a company studied the number of lost - time accidents occurring at its brownsville, texas, plant. historical…

a company studied the number of lost - time accidents occurring at its brownsville, texas, plant. historical records show that 6% of the employees suffered lost - time accidents last year. management believes that a special safety program will reduce such accidents to 5% 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.\na. what percentage of the employees will experience lost - time accidents in both years (to 2 decimals)?\nb. what percentage of the employees will suffer at least one lost - time accident over the two - year period (to 2 decimals)?

a company studied the number of lost - time accidents occurring at its brownsville, texas, plant. historical records show that 6% of the employees suffered lost - time accidents last year. management believes that a special safety program will reduce such accidents to 5% 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.\na. what percentage of the employees will experience lost - time accidents in both years (to 2 decimals)?\nb. what percentage of the employees will suffer at least one lost - time accident over the two - year period (to 2 decimals)?

Answer

Explanation:

Step1: Calculate percentage of employees with accidents in both years

Let $A$ be the event of having an accident last year and $B$ be the event of having an accident this year. We know $P(A)=0.06$ and the probability of having an accident this year given having an accident last year $P(B|A) = 0.15$. By the formula for conditional - probability $P(A\cap B)=P(B|A)P(A)$. $P(A\cap B)=0.15\times0.06 = 0.0090 = 0.90%$

Step2: Calculate percentage of employees with at least one accident

We know $P(A) = 0.06$, $P(B)=0.05$. Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. $P(A\cup B)=0.06 + 0.05-0.009$ $P(A\cup B)=0.1010=10.10%$

Answer:

a. $0.90%$ b. $10.10%$