Answer :
Answer:
a) [tex] P(B|A) = \frac{P(B \cap A)}{P(A)}[/tex]
And if we solve for [tex] P(B \cap A) [/tex] we got:
[tex] P(B \cap A) = P(B|A) *P(A) = 0.15*0.06= 0.009[/tex]
And that represent the probability of the employees will experience lost-time accidents in both years and in percentage is 0.9%
b) [tex] P(A \cup B) = P(A) +P(B) -P(A \cap B) [/tex]
And replacing we got:
[tex] P(A \cup B) = 0.06+ 0.05 -0.009=0.101[/tex]
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:
[tex] P(A)= 0.06 , P(B)= 0.05[/tex]
We also know the following conditional probability: [tex] P(B|A) = 0.15[/tex]
Using the conditional probability definition from Bayes we know that:
[tex] P(B|A) = \frac{P(B \cap A)}{P(A)}[/tex]
And if we solve for [tex] P(B \cap A) [/tex] we got:
[tex] P(B \cap A) = P(B|A) *P(A) = 0.15*0.06= 0.009[/tex]
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 [tex] P(A \cup B)[/tex]
And using the total probability rule defined by:
[tex] P(A \cup B) = P(A) +P(B) -P(A \cap B) [/tex]
And replacing we got:
[tex] P(A \cup B) = 0.06+ 0.05 -0.009=0.101[/tex]
And in % would be 10.1 %