Question

A company wanted to determine what percentage of its employees stayed for at least 1, 2, and 3 years. The data they compiled is given below and covers 200 male employees and 300 female employees. Male 1 year = 0.67, 2 year = 0.45, 3 year = 0.20. Female 1 year = 0.73, 2 years = 0.64, 3 years = 0.39.

What is the probability that a male employee stayed at least 3 years given that they stayed 1 year? Round to the nearest hundredth.

Answer 1

0.298

Explanation:

A company wanted to determine what percentage of its employees stayed for at least 1, 2, and 3 years.

The data they compiled is given below and covers 200 male employees and 300 female employees.

Male

1 year = 0.67

2 year = 0.45

3 year = 0.20

Female

1 year = 0.73

2 years = 0.64

3 years = 0.39.

Let A be the event that a male employee stayed at least 3 years

So,P(A) = 0.20

Let B be the event that they stayed 1 year

So, P(B) = 0.67

Now we are supposed to find the probability that a male employee stayed at least 3 years given that they stayed 1 year

So, the probability that a male employee stayed at least 3 years given that they stayed 1 year =
`(0.20)/(0.67)`

=
`0.298`

Hence the probability that a male employee stayed at least 3 years given that they stayed 1 year is 0.298.

Answer 2

Answer: 0.58209 (approx)

Explanation:

Let A is the event of a male employee stayed at least 3 years.

And, B is the event of a female employee stayed at least 1 year.

Then, According to the question,

P(A) = 0.20

P(B) = 0.67

Thus,
`P(A\cap B) = 0.20` ( because the one who having at least 3 years experience already having at least 1 year of experience.)

⇒ The probability that a male employee stayed at least 3 years given that they stayed 1 year =
`(P(A\cap B))/(P(B))`

=
`(0.39)/(0.67)`

= 0.58208955223 ≈ 0.58209