Question

The general manager, marketing director, and 3 other employees of CompanyAare hosting a visitby the vice president and 2 other employees of CompanyB. The eight people line up in a randomorder to take a photo. Every way of lining up the people is equally likely.Required:a. What is the probability that the bride is next to the groom?b. What is the probability that the maid of honor 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

Answer:

Following are the answer to this question:

Step-by-step explanation:

Let, In the Bth place there are 8 values.

In point a:

There is no case, where it generally manages its next groom is = 7 and it will be arranged in the 2, that can be arranged in 2! ways. So, the total number of ways are: $\to 7 \times 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$ required probability:

$= \frac{14}{8!}\\\\= \frac{14}{8\times7 \times6 \times 5 \times 4 \times 3\times 2 \times 1 }\\\\= \frac{1}{8\times6 \times5 \times 4 \times 3}\\\\= \frac{1}{8\times6 \times5 \times 4 \times 3}\\\\=\frac{1}{2880}\\\\=0.00034$

In point b:

Calculating the leftmost position:

$\to \frac{7!}{8!}\\\\\to \frac{7!}{8 \times 7!}\\\\\to \frac{1}{8}\\\\\to 0.125$

In point c:

This option is false because

$\to P(A \cap B) \neq P(A) \times P(B)\\\\\to \frac{12}{8!} \neq \frac{14}{8!}\times \frac{1}{8}\\\\\to \frac{12}{8!} \neq \frac{7}{8!}\times \frac{1}{4}$