ECI 114
Fall 2009
Name
Solutions
Midterm 1 (100 points)
Wednesday, October 28, 2009
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. Put your name on the back of the last
page.
GOOD LUCK!
1.
For each of the following statements, circle the letter “T” if it is true, and “F” if it is
false. (3 * 8 =24 points)
T
F
a. If events A and B are mutually exclusive, then A and B are independent.
False.
If events A and B are mutually exclusive, then A and B are NOT independent. See
course notes for details.
T
F
b. If events A and B are collectively exhaustive, then Pr[A]+Pr[B] must equal
Pr[S]=1, where S represents the entire sample space.
False. If events A and B are collectively exhaustive, Pr[A
∪
B] =
Pr[S] = 1, and Pr[A]+Pr[B] is
no less than 1. But unless they are also mutually exclusive (such that Pr[A
∩
B]=0), then Pr[A] +
Pr[B] (= Pr[A
∪
B] + Pr[A
∩
B] = 1 + Pr[A
∩
B]) will be > 1.
T
F
c. If events A, B and C are independent, then Pr[(A
∩
B)
∩
(B
∩
C)] =
Pr[A] Pr[B]
2
Pr[C].
False. If A, B and C are independent, then Pr[(A
∩
B)
∩
(B
∩
C)] = Pr[(A
∩
B
∩
C)] =
Pr[A] Pr[B] Pr[C].
Pr[(A
∩
B)
∩
(B
∩
C)] does not equal Pr(A
∩
B) Pr(B
∩
C), because A
∩
B and B
∩
C are NOT
independent.
For example, if A, B, and C are the events of getting a Head on the first, second,
and third tosses of a fair coin, respectively, then the three events are independent, and you can
easily show (by enumerating the 8 equally likely outcomes of the three tosses) that Pr(A
∩
B) =
Pr[HHH or HHT] = ¼ and Pr(B
∩
C) = Pr[HHH or THH] = ¼, but
Pr[(A
∩
B)
∩
(B
∩
C)] = Pr[HHH] =
1/8, which does not equal ¼ x ¼.
T
F
d. (A
∩
B)
∪
(C
∩
B)
∪
(A
∩
C) = (A
C
∩
C
∩
B)
∪
(B
C
∩
A
∩
C)
∪
(C
C
∩
A
∩
B)
∪
(A
∩
C
∩
B).
True. (A
∩
B)
∪
(C
∩
B)
∪
(A
∩
C) represents “among A, B and C, at least two events occur”,
which is equivalent to (A
C
∩
C
∩
B)
∪
(B
C
∩
A
∩
C)
∪
(C
C
∩
A
∩
B), where exactly two events occur
simultaneously, union (A
∩
C
∩
B), where all three events occur simultaneously.
T
F
e. 6
distinct
objects are partitioned into 3
identical
groups, with 2 objects in
each group. There are in all 90 ways to make such a partition.
Midterm 1, ECI 114, FALL 2009
Page
1
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View Full DocumentFalse. There are in all 6!/2!/2!/2!=90 ways that 6
distinct
objects can be partitioned into 3
distinct
groups with 2 objects in each group. If those three groups are identical, then there are
in all 90/3!=15 ways to make such a partition.
T
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 Fall '08
 Mocktarian
 Probability theory, probability density function, Type I and type II errors, ECI

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