Question

A certain multinational corporation employs many people from all over the world. It is known that 54% of all people employed by this corporation are from Canada. Of all Canadian employees, 31% hold management positions. Of employees who are not Canadian, 21% hold management positions. Let C represent the event that an employee is Canadian, and M represent the event that an employee holds a management position. In each part to this question, you must first express each probability in terms of the events C and M and justify any probability calculation using a formula.(a) Express each of the three probabilities listed above as the probability of an event involving C and/or M.(b) What is the probability that a randomly selected employee is not from Canada?(c) What is the probability that a randomly selected employee is from Canada and holds a management position?(d) What proportion of employees hold management positions?(e) What is the probability that a randomly selected employee is Canadian, or holds a management position, or both?(f) What proportion of people in management positions are Canadian?

Answer 1

Given:

Percent of people employed from Canada = 54%

Perecent of Canadian employees who hold management positions = 31%

Percent of non-Canadian emloyees who hold management positions = 21%

Where:

C represents the event that an an employee is Canadian, and M represent the event that an employee holds a management position.

Let's solve for the following:

• (a). Express each of the three probabilities listed above as the probability of an event involving C and/or M.

Here, we have:

• P(C) = 54% ==> 0.54

,

• P(M/C) = 31% ==> 0.31

• P(M/C') = 21% ==> 0.21

• (b). What is the probability that a randomly selected employee is not from Canada?

To find the probabilty, we have:

P(not from Canada) = 1 - P(from Canada)

P(C') = 1 - P(C)

P(C') = 1 - 0.54

P(C') = 0.46

Therefore, the probability that a randomly selected employee is not from Canada is 0.46

• (c). What is the probability that a randomly selected employee is from Canada and holds a management position?

Here, we have:

P(C ∩ M) = P(C) x P(M/C)

P(C ∩ M) = 0.54 x 0.31

P(C ∩ M) = 0.1674

Therefore, the probability that a randomly selected employee is from Canada and holds a management position is 0.1674.

• (d). What proportion of employees hold management positions?

Apply the conditional probability formula:

P(M) = P(C' ∩ M) + P(C ∩ M)

P(M) = (P(C') x P(M/C')) + (P(C) x P(M/C))

P(M) = (0.46 x 0.21) + (0.54 x 0.31)

P(M) = 0.0966 + 0.1674

P(M) = 0.264

Therefore, the proportion of employees who hold management positions is 0.264

• (,e). What is the probability that a randomly selected employee is Canadian, or holds a management position, or both?

Apply the formula:

P(C ∪ M) = P(C) + P(M) - P(C ∩ M)

P(C ∪ M) = 0.54 + 0.264 - 0.1674

P(C ∪ M) = 0.6366

Therefore, the probability that a randomly selected employee is Canadian, or holds a management position, or both is 0.6366

• (f,). What proportion of people in management positions are Canadian?​

Apply the formula:

`\begin{gathered} P(C\text{ /M)=}(P(C\cap M))/(P(M)) \\ \\ \text{ P(C/M) = }(0.1674)/(0.264) \\ \\ \text{ P(C/M) = }0.634 \end{gathered}`

Therefore, the proportion of people in management positions are Canadian is 0.634.

ANSWER:

(a).

• P(C) = 54% ==> 0.54

,

• P(M/C) = 31% ==> 0.31

,

• P(M/C') = 21% ==> 0.21

• (b). 0.46

• (c). 0.1674

• (d). 0.264

• (e). 0.6366

• (f). 0.634