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UC Berkeley
Department of Electrical Engineering and Computer Science
EE 126: Probablity and Random Processes
Problem Set 1: Solutions
Fall 2007
Issued: Thursday, August 30, 2007
Due: Friday, September 7, 2007
Reading: Berstsekas & Tsitsiklis, §1.1, §1.2. and §1.3
Problem 1.1
For three tosses of a fair coin, define the sample space. Find the probabilities of the following
events:
(a) the sequence HHH?
(b) the sequence HT H?
(c) seeing two heads and one tail?
(d) the outcome “More heads than tails”?
Solution:
The sample space is {H, T }3 = {H, T } × {H, T } × {H, T }. The size of the sample space
is |{H, T }3 | = 8.
(a) This is a single outcome in the sample space under a uniform probability law, so the
probability is 1/|Ω| = 1/8
(b) 1/8
(c) This event contains three outcomes: {T HH, HT H, HHT }. Therefore, the probability
is 3/8 .
(d) This event can be expressed more explicitly as either two heads and one tails or all
three heads. The probability of the two heads cases is given in (c), and so we add the
probability of the three heads outcome to get 3/8 + 1/8 = 1/2 .
Problem 1.2
In this exercise, we prove Bonferroni’s inequality:
(a) Show that for any two events A and B, we have
P(A ∩ B) ≥ P(A) + P(B) − 1.
(b) Generalize to the case of n events A1 , A2 , . . . , An by showing that
P(A1 ∩ A2 ∩ · · · ∩ An ) ≥ P(A1 ) + P(A2 ) + · · · + P(An ) − (n − 1).
Hint: To generalize in part (b), use de Morgan’s law.
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Solution:
We have P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A ∪ B) ≤ 1, which implies part (a).
For part (b), we use de Morgan’s law to obtain
1 − P(A1 ∩ · · · ∩ An ) = P (A1 ∩ · · · ∩ An )c
= P(Ac1 ∪ · · · ∪ Acn )
≤ P(Ac1 ) + · · · + P(Acn )
= 1 − P(A1 ) + · · · + 1 − P(An )
= n − P(A1 ) − · · · − P(An ).
Problem 1.3
Alice and Emily each choose at random a number between zero and one according to the
uniform probability law. Consider the following events:
A = {The magnitude of the difference of the two numbers is greater than 1/3.}
B = {At least one of the numbers is greater than 2/3.}
C = {The sum of the two numbers is 1.}
D = {Alice’s number is greater than 2/3.}
Find the following probabilities: P(A), P(B), P(A ∩ B), P(C), P(D), P(A ∩ D).
Solution:
See Figure 1 for a the geometry of the solutions.
122
233
=
1
3
−
1
9
3. P(A ∩ B) = 2
1
2
1. P(A) = 2
2. P(B) =
1
3
+
4
9
=
5
9
1
3
2
3
+
1
3
=
1
3
4. P(C) = 0 since the event has no area.
5. P(D) =
1
3
6. P(A ∩ D) = P(A ∩ B)/2 =
1
2
1
3
+
2
3
1
3
=
1
6
Problem 1.4
Given two events A and B on a common sample space, give expressions for the following
probabilities in terms of P(A), P(B) and P(A ∩ B).
(a) The probability that at least one of A or B occurs?
2
(a) A
(b) B
(c) C
(d) D
Figure 1: The shaded areas in each diagram represent each respective event, where
the horizontal axis is Alice’s number and the vertical axis is Emily’s number.
Each square is [0 1] × [0 1].
(b) The probability that exactly one of A and B occurs?
c
c
Which one of these probabilities is equal to P (A ∩ B ) ∪ (A ∩ B) ?
Solution:
(a) This event is the union of A and B, and so by the improved additivity axiom in class
we have P(A ∪ B) = P(A) + P(B) − P(A ∩ B) . The intersection must be subtracted
out so as to avoid double-counting its probability.
(b) This event can also be expressed as “A and not B OR B and not A,” and we can
translate it into set notation as P((A ∩ B c ) ∪ (Ac ∩ B)), which is the expression given.
The event can also be given by (A ∪ B) \ (A ∩ B), and so we can use part (a) to express
this probability as P(A) + P(B) − 2 · P(A ∩ B) .
Problem 1.5
Barbara has a peculiar pair of four-sided dice. When she rolls the dice, the probability of any
particular outcome (pair of numbers) is proportional to the product of the outcomes from
each die. All outcomes that result in a particular product are equally likely.
(a) What is the probability of the product being even?
(b) What is the probability of Barbara rolling a 2 and a 3?
Solution:
There are two possible interpretations of this question, which correspond to whether the
sample space is defined as {1, 2, 3, 4}2 , where order matters and (2, 3) is distinct from (3, 2),
or if the sample space is defined as {{i, j} | i, j ∈ {1, 2, 3, 4}}, where order does not matter.
The two cases are not the same because the implicit definition of the probability law given
will differ when applied to each case (in particular, the normalizing constant that is calculated
will be different). Both interpretations are valid, and both solutions are presented below.
Assuming Ω = {1, 2, 3, 4}2
3
The problem states that the probability of some pair (a, b) being rolled is P({(a, b)}) =
c · a · b for some proportionality constant c. The values of a · b for all elements in the sample
space is given in the table below:
1
1
2
3
4
1
2
3
4
2
2
4
6
8
3
3
6
9
12
4
4
8
12
16
To find our probability law, we must multiply the elements in the table above by the
normalizing constant c. To find c we use the normalization axiom, which requires:
X
X
P({(a, b)}) = 1
a∈{1,2,3,4} b∈{1,2,3,4}
And so we see that c =
1
P P
a
b
a·b .
Thus c =
1
100 ,
and we can express our probability law
on each singleton subset of the sample space in closed form as P({(a, b)}) =
(a)
2+4+2+4+6+6+12+4+8+12+16
100
=
a·b
100 .
84
100
(b) The event is {(2, 3) ∪ (3, 2)}
6
100
+
6
100
=
3
25
Assuming Ω = {{i, j} | i, j ∈ {1, 2, 3, 4}}
The problem states that the probability of an element in the sample space is P({{a, b}}) =
c · a · b for some proportionality constant c. The values of a · b for all elements in the sample
space is given in the table below:
1
2
3
4
1
1
2
3
4
2
3
4
4
6
8
9
12
16
Note the matrix now has a triangular form.
4
To find our probability law, we want to calculate the constant c. To find c we use the
normalization axiom, which requires:
X
X
P({{a, b}}) = 1
a∈{1,2,3,4} b≤a
And so we see that c = 65. The probability law on each singleton subset of the sample
space is P({{a, b}}) = a·b
65 .
(a)
(b)
2+4+6+4+8+12+16
65
=
52
65
6
65
Problem 1.6
Mike and John are playing a friendly game of darts where the dart board is a disk of a radius
of 10 inches. Whenever a dart falls within 1 inch of the center (the bullseye), then 50 points
are scored. If the point of impact is between 1 and 3 inches from the center, then 30 points
are scored, if it is at a distance of 3 to 5 inches, then 20 points are scored and if it is further
than 5 inches, then 10 points are scored. Assume that both players are skilled enough to be
able to throw the dart within the boundaries of the board.
Mike can place the dart uniformly on the board (i.e. the probability of the dart falling in
a given region is proportional to its area.)
(a) What is the probability that Mike scores 50 points on one throw?
(b) What is the probability of him scoring 30 points?
(c) John is right handed and is twice more likely to throw in the right half of the board
than in the left half. Across each half, the dart falls uniformly in that region. Answer
the previous questions for John’s throw.
Solution:
(a) The probability of the bullseye is the ratio of its area to the dartboard area, so it is
π·12
given by π·10
2 = 1/100 .
(b) To score exactly thirty points, we need to fall in the region 1 to 3 inches from the center,
2 −π·12
so we have π·3π·10
= 8/100 .
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(c) By symmetry, John’s bias does not affect his probability scores, and so John’s probabilities are the same as Mike’s. To be more precise, we can use total probability:
PJohn (50pts) = PJohn (50pts | right) · PJohn (right) + PJohn (50pts | lef t) · PJohn (lef t)
= PJohn (50pts) · PJohn (right) + PJohn (50pts) · PJohn (lef t)
= PM ike (50pts) · PJohn (right) + PM ike (50pts) · PJohn (lef t)
= PM ike (50pts)
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The first equality is due to total probability, and the second equality holds because
conditioning on the left- or right-side of the board gives a uniform density on that half,
and so John’s probability of scoring 50 points is independent of which half he throws
onto. Since the rule on each half is uniform and the ratio of the areas are the same as
the case of the whole dartboard, we can replace John’s probability of scoring 50 points
on a given half with the equivalent Mike’s probability of scoring 50 points. The final
equality holds by factoring out PM ike(50pts) and using total probability.
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