A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 5% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 4% 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. a. What percentage of the employee will experience a lost-time accident in both years (to 1 decimal)?

The percentage of the employee will experience a lost-time accident in both years is 0.0%

Step-by-step explanation:

Let A denote events that employees suffered lost-time accidents during the last year

Let B denote events that employees suffered lost-time accidents during the current year

P(A) = 5% = 0.05

P(B) = 4% = 0.04

P(B | A) = 15% = 0.15

(a) P (A ∩ B) = P(B | A) × P(A)

= 0.15 × 0.05

= 0.0075

= 0.0 (1 decimal place)

The probability that an employee will experience a lost- time accident in both years is 0.0

-60.75

step-by-step explanation:

just add the numbers then put the negative sign on it