109 6.5. Independent Events — Mathematics for Public and Occupational Health Professionals

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Independent Events

 

 

 

 

 

E\cap G = {HHT}, P(E\cap G) = 1/8

a. In order for E and F to be independent, we must have:
P(E\cap F) = P(E)\,P(F).
But 3/8\ne 1/2\cdot 1/2
Therefore, E and F are not independent.
b.F and G will be independent if:
P(F\cap G) = P(F)\,P(G)
Since 1/4\ne 1/2\cdot 1/4, F and G are not independent.
c. We look at P(E\cap G)=P(E)\,P(G):
1/8 = 1/2\cdot 1/4
Therefore, E and G are independent events.

 

 

 

 

 

Practice questions

1. In a survey of 100 people, 40 were casual drinkers, and 60 did not drink. Of the ones who drank, 10 had minor headaches. Of the non-drinkers, 5 had minor headaches. Are the events “drinkers” and “had headaches” independent?

2. Suppose that 80% of the people wear seat belts, and 5% of the people quit smoking last year. If 4% of the people who wear seat belts quit smoking, are the events wearing a seat belt and quitting smoking independent?

3. If P(E)=0.9, P(F\,|\, E)=0.36, and E and F are independent, find P(F).

4. John’s probability of passing Data Management is 40%, and Linda’s probability of passing the same course is 70%. If the two events are independent, find the following probabilities:

5. The table below shows the distribution of employees in a company that reported a previous workplace injury based on their years of working experience at the company.

Less than 10 years of experience (L) 10 or more years of experience (E) Total
Did not report a workplace injury (N) 300 100 400
Reported a workplace injury (Y) 150 50 200
450 150 600
Use this table to determine the following probabilities:

6. Given P(A)=0.3 , P(A\cup B) = 0.65, if A and B are independent, find P(B).

 

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