Answer:
P(W∩M)= 20/39= 0.51
Explanation:
Hello!
You have a total of 13 employees in the office, 5 are men and 8 are women.
Be the events:
M: The employee is a male
W: the employee is female
If you were to choose only one employee, the probability of the employee being a man will be P(M)= 5/13 and the probability of the employee being a woman will be P(W)= 8/13.
Now you are in a situation where you have to choose two employees at random, without replacement, and need to calculate the probability of choosing one woman and one man. There is no order specified in the question so when choosing the two employees there are two possible scenarios:
"The first employee is a woman and the second employee is a man"
or
"A first employee is a man and the second employee is a woman"
Symbolically:
P(W∩M)= P((W₁∩M₂)∪(M₁∩W₂))= (P(W₁∩M₂))+P(P(M₁∩W₂))
The subindex marks the order the events appeared.
Remember "and" in the sentence marks the intersection between two events "∩" and "or" mark the union between two events "∪"
First I'll solve both intersections:
P(W₁∩M₂)= P(W₁) * P(M₂)= (8/13) * (5/12)= 10/39
As I said before, this exercise is without replacement, so the first person chosen will be out of the 13 employees of the office, but when choosing the second one the total number of employees will be reduced by one (you count out the already chosen employee)
P(M₁∩W₂)= P(M₁) * P(W₂)= (5/13)*(8/12)= 10/39
In this item is the same, for the second draft, the total of employees is reduced by one since one male has already been chosen.
⇒ Mathematically both intersections are the same so you can express their union as:
(P(W₁∩M₂))+P(P(M₁∩W₂))= 2* (P(W₁∩M₂))= 2*10/39= 20/39= 0.51
I hope you have a SUPER day!