7. An industrial plant is conducting a study to determine how quickly injured workers are back on the job following injury. Records show that 10% of all injured workers are admitted to the hospital for treatment and 15% are back on the job the next day. In addition, studies show that 2% are both admitted for hospital treatment and back on the job the next day.

a. If a worker is injured, what is the probability that the worker will either be admitted to a hospital or be back on the job the next day or both?

b. What is the probability that an admitted worker not to be back on job the next day?

Question

Maci Mcpherson

Mathematics

25 March, 02:38

7. An industrial plant is conducting a study to determine how quickly injured workers are back on the job following injury. Records show that 10% of all injured workers are admitted to the hospital for treatment and 15% are back on the job the next day. In addition, studies show that 2% are both admitted for hospital treatment and back on the job the next day.

a. If a worker is injured, what is the probability that the worker will either be admitted to a hospital or be back on the job the next day or both?

b. What is the probability that an admitted worker not to be back on job the next day?

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Answers (1)

Know the Answer?

Izabelle Evans · Answer 1

P (A ∪ B) = 0.23

Step-by-step explanation:

When 2 events are non mutually exclusive you have the following formula for one of them or both:

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

for this problem using A for injured workers admitted to the hospital and B for the injured workers back on the next day. you have:

P (A) = 0.1

P (B) = 0.15

P (A ∩ B) = 0.02

The probability that a injured worker is admitted to the hospital and is back the next day represents the intersection of both events

P (A ∪ B) = 0.1 + 0.15 - 0.02 = 0.23