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.