224fa21exam1sol
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Rumson Fair Haven Reg H *
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Course
224
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
4
Uploaded by CoachRiverTiger30
Math 224
Exam 1
Fall 2021
Thurs Sept 23
This is a closed book test.
No electronic devices are allowed.
There are 5 pages and 5
problems, each worth 6 points, for a total of 30 points.
Problem 1.
Let
S
be the sample space of an experiment and consider two events
A, B
⊆
S
with the following properties:
P
(
A
) = 3
/
7
,
P
(
B
) = 4
/
7
and
P
(
A
∩
B
) = 1
/
7
.
(a) Compute the probability
P
(
A
∪
B
) that
A
or
B
happens.
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
)
-
P
(
A
∩
B
) =
3
7
+
4
7
-
1
7
=
6
7
.
(b) Compute the probability
P
(
A
∩
B
0
) that
A
happens but
B
does not happen.
P
(
A
∩
B
) +
P
(
A
∩
B
0
) =
P
(
A
)
P
(
A
∩
B
0
) =
P
(
A
)
-
P
(
A
∩
B
) =
3
7
-
1
7
=
2
7
.
(c) Compute the probability
P
(
A
|
B
) that
A
happens, assuming that
B
happens.
P
(
A
|
B
) =
P
(
A
∩
B
)
P
(
B
)
=
1
/
7
4
/
7
=
1
4
.
Problem 2.
Consider a biased coin with
P
(
H
) =
p
and
P
(
T
) =
q
. Suppose that the coin
is flipped 5 times and let
X
be the number of heads.
(a) Compute the probability
P
(
X
= 3) in terms of
p
and
q
..
P
(
X
= 3) =
5
3
p
3
q
2
=
5!
3!2!
·
p
3
q
2
= 10
p
3
q
2
(b) Compute the probability
P
(
X
≥
1) in terms of
p
and
q
.
P
(
X
≥
1) = 1
-
P
(
X
= 0) = 1
-
5
0
p
0
q
5
= 1
-
q
5
(c) Use your formula from part (b) to compute the probability of getting at least one
“six” in five rolls of a
fair
six-sided die.
Let
H
= “we get six” and
T
= “we don’t get six”, so that
p
=
P
(
H
) = 1
/
6 and
q
=
P
(
T
) = 5
/
6. Then the formula from part (b) gives
P
(
X
≥
1) = 1
-
q
5
= 1
-
5
6
5
(or 59
.
8%)
Problem 3.
An urn contains 2 red and 5 green balls. Suppose that 3 balls are drawn
without replacement.
(a) What is the size of the sample space?
#
S
=
7
3
=
7!
3!4!
=
7
·
6
·
5
3
·
2
·
1
= 7
·
5 = 35
(b) What is the probability of getting one red and two green balls?
Since the outcomes are equally likely, the probability is #
E/
#
S
where #
E
is the
number of ways to choose 1 red and 2 green balls:
P
(1 red and 2 green) =
#
E
#
S
=
(
2
1
)(
5
2
)
(
7
3
)
=
2
·
10
35
=
4
7
.
(c) What is the probability of getting no green balls?
If you get no green balls then you must get 3 red balls, which is impossible because
there are only 2 red balls in the urn. If we write
(
2
3
)
= 0 then
P
(no green balls) =
(
2
3
)(
5
0
)
(
7
3
)
=
0
(
5
0
)
(
7
3
)
= 0
.
Problem 4.
A
fair
four-sided die has sides labeled
a, b, c, d
. Roll the die 3 times and let
A, B, C, D
be the number of times that sides
a, b, c, d
show up.
(a) What is the size of the sample space?
#
S
=
4
|{z}
1st roll
×
4
|{z}
2nd roll
×
4
|{z}
3rd roll
= 4
3
= 64
(b) What is the number of words of length 3 that can be made by using each of the
letters
a, b, c
exactly once?
Use your answer to compute
P
(
A
= 1
, B
= 1
, C
=
1
, D
= 0).
The number of words of length 3 using the letters
a, b, c
(
d
does not appear) is
3!
1!1!1!0!
= 3! = 6
,
hence probability of getting
a, b, c
in some order is
P
(
A
= 1
, B
= 1
, C
= 1
, D
= 0) =
6
64
=
3
32
.
(c) What is the number of words of length 3 that can be made from two copies of “
a
”
and one copy of “
b
”? Use your answer to compute
P
(
A
= 2
, B
= 1
, C
= 0
, D
= 0).
The number of words of length 3 using the letters
a, a, b
(
c
and
d
do not appear) is
3!
2!1!0!0!
= 3
hence probability of getting
a, a, b
in some order is
P
(
A
= 2
, B
= 1
, C
= 0
, D
= 0) =
3
64
.
Remark: It is also possible to solve these problems with the multinomial formula. For all
i
+
j
+
k
+
‘
= 3 we have
P
(
A
=
i, B
=
j, C
=
k, D
=
‘
) =
3!
i
!
j
!
k
!
‘
!
1
4
i
1
4
j
1
4
k
1
4
‘
=
3!
i
!
j
!
k
!
‘
!
1
4
i
+
j
+
k
+
‘
=
3!
i
!
j
!
k
!
‘
!
1
4
3
=
3!
i
!
j
!
k
!
‘
!
4
3
.
Problem 5.
A diagnostic test is administered to a random person to determine if they
have a certain disease. Consider the events
D
= “the person has the disease” and
T
=“the
test returns positive”. Suppose that
P
(
T
|
D
0
) = 1%
,
P
(
T
0
|
D
) = 2%
and
P
(
D
) = 30%
.
(a) Compute the probabilities
P
(
T
0
|
D
0
) and
P
(
T
|
D
).
For all events
A, B
we have
P
(
A
|
B
) +
P
(
A
0
|
B
) = 1. In our case:
P
(
T
0
|
D
0
) = 1
-
P
(
T
|
D
0
) = 99%
,
P
(
T
|
D
) = 1
-
P
(
T
0
|
D
) = 98%
.
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(b) Compute the probability
P
(
T
). [Hint: Law of Total Probability.]
The law of total probability and the definition of conditional probability give
P
(
T
) =
P
(
D
∩
T
) +
P
(
D
0
∩
T
)
=
P
(
D
)
P
(
T
|
D
) +
P
(
D
0
)
P
(
T
|
D
0
)
= (0
.
3)(
.
98) + (0
.
7)(0
.
01)
= (3
/
10)(98
/
100) + (7
/
10)(1
/
100)
= (294
/
1000) + (7
/
1000)
= 301
/
1000
= 30
.
1%
(c) Compute the probability
P
(
D
|
T
). [Hint: Bayes’ Theorem.]
In part (b) we saw that
P
(
D
∩
T
) = 294
/
1000 and
P
(
T
) = 301
/
1000. Then the
definition of conditional probability gives
P
(
D
|
T
) =
P
(
D
∩
T
)
P
(
T
)
=
294
/
1000
301
/
1000
=
294
301
(or 97
.
7%)
.
Remark: The hint was to use Bayes’ Theorem but it wasn’t really necessary because
I guided you through all the steps. To use Bayes’ Theorem explicitly, write
P
(
T
)
P
(
D
|
T
) =
P
(
D
)
P
(
T
|
D
)
P
(
D
|
T
) =
P
(
D
)
P
(
T
|
D
)
P
(
T
)
,
and then continue from there.