SOLUTIONS
STAT W3105
INTRODUCTION TO PROBABILITY
MIDTERM 1 FALL 2013
Professor: Shaw-Hwa Lo
Problem
1
Credits
/10
2
/ 15
3
/ 15
4
/ 10
5
/ 15
6
/ 15
7
/20
Bonus
/10
Total
/100
1. Next year, Columbia University will change the format of the unique identifi
3105 Midterm 1 Practice Exam Problems Fall 2013
1. How many different letter arrangements can be made from the letters?
(a) October
(b) Pumpkin
(c) Halloween
2. A certain town with a population of 100,000 has 3 newspapers: I, II, and III. The proportions
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Stat 3105 HW #4 Solutions
44. The jailers reasoning is faulty. This is similar to the Monty Hall problem. By revealing that one of his
fellow prisoners is to be set free, the probability of his other fellow prisoner being executed has risen to
2/3; wherea
W3105 HW #8 solution
HW #11 Solutions
Stat 3105
1.
a) p(i, j) = # of rolls that have i as the largest number and j as the sum of the two rolls/ 36
for example, p(6,7) = 2/36 (6,1) and (1,6)
p(5,7) = 2/36 (5,2) and (2,5)
p(1,3) = 0, etc.
b) enumerate all p
STAT 3105: MIDTERM 2 REVIEW QUESTIONS:
1. A doctor is studying the relationship between blood pressure and heartbeat abnormalities in
her patients. She tests a random sample of her patients and notes their blood pressures (high, low,
or normal) and their
SStat3105 HW #5 Solutions
TAT 3105 HW #6 Solutions
10. We are interested in P(X = i | X>0) for i = 1,2,3
= P(X=i and X > 0) / P(X>0) = P(X = i)/P(X>0) since all i > 0
= P(X = i) / (P(X=1) + P(X=2) + P(X=3)
We see P(X>0 = 39/165 + 15/165 + 1/165 = 55/165 =
3105 HW # 3 Solutions
10.
11/50 Assume the second and third pulls are the first and second by symmetry, and imagine the
3rd pull being after 2 spades (and 2 out of 52 cards) are gone. Thus there are 11 spades left out
of 50 cards.
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3105 Midterm 1 Practice Exam Problems
Solutions
1. How many different letter arrangements can be made from the letters?
(a) October
(b) Pumpkin
(c ) Halloween
Solution:
(a) 7! = 2520
2!
(b)
7! = 2520
2!
(c) 9! = 90720
2!*2!
2. A certain town with a popula