Fall 2011
Name
Solutions
Midterm 1 (100 points)
Wednesday, October 19, 2011
Show your work. It’s not necessary to give a final numerical answer unless specifically asked, but you need to
provide all the information necessary to compute the final answer. Please circle or otherwise highlight your
answer. Turn in your “crib sheet” with your exam. GOOD LUCK!
1.
For each of the following statements, circle the letter “T” if it is true, and “F” if it is false.
(3 * 9 =27 pts)
The correct answer has been bolded.
T
F
a.
S = {… 2.0, 1.6, 1.2, 0.8, 0.4, 0.0, 0.4, 0.8, 1.2, 1.6, 2.0, …} is an example
of a continuous sample space.
There are still only a countably infinite number of points in the sample space.
T
F
b.
For any event A, (A
∩
S)
∪
(A
C
∩
S) = S, where S represents the entire sample
space.
A
∩
S=A, A
C
∩
S= A
C
, S=A
∪
A
C
by definition of complement.
T
F
c.
Pr[AB
c
] = 1 – Pr[AB] .
For it to be true, the complement should be on the A:
Pr[A
c
B] = 1 – Pr[AB].
T
F
d.
If two events form a partition, then they are always independent.
Because they are M.E., and M.E. events are
dependent
– knowing one occurs
affects the probability of the other occurring (you know it cannot occur, since
they are M.E.).
T
F
e.
The mean and median are identical for symmetric distributions.
T
F
f.
If Pr[A] + Pr[B] + Pr[C] = 1, then the set of events {A, B, C} forms a partition.
The converse of this is true.
But suppose, e.g., B=C, and Pr[A] = Pr[B] =
Pr[C] = 1/3. Then the sum = 1, but {A, B, C} is not a partition since B and C
are not M.E. (nor are {A, B, C} C.E.).
You could also devise an example in
which B and C overlap but are not equal.
T
F
g.
1
7
7
6
n
n
n

×
= ×
÷
÷
LHS:
7
!
!
7!(
7)!
6!(
7)!
n
n
n
n
×
=


, RHS:
(
1)!
!
6!(
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Mocktarian
 Conditional Probability, Probability, Probability theory, speci fically

Click to edit the document details