two key concepts that help solve selection-related problems. While permutations consider the order of selection important, combinations disregard order, focusing solely on the selection itself.
For the game show problem, combinations are more appropriate because choosing prizes \(A, B,\) and \(C\) is the same as choosing \(B, C,\) and \(A\). The concepts might overlap, but understanding that combinations are used when order does not matter is crucial.
This distinction helps simplify complex selection problems by removing unnecessary considerations of sequence, allowing for more straightforward mathematical problem-solving.

Mathematical Problem Solving

Mathematical problem solving in combinatorics involves careful application of formulas and logical reasoning. When faced with a problem like the game show, it is important to methodically break down each requirement:

• First, identify if repetitions matter and if the order is important.
• Apply the combination formula to calculate the initial sample space.
• Use combinations to handle specific conditions, such as including certain prizes.
• Carefully allocate outcomes to avoid double counting, especially in overlapping cases.

By considering each of these steps, students can tackle similar combinatorial problems effectively, developing a strong foundation in mathematical reasoning and problem-solving skills.