55
Let n be the number of sophomore girls. We need to nd n such that
P (f reshman, boy) = P (f reshman)P (boy)
(4)
P (f reshman, girl) = P (f reshman)P (girl)
(5)
P (sophomore, boy) = P (sophomore)P (boy)
(6)
P (sophomore, girl) = P (sophomore)P (girl)
(7
3
Alternatively, one could calculate P (ace|interchanged not selected) = 51 by conditioning
on the number of aces in our second half deck. Let Dk be the event that we have
k = 1, 2, 3 aces in our second half deck. Also, to simplify notation, let I c be th
When the randomly selected child is the youngest child of the family,
P (F |C1 )P (C1 )
P (F )
P (F |C1 )P (C1 )
= P4
i=1 P (F |Ci )P (Ci )
1(.1)
=
1
1(.1) + 2 (.25) + 1 (.35) + 1 (.3)
3
4
P (C1 |F ) =
Ch 3 Problem 32 b.
P (E|C4 )P (C4 )
P (E)
P (E|C4 )P
17 b.
The desired probability is
P (D|C) =
P (CD)
.22(.36)
=
P (C)
.3
37 a.
P (H|f air)P (f air)
P (H)
P (H|f air)P (f air)
=
P (H|f air)P (f air) + P (H|f airc )P (f airc )
1 1
2
= 1 1 2 1
2 2 +1 2
1
=
3
P (f air|H) =
37 b.
P (HH|f air)P (f air)
P (HH)
Solving for P (A) we get the desired result:
P (A) =
1
(1
P (E)
P (E)
P (F )
=
P (E)
P (E) + P (F )
78 a.
If a total of 4 games are played, each player must win one of the rst two games and then
one of them wins the next two. That is, the mutually exclusi
The probability their rst ospring is albino depends on how likely the rst parent is a
carrier. Let E1 be the event that their rst child is albino. Then
c
c
P (E1 ) = P (E1 |C1 )P (C1 ) + P (E1 |C1 )P (C1 )
1 2
1
= +0
4 3
3
1
=
6
61 b.
Let E2 be the event
63 b.
Let D be the event that both duelists are hit. Conditioning on the outcome of the rst
dual as we did in part A,
P (D) = P (D|I)P (I) + P (D|II)P (II) + P (D|III)P (III) + P (D|IV )P (IV )
= 0 pB (1
pA ) + 0 pA (1
= pA pB + p(D)(1
pA )(1
pB ) + 1 pA
We can nd each of these probabilities using the Law of Total Probability, i.e.
c
c
c
P (J3 J1 J2 ) = P (J3 J1 J2 |G)P (G) + P (J3 J1 J2 |Gc )P (Gc )
= (.7)2 (.3)(.7) + (.2)2 (.8)(.3)
c
c
c
P (J3 J1 J2 ) = P (J3 J1 J2 |G)P (G) + P (J3 J1 J2 |Gc )P (Gc )
=
Problem 3 c.
P (A, pos)
P (pos)
P (A)
P (pos)
P (A)
P (pos|Ac )
P (A|pos) =
(8)
Where the rst step is since P (A, pos) = P (A \ pos) P (A) as (A \ pos) A and the
P (A)
second step is since P (pos) P (pos|Ac ). Thus, P (A|pos) P (pos|Ac )
Problem 3 d.
P (p
Solution to the Swine Flu Problem
IEOR 172
Riley Murray
September 4, 2015
We are asked whether Drug B is better than Drug A given the following:
P (C|M, B) > P (C|M, A) and P (C|F, B) > P (C|F, A)
where C is the event that a patient is cured, F is the eve
Home Work 2 Solutions
IEOR 172, Fall 2015
Ch 3 Problem 5 a.
Using the multiplication rule, the probability is
6
15
5
14
9
13
8
12
Ch 3 Problem 5 b.
To nd the probability that the rst two are black, we do not care about the outcome
9
8
of the last two sele
Homework 11 solution
Theoretical Exercises:
Problem 7.1
Problem 7.17
2
Problem 7.19
Problem 7.34
3
Problem 7.40
4
Extra Problem 1:
Extra Problem 2:
5
Extra Problem 3:
4
6
7
Extra problem 7
6
d.
e.
f.
9
g.
h.
(, ) = () = [(|)] = 2
Extra Problem 6:
8
1
= (
Solution to the Swine Flu Problem
IEOR 172
Riley Murray
September 4, 2015
We are asked whether Drug B is better than Drug A given the following:
P (C|M, B) > P (C|M, A) and P (C|F, B) > P (C|F, A)
where C is the event that a patient is cured, F is the eve
Home Work 1 Solutions
IEOR 172, Fall 2015
10 a.
From Permutation Rule 1, there are 8! = 40320 possible seating arrangements
10 b.
If A is always in front of B, and A and B must sit next to each other, then there are 7!
possible seating arrangements. We al
By the premise given at the beginning of the problem, both P (C|M, B) P (C|M, A) > 0
and P (C|F, B) P (C|F, A) > 0. This implies that Drug B is better than Drug A if
(1
p)x + py > 0
where x, y > 0 - which we know is true!
So if we force P (M |A) = P (M |B
Ch 3 Problem 66 a.
Let E be the event that current ows from A to B. Then
P (E) = P (E|5 closed)p5
= P (1 and 2 closed [ 3 and 4 closed|5 closed)p5
= [P (1 and 2 closed) + P (3 and 4 closed)
= [p1 p2 + p3 p4
p1 p2 p3 p4 ]p5
P (1 and 2 closed \ 3 and 4 clos
Ch 3 Problem 10
Let Si be the event that the i-th card is a spade. Then the desired probability can be
found using
P (S1 S2 S3 )
P (S1 |S2 S3 ) =
P (S2 S3 )
where P (S1 S2 S3 ) =
13
52
12
51
11
50
and
c
c
P (S2 S3 ) = P (S2 S3 |S1 )P (S1 ) + P (S2 S3 |S1
To prove P (pos|A)/P (pos|Ac )
P (pos|Ac ), assume
1 ) P (pos|A)
P (pos|A)
P (pos|Ac )
1
we can multiply both sides by P (pos|Ac ) without changing the direction of the inequality,
thus,
P (pos|A) P (pos|Ac )
To prove P (pos|A)/P (pos|Ac )
1 ) P (A|pos)/P
Let E be the event that a randomly chosen applicant is male. Let F be the event that a
randomly chosen applicant is admitted. The total number of applicants is 220.
Gender and admission rate are not independent since
91
(91 + 19) (91 + 19)
= .25 6= .41 =
on the day before,
Pn = P n
= pPn
= (2p
1
p + (1
1
+1
1)Pn
Pn
p
1
Pn
+ (1
1)
1
(1
p)
+ pPn
1
p)
Now, the identity holds for P0 . Assume by induction that the identity holds for n 1.
We will show that the identity then holds for n and thus holds for all
Home Work 4 Solutions
IEOR 172, Fall 2015
Problem 3.74
For player A to roll a 9, they must roll cfw_6, 3, cfw_3, 6, cfw_5, 4, or cfw_4, 5, thus, the probability
4
A will roll a 9 is pA = 36
= 19 . For player B to roll a 6, they must roll cfw_1, 5, cfw_5,
Home Work 5 Solutions
IEOR 172, Fall 2015
Problem 4.20 a.
Let p = 18
38 =
9
19
and q =
20
38
=
10
19
P (X > 0) = p + p2 q =
9
10
+
19 19
9
19
2
= 0.5917
Problem 4.20 b.
No, there are two paths where we win but three where we loose. One of these paths has
IEOR 172 MIDTERM 1
NAME ) 120 points
Please show all your work on these pages. Good luck!
1. (3 points each) in how many ways can 8 people be seated in a row if
(at) There are no restrictions on the seating arrangement?
(h) Persons A and B must sit next t
IEOR 172 MIDTERM 2
100 points
NAME .
Please show all your work on these pages. Only go as far as is reasonable without a calculator
in terms of simplifying your answers. Good luck!
1. Aircraft failures occur about once in every 5000 ﬂight hours, following