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alexdok [17]
1 year ago
6

Although the rules of probability are just basic facts about percents or proportions, we need to be able to use the language of

events and their probabilities. Choose an American adult aged 20 years and over at random. Define two events:
A= the person chosen is obese
B= the person chosen is overweight, but not obese
According to the National Center for Health Statistics, P(A) = 0.38 and P(B) =0.33.

a. Select the correct description stating what the event A or B is

a. A or B is the event that the person chosen is not obese or not overweight.
b. A or B is the event that the person chosen is overweight or obese or both.
c. A or B is the event that the person chosen is overweight and obese.
d. A or B is the event that the person is overweight or obese

What is P(A or B)?

a. P(A or B)= 0.02
b. P(A or B)= 0.34
c. P(A or B)= 0.71
d. P(A or B)= 0.55

c. If C is the event that the person chosen has normal weight or less, what is P(C)?

a. P(C)= 0.29
b. P(C)= 0.66
c. P(C) = 0.68
d. P(C)= 0.45
Mathematics
1 answer:
netineya [11]1 year ago
6 0

Answer:

a) Option D is correct for this question.

That is, A or B is the event that the person is overweight or obese.

But for other questions whose sets aren't disjoint, event A or B usually means all the elements that are in either A or B or both sets.

b) Option C.

P(A or B) = 0.71

c) Option A

P(C) = 1 - P(A or B) = 0.29

Step-by-step explanation:

b) Event B specifically rules out obesity, meaning, set A and set B have no elements in common.

In a normal probability question, event A or B usually means all the elements that are in either A or B and elements that are in the two sets.

But for this question, since, it has been made clear that there are no common elements in the two sets, event A or B is event that the person is overweight or obese. Option D.

b) For disjoint sets, P(A or B) = P(A) + P(B) = 0.38 + 0.33 = 0.71. Option C.

c) P(C) is the set of all elements that are not in either A or B.

P(C) = 1 - P(A or B) = 1 - 0.71 = 0.29. Option A.

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