2 years ago

Can anyone please help me?

Mathematics

1 answer:

otez555 [7]2 years ago

5 0

Wait which 1 are yhu answering?

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The general manager, marketing director, and 3 other employees of CompanyAare hosting a visitby the vice president and 2 other e

zubka84 [21]

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

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