1. A new intern at a major corporation was asked to pull 2 chairs from an adjacent conference room for an important meeting. The intern knows that two of the six chairs in that adjacent conference room are broken, but he does not have time to check each chair. What is the probability that, if the intern selects two chairs at random, both will be broken?
2. 350 Individuals are surveyed regarding smoking habit and marital status.
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Smoker |
Non-smoker |
Total |
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Married |
54 |
146 |
200 |
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|
|
Divorced |
38 |
62 |
100 |
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|
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Never Married |
11 |
39 |
50 |
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Total |
103 |
247 |
350 |
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a. If one subject is randomly selected, what is the probability of selecting a smoker?
P(smoker) =
b. If one subject is randomly selected, what is the probability of selecting someone who is both a smoker and divorced?
P(smoker AND divorced) =
c. If one subject is randomly selected, what is the probability of selecting someone who is either a smoker or is divorced?
P(smoker OR divorced) =
d. From the set of smokers, what is the probability that a selected individual is divorced?
P(divorced | smoker) =
3. 11 locations in a defined oil field are believed likely to contain profitable drilling sites. If the oil company can afford to drill only 4 of these sites, how many different combinations of drill sites are possible?
4. A clinic's test for the flu results in 5% of patients testing positive. The medical supply company behind this test claims that 90% of patients who test positive for flu actually have the disease, and 15% of patients testing negative also have the flu. What is the probability that a patient being tested for flu actually has the disease.
Using the following probability tree, calculate the probability of a patient actually having the flu, P(flu) = ?
5. An outdoor event organizer wishes to estimate the likelihood of a profitable event based on his experience of high, average, and low attendance when it rains versus doesn't rain. The weather forecast calls for a 60% chance of rain. Based on the event organizer's experience:
P(high attendance | rain) = 0.15 P(average attendance | rain) = 0.25
P(high attendance | no rain) = 0.50 P(average attendance | no rain) = 0.35
a. Draw an appropriate probability tree for this problem.
b. Label every branch with its probability value. (You do not need to calculate any joint probabilities or solve the problem.)
Attachment:- Assign.rar