161A_HW3_Solution
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Feb 20, 2024
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Homework 3 Solution
MATH 161A
Professor Gottlieb
45. The population of a particular country consists of three ethnic groups. Each individual belongs to
one of the four major blood groups. The accompanying joint probability table gives the proportions
of individuals in the various ethnic group-blood group combinations.
Suppose that an individual is randomly selected from the population, and define events by A =
{
type A selected
}
, B =
{
type B selected
}
, and C =
{
ethnic group 3 selected
}
(a) Calculate
P
(
A
),
P
(
C
),
P
(
A
∩
C
)
P
(
A
) = 0
.
106 + 0
.
141 + 0
.
200 =
0.447
.
P
(
C
) = 0
.
215 + 0
.
200 + 0
.
065 + 0
.
020 =
0.500
.
P
(
A
∩
C
) =
0.200
.
(b) Calculate both
P
(
A
|
C
) and
P
(
C
|
A
), and explain in context what each of these probabilities
represents.
P
(
A
|
C
) =
P
(
A
∩
C
)
P
(
C
)
=
0
.
200
0
.
500
=
0.4
.
Given that the selected individual is in ethnic group 3, the probability that he or she has
type A blood would be 0.4
P
(
C
|
A
) =
P
(
A
∩
C
)
P
(
A
)
=
0
.
200
0
.
447
=
0.447
.
Given that the selected individual has type A blood, the probability that he or she is from
ethnic group 3 would be 0.447
(c) If the selected individual does not have type B blood, what is the probability that he or she
is from ethnic group 1?
D =
{
ethnic group 1 selected
}
. Want to find
P
(
D
|
B
0
)
.
P
(
B
) = 0
.
008 + 0
.
018 + 0
.
065 = 0
.
091
.
P
(
B
0
) = 1
-
0
.
091 = 0
.
909
.
P
(
D
|
B
0
) =
P
(
D
∩
B
0
)
P
(
B
0
)
=
0
.
082 + 0
.
106 + 0
.
004
0
.
909
=
0.211
.
46. Suppose an individual is randomly selected from the population of all adult males living in the
United States. Let A be the event that the selected individual is over 6 ft in height, and let B be
the event that the selected individual is a professional basketball player. Which do you think is
larger,
P
(
A
|
B
) or
P
(
B
|
A
)? Why?
We know that
P
(
A
|
B
) =
P
(
A
∩
B
)
P
(
B
)
, and
P
(
B
|
A
) =
P
(
A
∩
B
)
P
(
A
)
. Since they have same numerator, we
only need to compare their denominators (
P
(
A
) and
P
(
B
)). The larger denominator leads to a
smaller fraction.
P(A) = the probability that the selected individual is over 6 ft in height.
P(B) = the probability that the selected individual is a professional basketball player.
Intuitively, there are more people who are taller than 6 ft than professional basketball players.
That is,
P
(
A
)
> P
(
B
). Therefore,
P
(
A
|
B
) =
P
(
A
∩
B
)
P
(
B
)
>
P
(
A
∩
B
)
P
(
A
)
=
P
(
B
|
A
)
.
49. The accompanying table gives information on the type of coffee selected by someone purchasing
a single cup at a particular airport kiosk. Consider randomly selecting such a coffee purchaser.
(a) What is the probability that the individual purchased a small cup? A cup of decaf coffee?
P
(small cup) = 14% + 20% =
34
%
.
P
(decaf coffee) = 20% + 10% + 10% =
40
%
(b) If we learn that the selected individual purchased a small cup, what now is the probability
that he/she chose decaf coffee, and how would you interpret this probability?
P
(decaf coffee
|
small cup) =
P
(decaf coffee
∩
small cup)
P
(small cup)
=
20%
34%
=
0
.
5882
The probability that a individual who purchased a small cup chose decaf coffee is 58.82%.
(c) If we learn that the selected individual purchased decaf, what now is the probability that
a small size was selected, and how does this compare to the corresponding unconditional
probability of (a)?
P
(small cup
|
decaf coffee) =
P
(small cup
∩
decaf coffee)
P
(decaf coffee)
=
20%
40%
=
0
.
50
The probability that a individual who purchased decaf coffee chose small cup is 50%.
60. Seventy percent of the light aircraft that disappear while in flight in a certain country are subse-
quently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas
90% of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disap-
peared.
(a) If it has an emergency locator, what is the probability that it will not be discovered?
Let
D
= the light aircraft is discovered.
L
= the light aircraft has an emergency locator.
We know that
P
(
D
) = 0
.
70
P
(
D
0
) = 1
-
0
.
7 = 0
.
3
P
(
L
|
D
) = 0
.
6
P
(
L
0
|
D
) = 1
-
0
.
6 = 0
.
4
P
(
L
0
|
D
0
) = 0
.
9
P
(
L
|
D
0
) = 1
-
0
.
9 = 0
.
1
P
(
D
0
|
L
) =
P
(
L
|
D
0
)
P
(
D
0
)
P
(
L
|
D
)
P
(
D
) +
P
(
L
|
D
0
)
P
(
D
0
)
=
0
.
1
×
0
.
3
0
.
6
×
0
.
7 + 0
.
1
×
0
.
3
=
0
.
067
(b) If it does not have an emergency locator, what is the probability that it will be discovered?
P
(
D
|
L
0
) =
P
(
L
0
|
D
)
P
(
D
)
P
(
L
0
|
D
)
P
(
D
) +
P
(
L
0
|
D
0
)
P
(
D
0
)
=
0
.
4
×
0
.
7
0
.
4
×
0
.
7 + 0
.
9
×
0
.
3
=
0
.
509
73. If A and B are independent events, show that
A
0
and B are also independent. [Hint: First establish
a relationship between
P
(
A
0
∩
B
)
, P
(
B
)
, andP
(
A
∩
B
)
.
]
Since A and B are independent,
P
(
A
∩
B
) =
P
(
A
)
P
(
B
) or
P
(
A
|
B
) =
P
(
A
)
.
P
(
A
0
|
B
) =
P
(
A
0
∩
B
)
P
(
B
)
=
P
(
B
)
-
P
(
A
∩
B
)
P
(
B
)
=
P
(
B
)
-
P
(
A
)
P
(
B
)
P
(
B
)
= 1
-
P
(
A
) =
P
(
A
0
)
Therefore, if A and B are independent events,
A
0
and B are also independent.
74. The proportions of blood phenotypes in the US population are as follows:
A
B
AB
O
0
.
40
0
.
11
0
.
04
0
.
45
Assuming that the phenotypes of two randomly selected individuals are independent of one an-
other, what is the probability that both phenotypes are O? What is the probability that the
phenotypes of two randomly selected individuals match?
P
(
O
∩
O
) = 0
.
45
×
0
.
45 =
0
.
2025
P
(two phenotypes matched) =
P
(
A
∩
A
) +
P
(
B
∩
B
) +
P
(
AB
∩
AB
) +
P
(
O
∩
O
)
= 0
.
40
2
+ 0
.
11
2
+ 0
.
04
2
+ 0
.
45
2
=
0
.
3762
77. An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is
defective. Suppose rivets are defective independently of one another, each with the same proba-
bility.
(a) If 20% of all seams need reworking, what is the probability that a rivet is defective?
Let p = probability that a rivet is defective.
(1
-
p
)
25
= 0
.
8
⇒
p
= 1
-
25
√
0
.
8 =
0
.
0089
(b) How small should the probability of a defective rivet be to ensure that only 10% of all seams
need reworking?
Likewise,
(1
-
p
)
25
= 0
.
9
⇒
p
= 1
-
25
√
0
.
9 =
0
.
0042
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79. Two pumps connected in parallel fail independently of one another on any given day. The proba-
bility that only the older pump will fail is 0.10, and the probability that only the newer pump will
fail is 0.05. What is the probability that the pumping system will fail on any given day (which
happens if both pumps fail)?
Let
A
= newer pump fails.
B
= older pump fails.
We want
P
(
A
∩
B
).
We know by independence that
P
(
A
∩
B
) =
P
(
A
)
P
(
B
).
So we must find P(A) and P(B). We
know that the probability that only the older one fails is 0.1. That is
P
(
B
∩
A
0
) = 0
.
1. Likewise,
P
(
A
∩
B
0
) = 0
.
05. By independence we get two equations with two unknowns
P
(
B
)
P
(
A
0
) = 0
.
1 and
P
(
A
)
P
(
B
0
) = 0
.
05
So we have
P
(
B
)[1
-
P
(
A
)] =
P
(
B
)
-
P
(
B
)
P
(
A
) = 0
.
1
(1)
P
(
A
)[1
-
P
(
B
)] =
P
(
A
)
-
P
(
A
)
P
(
B
) = 0
.
05
(2)
We use (1) - (2) to get
P
(
B
)
-
P
(
A
) = 0
.
05
⇒
P
(
B
) = 0
.
05 +
P
(
A
)
Now substitution in (2)
P
(
A
)
-
P
(
A
)[
P
(
A
) + 0
.
05] = 0
.
05
⇒
P
(
A
)
2
-
0
.
95
P
(
A
) + 0
.
05 = 0
⇒
P
(
A
) =
0
.
95
±
p
(0
.
95)
2
-
4(0
.
05)
2
⇒
P
(
A
) = 0
.
0559 or 0
.
8941
⇒
P
(
B
) = 0
.
1059 or 0
.
947
Therefore,
P
(
A
∩
B
) =
P
(
A
)
P
(
B
) = 0
.
0559
×
0
.
1059 =
0
.
00592
89. Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then
let loose for a period of time. Consider the two events
C
1
=
{
left eartag is lost
}
and
C
2
=
{
right
eartag is lost
}
. Let
π
=
P
(
C
1
) =
P
(
C
2
), and assume
C
1
and
C
2
are independent events. Derive
an expression (involving
π
) for the probability that exactly one tag is lost, given that at most one
is lost.
P
(exactly one tag is lost
|
at most one tag is lost)
=
P
(exactly one tag is lost
∩
at most one tag is lost)
P
(at most one tag is lost)
=
P
(exactly one tag is lost)
P
(at most one tag is lost)
=
P
((
C
1
∩
C
0
2
)
∪
(
C
0
1
∩
C
2
))
P
((
C
1
∩
C
0
2
)
∪
(
C
0
1
∩
C
2
)
∪
(
C
0
1
∩
C
0
2
))
=
P
(
C
1
∩
C
0
2
) +
P
(
C
0
1
∩
C
2
)
P
(
C
1
∩
C
0
2
) +
P
(
C
0
1
∩
C
2
) +
P
(
C
0
1
∩
C
0
2
)
=
2
π
(1
-
π
)
2
π
(1
-
π
) + (1
-
π
)
2
=
2
π
(1
-
π
)
1
-
π
2
=
2
π
1
-
π
Note:
E
1
= exactly one tag is lost =
{
C
1
∩
C
0
2
, C
0
1
∩
C
2
}
E
2
= at most one tag is lost =
{
C
1
∩
C
0
2
, C
0
1
∩
C
2
, C
0
1
∩
C
0
2
}
So
E
1
⊂
E
2
. Then
P
(
E
1
|
E
2
) =
P
(
E
1
∩
E
2
)
P
(
E
2
)
=
P
(
E
1
)
P
(
E
2
)
,
since
E
1
⊂
E
2