Question

A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to during the current year. In addition, it estimates that 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)

Answer 1

Answer:

a) $P(B|A) = \frac{P(B \cap A)}{P(A)}$

And if we solve for $P(B \cap A)$ we got:

$P(B \cap A) = P(B|A) *P(A) = 0.15*0.06= 0.009$

And that represent the probability of the employees will experience lost-time accidents in both years and in percentage is 0.9%

b) $P(A \cup B) = P(A) +P(B) -P(A \cap B)$

And replacing we got:

$P(A \cup B) = 0.06+ 0.05 -0.009=0.101$

And in % would be 10.1 %

Step-by-step explanation:

Assuming the following question:

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?

For this case we define the following events:

A= Represent the case last year

B= Represent the current year

We know the following probabilities from the problem:

$P(A)= 0.06 , P(B)= 0.05$

We also know the following conditional probability: $P(B|A) = 0.15$

Using the conditional probability definition from Bayes we know that:

$P(B|A) = \frac{P(B \cap A)}{P(A)}$

And if we solve for $P(B \cap A)$ we got:

$P(B \cap A) = P(B|A) *P(A) = 0.15*0.06= 0.009$

And that represent the probability of the employees will experience lost-time accidents in both years and in percentage is 0.9%

b. What percentage of the employees will suffer at least one lost-time accident over the  two-year period?

For this case we can find this probability with $P(A \cup B)$

And using the total probability rule defined by:

$P(A \cup B) = P(A) +P(B) -P(A \cap B)$

And replacing we got:

$P(A \cup B) = 0.06+ 0.05 -0.009=0.101$

And in % would be 10.1 %