Will Landau
September 4, 2011
STAT 231 Problem Set 2 Solutions
Exercise 2.1 (Devore 2.1).
a. If team 1 places first, team 3 places second, team 2 places third, and
team 4 places last, then we denote the outcome by 1324. Since the
competition is a tournament and teams 1 and 2 play first, either:
i. Team 1 beats team 2, so team 1 goes to the championship game
and team 2 does not. 1 is among the left two digits and 2 is among
the right two digits of the outcome, or:
ii. Team 2 beats team 1, so team 2 goes to the championship game
and team 1 does not. 2 is among the left two digits and 1 is among
the right two digits of the outcome
We place a similar constraint on numbers 3 and 4 in
the outcome. These constraints leave us with the following outcomes in S:
{1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 4123, 4132, 3214, 3241, 4213, 4231}
b. A is the subset of S with outcomes having a left-most digit of 1. Hence,
A = {1324, 1342, 1423, 1432}.
c. B is the subset of S with a 2 as one of the left-most two digits. Hence,
B = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}.
d.
i. A ∪ B = outcomes in either A or B or both =
{1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}.
ii. A ∩ B = ∅. This makes sense because if team 1 won the tournament,
then team 2 wouldn’t have made it to the championship game.
iii. A0 = S − A =
{2314, 2341, 2413, 2431, 3124, 3142, 4123, 4132, 3214, 3241, 4213, 4231}
Exercise 2.2 (Devore 2.3).
a. Be careful about the notation you pick for your outcomes. It’s important
to translate the problem into symbols in a way that works. I will denote
a success in component 1, a failure in component 2, and a failure in
component 3 by the symbol, SFF. Then, A = {SSF, SF S, F SS}
b. B = {SSS, SSF, SF S, F SS}
c. C = {SSS, SSF, SF S}, since component 1 must function and at least
one of components 2 and 3 must function.
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d.
C 0 = {SF F, F SS, F SF, F F S, F F F }
A ∪ C = {SSS, SSF, SF S, F SS}
A ∩ C = {SSF, SF S}
B ∪ C = {SSS, SSF, SF S, F SS}
B ∩ C = {SSS, SSF, SF S}
Exercise 2.3 (Devore 2.4).
a. Pay attention to the pattern I used to construct this table.
Outcome
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Mort. 1
F
F
F
F
F
F
F
F
V
V
V
V
V
V
V
V
Mort. 2
F
F
F
F
V
V
V
V
F
F
F
F
V
V
V
V
Mort 3.
F
F
V
V
F
F
V
V
F
F
V
V
F
F
V
V
Mort. 4
F
V
F
V
F
V
F
V
F
V
F
V
F
V
F
V
b. Outcome numbers 2, 3, 5, 9
c. Outcome numbers 1, 16
d. Outcome numbers 1, 2, 3, 5, 9
e. The union is the event that either:
i. all of the mortgages are variable, or:
ii. at most at most one of them is variable
which leaves us with outcomes 1,2,3,5,9,16.
The intersection is the event that all of the mortgages are fixed: outcome
1.
2
f. The union is the event that either:
i. exactly three mortgages are fixed, or:
ii. all four of them are the same
which leaves us with outcomes 1,2,3,5,9,16.
The intersection is the event that exactly three mortgages are fixed AND
that all four mortgages are the same, which is impossible; if all four
mortgages are the same, then we have either FFFF or VVVV, neither of
which have exactly three fixed mortgages. Hence, the intersection is ∅.
Chapter 2: Probability
8.
Exercise 2.4 (Devore 2.8).
a.
A1 ! A2 ! A3
b.
A1 " A2 " A3
c.
A 1 " A 2# " A 3#
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Chapter 2: Probability
d.
(A1 ! A2"! A3 ") # (A1" ! A2 ! A3 ") # (A1 "! A2 "! A3 )
e.
A1 # (A2 ! A3)
Exercise 2.5 (Devore 2.11).
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a. 0.07: the value under ”Balanced”
b. Since we’re dealing with disjoint events, add the probabilities: .15 + .10
+ .05 = .30
c. Let A be the event that the customer owns stocks. Then, A = {high-risk
stock, moderate-risk stock }. Since the two outcomes in A are disjoint,
we add their probabilities to find that P (A) = .18 + .25 = .43. Now, A0
is the event that a customer does not own stocks. By complementation,
P (A0 ) = 1 − P (A) = 1 − .43 = .57.
Exercise 2.6 (Devore 2.12).
a. P (A ∪ B) = .50 + .40 − .25 = .65
b. P ((A ∪ B)0 ) = 1 − P (A ∪ B) = 1 − .65 = .35
c. A ∩ B 0 ; P (A ∩ B 0 ) = P (A) − P (A ∩ B) = .50 − .25 = .25
Exercise 2.7 (Devore 2.14).
Let C be the event that a randomly selected adult consumes coffee, and S be
the event that she regularly consumes soda.
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a. P (C ∪ S) = P (C) + P (S) − P (C ∩ S), so
P (C ∩ S) = P (C) + P (S) − P (C ∪ S) = .55 + .45 − .7 = .3
b. P ((C ∪ S)0 ) = 1 − P (C ∪ S) = 1 − .7 = .3
Exercise 2.8 (Devore 2.15).
a. If E is the event that at most one customer buys an electric dryer, then
E’ is the even that at least two customers buy electric dryers. By
complementation, P (E 0 ) = 1 − P (E) = 1 − .428 = .572 .
b. Let A be the event that all five customers buy gas dryers and let B be
the event that all five customers buy electric dryers. Then, A ∪ B is the
event that all five customers purchase the same kind of dryer. Hence,
(A ∪ B)0 is the event that all five customers do not purchase the same
kind of dryer: in other words, the event that at least one of each kind of
dryer is purchased. We want to find P ((A ∪ B)0 ). Since A and B are
disjoint events,
P ((A ∪ B)0 ) = 1 − P (A ∪ B) = 1 − P (A) − P (B) = 1 − .116 − .005 = .879 .
Exercise 2.9 (Devore 2.19).
a. Let A be the event that inspector A finds a randomly selected joint to be
defective. Let B be the analogous event for inspector B. A ∪ B is the
even that both inspectors find the joint defective, so (A ∪ B)0 is the
event that neither finds the joint defective. We calculate
1159
8841
P ((A ∪ B)0 ) = 1 − P (A ∪ B) = 1 − 10,000
= 10,000
= 0.8841
b. We want P (A0 ∩ B). Now, B = (A0 ∩ B) ∪ (A ∩ B). Since (A0 ∩ B) and
(A ∩ B) are disjoint, P (B) = P (A0 ∩ B) + P (A ∩ B), so
P (A0 ∩ B) = P (B) − P (A ∩ B). Since
P (A ∪ B) = P (A) + P (B) − P (A ∩ B), we get
P (A0 ∩ B) = P (B) − [P (A) + P (B) − P (A ∪ B)] = P (A ∪ B) − P (A).
Now, to plug in numbers:
P (A0 ∩ B) = P (A ∪ B) − P (A) =
1159
10,000
−
724
10,000
=
435
10,000
= 0.0435
Exercise 2.10 (Devore 2.20).
a. Let S1, S2 and S3 represent the day, swing, and night shifts, respectively.
Let C1 and C2 represent unsafe conditions and unrelated to conditions,
respectively. Then, the simple events are
{S1, C1}, {S2, C1}, {S3, C1}, {S1, C2}, {S2, C2}, {S3, C2}.
b. Using the table, we calculate P ({{S1, C1}, {S2, C1}, {S3, C1}}) =
P ({S1, C1}) + P ({S2, C1}) + P ({S3, C1}) = .1 + .08 + .05 = .23
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c. In the calculation, we consider all simple events without the day shift
symbol, S1.
P ({{S2, C1}, {S3, C1}, {S2, C2}, {S3, C2}}) = P ({S2, C1}) +
P ({S3, C1}) + P ({S2, C2}) + P ({S3, C2}) = .08 + .05 + .20 + .22 = .55
Exercise 2.11 (Devore 2.26).
a. P (A01 ) = 1 − P (A1 ) = 1 − .12 = .88
b. P (A1 ∩ A2 ) = P (A1 ) + P (A2 ) − P (A1 ∪ A2 ) = .12 + .07 − .13 = .06 (see
the proposition on page 60.)
c. P (A1 ∩ A2 ∩ A3 ?) = P (A1 ∩ A2 ) − P (A1 ∩ A2 ∩ A3 ) = .06 - .01 = .05 (if
this doesn’t make sense, draw the Venn Diagram.)
d. P (at most two errors) = 1P (all three types) =
1 − P (A1 ∩ A2 ∩ A3 ) = 1 − .01 = .99
Exercise 2.12 (Devore 2.47).
a. P (B | A) =
b. P (B 0 | A) =
c. P (A | B) =
d. P (A0 | B) =
P (A∩B)
P (B)
=
P (A∩B 0 )
P (A)
P (A∩B)
P (B)
e. P (A | A ∪ B) =
= .5
.25
.5
=
=
P (A0 ∩B)
P (B)
.25
.50
.25
.4
= .6125
.15
.4
=
= .5
= .3875
P (A∩(A∩B))
P (A∪B)
=
P (A)
P (A∪B)
=
.5
.65
= .7692
Exercise 2.13 (Devore 2.48).
a. P (A2 | A1 ) =
P (A1 ∩A2 )
P (A1 )
=
b. P (A1 ∩ A2 ∩ A3 | A1 ) =
.06
.12
.01
.12
= .5
= .0833
c. P ((exactly one) | (at least one))
= P ((A1 ∩ A02 ∩ A03 ) ∪ (A01 ∩ A2 ∩ A03 ) ∪ (A01 ∩ A02 ∩ A3 ) | (A1 ∪ A2 ∪ A3 ))
P ([(A1 ∩ A02 ∩ A03 ) ∪ (A01 ∩ A2 ∩ A03 ) ∪ (A01 ∩ A02 ∩ A3 )] ∩ (A1 ∪ A2 ∪ A3 ))
P (A1 ∪ A2 ∪ A3 )
0
0
0
P ((A1 ∩ A2 ∩ A3 ) ∪ (A1 ∩ A2 ∩ A03 ) ∪ (A01 ∩ A02 ∩ A3 ))
=
P (A1 ∪ A2 ∪ A3 )
0
0
P ((A1 ∩ A2 ∩ A3 )) + P ((A01 ∩ A2 ∩ A03 )) + P ((A01 ∩ A02 ∩ A3 ))
=
P (A1 ∪ A2 ∪ A3 )
=
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Now, we have to do some intermediate calculations:
P (A1 ∩ A2 ) = P (A1 ) + P (A2 ) − P (A1 ∪ A2 ) = .12 + .07 − .13 = .06
P (A1 ∩ A3 ) = P (A1 ) + P (A3 ) − P (A1 ∪ A3 ) = .12 + .05 − .14 = .03
P (A2 ∩ A3 ) = P (A2 ) + P (A3 ) − P (A2 ∪ A3 ) = .07 + .05 − .1 = .02
P (A01 ∩ A2 ∩ A3 ) = P (A2 ∩ A3 ) = P (A1 ∩ A2 ∩ A3 ) = .06 − .01 = .05
P (A1 ∩ A02 ∩ A3 ) = P (A1 ∩ A3 ) = P (A1 ∩ A2 ∩ A3 ) = .03 − .01 = .02
P (A1 ∩ A2 ∩ A03 ) = P (A1 ∩ A2 ) = P (A1 ∩ A2 ∩ A3 ) = .02 − .01 = .01
From the above, we can make the Venn Diagram:
By calculating:
P (A1 ∩ A02 ∩ A03 ) = .12 − .05 − .02 − .01 = .04
P (A01 ∩ A2 ∩ A03 ) = .07 − .05 − .01 − .01 = 0
P (A01 ∩ A02 ∩ A3 ) = .05 − .02 − .01 − .01 = .01
And we plug in numbers:
P ((exactly one) | (at least one)) =
d. Use the results from exercise 26.0
1 ∩A2 ∩A3 )
=
P (A03 | A1 ∩ A2 ) = P (A
P (A1 ∩A2 )
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.04 + 0 + .01
= .3571
.04 + .05 + .01 + .02 + .01 + .01
.05
.06
= .833
Exercise 2.14 (Devore 2.52).
Let F1 be the event that the first pump fails and F2 be the event that the
second one fails. Since the pumps are identical, P (F1 ) = P (F2 ) (assume they
are under equal stresses too). Then:
P (F1 ∪ F2 ) = P (F1 ) + P (F2 ) − P (F1 ∩ F2 )
.07 = 2P (F1 ) − .01
.04 = P (F1 )
Exercise 2.15 (Devore 2.53).
P (B)
P (B | A) = P P(B∩A)
(A) = P (A) (since event B is contained in event A) =
.05
.6 = .0833
Exercise 2.16 (Devore 2.63).
a.
b. P (A ∩ B ∩ C) = .75 · .9 · .8 = .5400
c. P (B ∩ C) = P (A ∩ B ∩ C) + P (A? ∩ B ∩ C) = .54 + .25 · .8 · .7 = .68
d. P (C) = P (A∩B ∩C)+P (A?∩B ∩C)+P (A∩B?∩C)+P (A?∩B?∩C) =
.54 + .045 + .14 + .015 = .74
e. P (A | B ∩ C) =
P (A∩B∩C
P (B∩C)
=
.54
.68
= .7941
Exercise 2.17 (Devore 2.67).
Let T be the event that the randomly selected person is actually a terrorist.
Write + if the screening determined the person to be a future terrorist, and
write − if the screening determined the person to not be a future terrorist. We
are given P (+ | T ) = .99, P (− | T 0 ) = .999, and
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P (T ) = 1, 000/300, 000, 000 = .0000033. Apply Bayes’ Theorem:
P (T | +) =
P (T )P (+ | T )
P (T )P (+ | T )
=
0
0
P (T )P (+ | T ) + P (T )P (+ | T )
P (T )P (+ | T ) + (1 − P (T ))(1 − P (− | T 0 ))
.0000033 · .99
=
= .003289 .
.00003 · .99 + (1 − .00033)(1 − .999)
In other words, about .3% of everyone flagged as terrorists would be actual
terrorists. This screening turned out to be pretty ineffective. Why? Because
the probability of randomly selecting a terrorist is so small. This is a common
problem in other screenings, like tests for rare diseases in large populations.
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