A company studied the number of lost-time accidents occurring at its Brownsville

Question

A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 10% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 6% during the current year. In addition, it estimates that 15% of employees who had lost-time accidents last year will experience a lost-time accident during the current year.

What is the probability an employee will experience a lost-time accident in both years (to 3 decimals)?

What is the probability an employee will experience a lost-time accident over the two-year period (to 3 decimals)?

Explanation / Answer

Here we are given that: Historical records show that 10% of the employees suffered lost-time accidents last year. Therefore, P( lost-time accidents previous year ) = 0.1

Also, we are given that: Management believes that a special safety program will reduce such accidents to 6% during the current year. Therefore, P( lost-time accidents current year ) = 0.06

Also, we are given that P( lost-time accidents current year | lost-time accidents previous year ) = 0.15

Therefore, if P( lost-time accidents current year | no lost-time accidents previous year ) = K. Then we get:

0.15*0.1 + (1 - 0.1)*K = 0.06

0.015 + 0.9K = 0.06

K = 0.05

This means that 5% of the employees who did not lost-time accidents last year will experience a lost-time accident during the current year.

a) Now the probability an employee will experience a lost-time accident in both years is computed as:

= P( lost-time accidents previous year )*P( lost-time accidents current year | lost-time accidents previous year )

= 0.1*0.15

= 0.015

Therefore 0.015 is the required probability here.

b) Now the probability an employee will experience a lost-time accident over the two-year period

= 1- Probability that the person will not experience a lost-time accident over the two-year period

= 1 - 0.9*(1 - 0.05)

= 0.145

Therefore 0.145 is the required probability here.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.