Question

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.
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?

Answer 1

Answer:

a

$\% E = 0.9 \%$

b

$\%E_1 = 10.1 \%$

Step-by-step explanation:

From the question we are told that

The probability that an employees suffered lost-time accidents last year is $P(e) = 0.06$

The probability that an employees suffered lost-time accident during the current year is

$P(c) = 0.05$

The probability that an employee will suffer lost time during the current year given that the employee suffered lost time last year is

$P(c | e) = 0.15$

Generally the probability that an employee will experience lost time in both year is mathematically represented as

$P(c \ n \ e) = P(e) * P(c \ |\ e)$

=> $P(c \ n \ e) = 0.06* 0.15$

=> $P(c \ n \ e) = 0.009$

Generally the percentage of employees that will experience lost time in both year is mathematically represented as

$\% E = P(e \ n \ c ) * 100$

=> $\% E = 0.009 * 100$

=> $\% E = 0.9 \%$

Generally the probability that an employee will experience at least one lost time accident over the two-year period is mathematically represented as

$P(e \ u \ c) = P(e) + P(c) - P(e \ n \ c)$

=> $P(e \ u \ c) = 0.06 + 0.05 - 0.009$

=> $P(e \ u \ c) = 0.101$

Generally the percentage of the employees who will suffer at least one lost-time accident over the two-year period is mathematically represented as

$\%E_1 = P(e \ u \ c) * 100$

=> $\%E_1 = 0.101* 100$

=> $\%E_1 = 10.1 \%$