Question

Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees are randomly​ selected, what is the probability that they were both hired on a Monday​?b. If two different employees are randomly​ selected, what is the probability that they were both hired on the same day of the week​?c. What is the probability that 7 people in the same department were all hired on the same day of the week​?d. Is such an event​ unlikely?

Answer 1

`P(A) = (1)/(9)`

`P(B) = (1)/(3)`

`P(C) = (1)/(729)`

Explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

`P(M) = (1)/(3)`.

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

`P(A) =P(M\ and\ M)\\\\P(A)=( (1)/(3))((1)/(3))\\\\P(A) = (1)/(9)`

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is
`P(A) = (1)/(9)`. We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

`P(B) = (1)/(9) + (1)/(9) + (1)/(9)\\\\P(B) = (1)/(3)`.

C. If the probability that two people have been hired on a specific day of the week is
`3((1)/(3)) ^ 2`, then the probability that 7 people have been hired on the same day is:

`P(C) = ((1)/(3)) ^ 7 + ((1)/(3)) ^ 7 + ((1)/(3)) ^ 7\\\\P(C) = (1)/(729)`

D. The probability is
`(1)/(729)`. This number is quite close to zero. Therefore it is an unlikely bastate event.