ACSC/MATH 347/447
Examples from Section 2.8 Lecture Notes
−
1
−
© John J Currano
02/05/2010
Slide 11: p. 61, #2.102.
In a certain population, 10% will contract disease I at some time in their
lifetime, 15% will contract disease II, and 3% will contract both.
Let
A
denote the event that a randomly chosen person contracts disease I and let
B
denote the event
that a randomly chosen person contracts disease II.
Then we are told
P
(
A
)
=
0.10
P
(
B
)
=
0.15
P
(
A
∩
B
)
=
0.03.
a.
Find the probability that a randomly chosen person will contract at least one of the diseases.
This asks for
P
(
A
∪
B
)
=
P
(
A
) +
P
(
B
)
−
P
(
A
∩
B
)
= 0.10 + 0.15 – 0.03 = 0.22.
b.
Given that a person has contracted at least one of the diseases, find the conditional probability that
s/he will contract both.
This asks for
()
.
)
(
)
(
)
(
)

(
B
A
P
B
A
B
A
P
B
A
B
A
P
∪
∪
∩
∩
=
∪
∩
Since
B
A
B
A
∪
⊆
∩
, it follows that
()
()
B
A
B
A
B
A
∩
=
∪
∩
∩
,
so we obtain
()
()
.
136
.
0
22
3
22
.
0
03
.
0
)
(
)
(
)
(
)
(
)

(
=
=
=
∪
∩
=
∪
∪
∩
∩
=
∪
∩
B
A
P
B
A
P
B
A
P
B
A
B
A
P
B
A
B
A
P
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Examples from Section 2.8 Lecture Notes
−
2
−
Slide 14: p. 60, #2.92.
We are told a lie detector concludes that a truthful person is lying with a
probability 0.05, and that lie detector tests are mutually independent. Three people are tested.
a.
What is the probability that the detector will conclude that all three are lying when all are telling
the truth?
Let
A
1
,
A
2
,
A
3
be the event that the detector concludes person 1, 2, 3, respectively, is lying. We are
told all three are truthful, so
P
(
A
i
) = 0.05 for
i
= 1, 2, 3.
We are asked for
P
(
A
1
∩
A
2
∩
A
3
)
=
P
(
A
1
)
P
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