IND ENG 172
Probability and Risk Analysis for Engineers
Prof. Mariana Olvera-Cravioto
Assignment #7
October 16, 2017
Page 1 of 1
Assignment #7 due October 20th, 2017
1. Alice passes through four traff
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 (
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 seco
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 )
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
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 o
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
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
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
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 (
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 (
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 i
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)
whe
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
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
Problem 7.75
Homework 13 Solutions
Problem 8.2
1
Problem 8.3
Problem 8.14
Problem 8.18: See the solution to Problem 8.2(a).
2
Problem 8.20
Problem 8.23
Theoretical Exercises:
Problem 7.51
3
Problem 8.
Extra Problem 1:
This is the solution from the last year. The
original probabilities in last year's assignment
was p(0 arrival)=0.4, p(1)=0.3, p(2)=0.2,
p(3)=0.1
5
6
7
Extra Problem 2:
8
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.
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)
whe
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,
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 - w
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 +
Problem 1 e.
P (E|F ) =
Thus, the odds are
P (EF )
91/220
91
=
=
= .83
P (F )
(91 + 19)/220
110
P (E|F )
P (E c |F )
P (E|F )
1 P (E|F )
=
.83
1 .83
=
Problem 2
Let E be the event that the 4th ball dr
20 a.
There are a total of 8 possible groups of friends. There are 2 6 sets of 5 with the
5
2
3
2
6
persons two feuding friends in them. Thus, there are 8
5
2 3 = 36 possible choices
20 b.
If we do no
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 s
Chase Aplin
IEOR 172 HW 7
1) Alice passes through four traffic lights on her way to work, and each light is equally
likely to be green or red, independently of the others.
a) What is the PMF, the mean
Chase Aplin
IEOR 172 HW 6
1) Consider four independent rolls of a 6-sided die. Let X be the number of 1s and let Y
be the number of 2s obtained. What is the joint PMF of X and Y.
4 1 5 4
PMF of Y from