Question

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.

Answer 1

Solution :

Let the three places be 1, 2, 3, 4, 5, 6, 7, 8

a). Number of the cases when a general manager is the next to a vice president is equal to 7 and the these 2 can be arranged in 21 ways. So the total number of ways = 7 x 2

                  = 14

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

Therefore the required probability is

  $=\frac{14}{8!}$

 = $\frac{14}{40320} = 0.000347$

b). The probability that the marketing director to be placed in the leftmost position is

   $=\frac{7!}{8!}$

  $=\frac{1}{8} = 0.125$

c). The two events are not independent because

   $P(A \cap B) \neq P(A) \times P(B)$

  $\frac{12}{8!} \neq \frac{14}{8!} \times \frac{1}{8}$

where A is the case a and B is the case b.

Answer 2

(a) The possibility of the general manager is next to the vice president is $\frac{1}{4}$.

(b) The possibility of the marketing director is in the leftmost position is $\frac{1}{8}$.

(c) So, the two events are dependent on each other.

Probability:

Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. The probability of all the events in a sample space adds up to 1.

Total people in company A and company B is $=8$

Overall ways in which these $8$ people can be lined up$=8!$

                                                                                        $=40320$

(a) The probability that the general manager is next to the vice president is$=P(A)$

Now, we can combine the general manager and vice president as one, then the total people in both the company will become $7$.

by arranging these $7$ people in one line $=7!$

                                                                $=5040$

Again, combine the general manager and vice president in one line$=2!$

                                                                                                                $=2$

Therefore, $P(A)=\frac{5040\times 2}{40320}$

                          $=\frac{10080}{40320}$

$P(A)=\frac{1}{4}$

(b) The probability that the marketing director is in the leftmost position is$=P(B)$

Now, fixing the position of marketing director in the leftmost.

arranging the $7$ other people in $7!$ ways $=5040$

Therefore,$P(B)=\frac{5040}{40320}$

                          $=\frac{1}{8}$

$P(B)=\frac{1}{8}$

(c) Assuming event B already occurred which means that the position of marketing director is already fixed in the leftmost position.

Now, trying to find out the probability of the general manager next to the vice president is event A. it comes different because we are not allowed to arrange rest $7$ people, we have to fix the position of one person that causes the repetition of probability.

So, the two events are dependent on each other.

Learn more about the topic of Probability: https://brainly.com/question/26959834