STA 220 – Fall 2009
Practice Exercises – Chapter 15
Exercise 15.9. Health. The probabilities that an adult American man has high blood pressure and/or high cholesterol are shown in
the following table:
Cholesterol
High (C1)
OK (C2)
Total
a.
Blood Pressure
High (B1) OK (B2)
0.11
0.21
0.16
0.52
0.73
Total
0.32
0.68
1.00
What is the probability that a man has both conditions?
P(B1 and C1) =
b.
What is the probability that he has high blood pressure?
P(B1) =
c.
What is the probability that a man has high cholesterol if it is known that he has high blood pressure?
𝑃(𝐶1 |𝐵1 ) =
𝑃(𝐶1 𝑎𝑛𝑑 𝐵1 )
𝑃(𝐵1 )
𝑃(𝐶1 |𝐵1 ) =
d.
What is the probability that a man has high blood pressure if it is known that he has high cholesterol?
𝑃(𝐵1 |𝐶1 ) =
𝑃(𝐵1 𝑎𝑛𝑑 𝐶1 )
𝑃(𝐶1 )
𝑃(𝐵1 |𝐶1 ) =
2nd Exercise: The American Medical Association compiles information on US physicians in Physician Characteristics and Distribution in
the US. Following is a contingency table for US surgeons cross-classified by specialty and base of practice.
a.
Office (B1) Hospital (B2) Other (B3)
Total
General Surgery (S1)
24,128
12,225
1,658
38,011
Ob/gyn (S2)
24,150
6,734
1,140
32,024
Orthopedics (S3)
13,364
4,248
414
18,026
Ophthalmology (S4)
12,328
2,694
518
15,540
Total
73,970
25,901
3,730
103,601
For a randomly selected surgeon, describe each of the following events in words: B 1, S3, B1 and S3.
B1:
S3:
B1 and S3:
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STA 220 – Fall 2009
b.
Compute the probability of each event in part (a).
𝑃(𝐵1 ) =
𝑃(𝑆3 ) =
𝑃(𝐵1 𝑎𝑛𝑑 𝑆3 ) =
c.
Compute the P(B1 or S3) using the general addition rule and your answers in part (b).
P(B1 or S3) =
P(B1 or S3) =
P(B1 or S3) =
d.
Obtain P(S3|B1) directly from the table.
𝑃(𝑆3 |𝐵1 ) =
e.
Obtain P(B1|S3) directly from the table.
𝑃(𝐵1 |𝑆3 ) =
f.
Answer questions (d) and (e) using the conditional probability rule.
𝑃(𝑆3 |𝐵1 ) =
From part (b) above, we know:
P(B1 and S3) =
P(B1) =
P(S3) =
𝑃(𝑆3 |𝐵1 ) =
𝑃(𝑆3 𝑎𝑛𝑑 𝐵1 )
𝑃(𝐵1 )
𝑃(𝑆3 |𝐵1 ) =
𝑃(𝐵1 |𝑆3 ) =
𝑃(𝐵1 𝑎𝑛𝑑 𝑆3 )
𝑃(𝑆3 )
𝑃(𝐵1 |𝑆3 ) =
g.
Use the multiplication rule to find the P(B1 and S3).
From (f) we have:
or
P(B1 and S3) = P(S3|B1) * P(B1)
(i)
P(B1 and S3) = P(B1|S3) * P(S3)
(ii)
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STA 220 – Fall 2009
Using (i):
P(B1 and S3) = P(S3|B1) * P(B1)
P(B1 and S3 ) =
P(B1 and S3 ) =
Using (ii):
P(B1 and S3) = P(B1|S3) * P(S3)
P(B1 and S3 ) =
P(B1 and S3 ) =
Exercise 15.25 Unsafe food. Early in 2007 Consumer Reports published the results of an extensive investigation of broiler chickens
purchased from food stores in 23 states. Tests for bacteria in the meat showed that 81% of the chickens were contaminated with
campylobacter, 15% with salmonella, and 13% with both.
Let C be the event that campylobacter is present.
Let S be the event that salmonella is present.
a.
What’s the probability that a tested chicken was not contaminated with either kind of bacteria?
P((C or S)C) = 1 – P(C or S)
P(C or S) = P(C) + P(S) – P(C and S)
P(C or S) =
P((C or S)C) = 1- P(C or B)
P((C or S)C) =
b.
Are contamination with the two kinds of bacteria disjoint? Explain?
c.
Are contamination with the two kinds of bacteria independent? Explain.
Need to determine if
𝑃(𝐶|𝑆) =
𝑃(𝑆 𝑎𝑛𝑑 𝐶)
𝑃(𝑆)
𝑃(𝐶|𝑆) =
P(C) =
So contamination with the two kinds of bacteria are
salmonella, it is
to also have campylobacter.
3
. If a chicken is contaminated with