Question

A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 20% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 10% during the current year. In addition, it estimates that 25% of employees who had lost-time accidents last year will experience a lost-time accident during the current year.
Required:
a. What percentage of the employees will experience lost-time accidents in both years?
b. What percentage of the employees will suffer at least one lost-time accident over the two-year period?

Answer 1

Answer:

a. 5% of the employees will experience lost-time accidents in both years

b. 24% of the employees will suffer at least one lost-time accident over the two-year period

Step-by-step explanation:

a. What percentage of the employees will experience lost-time accidents in both years?

20% last year, of those who suffered last year, 25% during this year. So

$p = 0.2*0.25 = 0.05$

5% of the employees will experience lost-time accidents in both years.

b. What percentage of the employees will suffer at least one lost-time accident over the two-year period?

5% during the two years.

10% during the current year. 25% of employees who had lost-time accidents last year will experience a lost-time accident during the current year.

So the 10% is composed of 5% during both years(25% of 20%) and 5% of the 80% who did not suffer during the first year.

First year yes, not on the second.

75% of 20%. So, total:

$0.05 + 0.05*0.8 + 0.75*0.2 = 0.24$

24% of the employees will suffer at least one lost-time accident over the two-year period