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M Question 8, 3.4.39, HW Score: \( 72.22 \%, 6.0 \) ui y puirris \( 1 \leftarrow \begin{array}{l}\text { A committee has twelve members. There are two members that currently serve as the board's chairman and ranking member. Each member is equally } \\ \text { likely to serve in any of the positions. Two members are randomly selected and assigned to be the new chairman and ranking member. What is the } \\ \text { probability of randomly selecting the two members who currently hold the positions of chairman and ranking member and reassigning them to their } \\ \text { current positions? }\end{array} \) The probability is \( \square \). (Round to four decimal places as needed.)

Ask by Edwards Hodges. in the United States
Feb 26,2025

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Tutor-Verified Answer

Answer

The probability is 0.0152.

Solution

To solve this problem, we need to determine the probability of randomly selecting the two members who currently hold the positions of chairman and ranking member from a committee of twelve members. ### Step 1: Identify the total number of ways to select 2 members from 12. The total number of ways to choose 2 members from a group of 12 can be calculated using the combination formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of members, and \( k \) is the number of members to choose. In this case, \( n = 12 \) and \( k = 2 \): \[ C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12!}{2! \cdot 10!} = \frac{12 \times 11}{2 \times 1} = 66 \] ### Step 2: Identify the number of favorable outcomes. The favorable outcome is selecting the two specific members who currently hold the positions of chairman and ranking member. There is only 1 way to select these two specific members. ### Step 3: Calculate the probability. The probability \( P \) of selecting the two specific members is given by the ratio of the number of favorable outcomes to the total number of outcomes: \[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{66} \] ### Step 4: Round the probability to four decimal places. Now, we will calculate the decimal value of \( \frac{1}{66} \): \[ P \approx 0.01515151515 \] Rounding this to four decimal places gives: \[ P \approx 0.0152 \] ### Final Answer The probability of randomly selecting the two members who currently hold the positions of chairman and ranking member and reassigning them to their current positions is: \[ \boxed{0.0152} \]

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To solve this, you'll first want to determine how many ways you can select any two members from the committee. Since the committee has 12 members, the total combinations of selecting 2 members out of 12 can be calculated using the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \). So, \( C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66 \). Now, there's only one specific combination in which the two selected members are the current chairman and ranking member. Hence, the probability of randomly selecting these two specific members is given by: \[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{66}. \] When calculated, this gives you: \[ P \approx 0.01515, \] Rounded to four decimal places: \[ P \approx 0.0152. \] So the answer you are looking for is approximately \( 0.0152 \).

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