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 9% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 8% 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

Given,

P(Suffered last year) = 0.09

P(Suffered in current year) = 0.08

P(Suffered current year | Suffered last year) = 0.15

Hence,

a) P(Suffered current year and last year)

= P(Suffered current year | Suffered last year) * P(Suffered last year)

= 0.09*0.15

= 0.014

b) P(Suffered current year or last year)

= 0.09 + 0.15 - 0.014

= 0.226

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