Hello, The site shows two different answers for (a) I would like clarification as to which is correct. I can see it both ways. The different ways it has been solved are added as images. Different answers are given.   The general manager, marketing director, and 3 other employees of Company A are hosting a visit by the vice president and 2 other employees of Company B. The eight people line up in a random order to take a photo. Every way of lining up the people is equally likely. (a) What is the probability that the general manager is next to the vice president? (b) What is the probability that the marketing director is in the leftmost position? (c) Determine whether the two events are independent. Prove your answer by showing that one of the conditions for independence is either true or false.

A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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The site shows two different answers for (a) I would like clarification as to which is correct. I can see it both ways. The different ways it has been solved are added as images. Different answers are given.

 

The general manager, marketing director, and 3 other employees of Company A are hosting a visit by the vice president and 2 other employees of Company B. The eight people line up in a random order to take a photo. Every way of lining up the people is equally likely.

(a) What is the probability that the general manager is next to the vice president?

(b) What is the probability that the marketing director is in the leftmost position?

(c) Determine whether the two events are independent. Prove your answer by showing that one of the conditions for independence is either true or false.

 

 

The expression for all the possible places when the general manager is next to the vice president is:

\[
\{(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5),(6,7),(7,6),(7,8),(8,7)\}
\]

So, there are \( f=14 \) places when the general manager is next to the president.

The expression for the probability that the general manager is next to the vice president is:

\[
P(A) = \frac{f}{N}
\]

Substitute the given values in the above expression.

\[
P(A) = \frac{14}{40320} = 0.0003472
\]

Thus, the probability that the general manager and vice president are adjacent to each other is 0.0003472.
Transcribed Image Text:The expression for all the possible places when the general manager is next to the vice president is: \[ \{(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5),(6,7),(7,6),(7,8),(8,7)\} \] So, there are \( f=14 \) places when the general manager is next to the president. The expression for the probability that the general manager is next to the vice president is: \[ P(A) = \frac{f}{N} \] Substitute the given values in the above expression. \[ P(A) = \frac{14}{40320} = 0.0003472 \] Thus, the probability that the general manager and vice president are adjacent to each other is 0.0003472.
### Step 1

- **Total Number of People = 8**
- **Total Number of Ways 8 People are Lined Up = 8! = 40,320**

#### (a) Probability that the General Manager is Next to the Vice President = P(A)

**Approach:** Combine the general manager and vice president as one unit, reducing the total number of people to 7.

- **Arrange the 7 People in a Line = 7! = 5,040**
- **Arrange the Two Combined People (General Manager and Vice President) = 2! = 2**

\[ P(A) = \frac{(5040 \times 2)}{40320} = \frac{1}{4} \]

The probability that the general manager is next to the vice president is \(\frac{1}{4}\).
Transcribed Image Text:### Step 1 - **Total Number of People = 8** - **Total Number of Ways 8 People are Lined Up = 8! = 40,320** #### (a) Probability that the General Manager is Next to the Vice President = P(A) **Approach:** Combine the general manager and vice president as one unit, reducing the total number of people to 7. - **Arrange the 7 People in a Line = 7! = 5,040** - **Arrange the Two Combined People (General Manager and Vice President) = 2! = 2** \[ P(A) = \frac{(5040 \times 2)}{40320} = \frac{1}{4} \] The probability that the general manager is next to the vice president is \(\frac{1}{4}\).
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