A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 8% 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 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?

a) percentage of the employees that will experience lost-time accidents in both years = 1.2%

b) percentage of the employees that will suffer at least one lost-time accident over the two-year period = 10.8%

Step-by-step explanation:

given

percentage of lost time accident last year

P(L) = 8% = 0.08 of the employees

percentage of lost time accident current year

P(C) = 4% = 0.04 of the employees

P(C/L) = 15% = 0.15

using the probability

P(L ∩ C) = P(C/L) × P(L)

= 0.08 × 0.15 = 0.012 = 1.2%

percentage of the employees will experience lost-time accidents in both years = 1.2%

b) Using the probability of the event

P(L ∪ C) = P(L) + P(C) - P(L ∩ C)

= 0.08 + 0.04 -0.012 = 0.108 = 10.8%

percentage of the employees will suffer at least one lost-time accident over the two-year period = 10.8%

41*67/2=1373.5

step-by-step explanation:

i just used a calculator

the answer is 0.000652422

but if you ment to ask what is 6131 divided by 4, it is 1532.75.

step-by-step explanation: