Question

{Exercise 4.57 (Algorithmic)} A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 9% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 7% 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)?

Answer 1

SOLUTION

Ans:

Given that

P(L)=0.09

P(C)=0.07

P(C/L)=0.15

a)P(both)=P(C and L)=P(C/L)*P(L)=0.15*0.09=0.0135

b)P(atleast one)=P(L or C)=P(L)+P(C)-P(L and C)

=0.09+0.07-0.0135

= 0.1465