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. Round your answers to three decimal places.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) $P(Current year and last year)=0.15x0.09=0.0135=1.35\%$

b) $P(Last year U Current year)=0.09+0.08-0.0135=0.1565=15.65\%$

Step-by-step explanation:

Conditional probability is the probability of one event occurring with some relationship to one or more other events.

The total probability rule (or the Law of Total Probability) breaks up probability calculations into distinct parts. P(AUB)=P(A)+P(B)-P(A and B)

The general multiplication rule formula is: P(A ∩ B) = P(A) P(B|A)=P(B)P(A|B)

We have the following probabilities:

$P(Lastyear)=0.09=9\%$

$P(Currentyear)=0.08=8\%$

$P(Currentyear | Lastyear)=0.15=15\%$

Part a

We can use this multiplication rule:

$P(A and B)=P(B|A)(A)=P(A|B)P(B)$

For this part we want this probability:

$P(Current year and last year)=P(Current year|Last year)P(Last year)$

And if we replace we got:

$P(Current year and last year)=0.15x0.09=0.0135=1.35\%$

Part b

For this part we want the probability of one event in last year OR one event on the current year, so we can use the probability addition of events:

$P(AUB)=P(A)+P(B)-P(A and B)$

And for our case we have:

$P(Last year U Current year) = P(Last year)+P(Current year)-P(Last year and Current year)$

$P(Last year U Current year)=0.09+0.08-0.0135=0.1565=15.65\%$