The basic idea in probability theory is a sample space with a collection of events and a probability for each of those events. We talk about constructing the sample space and various events.

1. V63.0233, Theory of Probability Solutions
Worksheet for Sections 2.1–2.3 : Sample Spaces and the Axioms of Probability July 1, 2009
K
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4
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. abs out on a call
L
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3
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O
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M
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N
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c
1
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0
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P
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. abs in operation
Figure 1: Sample space from Problems 1 and 2 with events circled.
1. A small taxicab company owns four cabs. Let X be the sample space of pairs (x, y), representing
the outcome that x cabs are in operative condition and y of those cabs are out on a call. Let K be
the event that at least two cabs are out on a call, L the event that exactly one of the operative cabs
is not out on a call, and M the event that only one of the cabs is operative.
(a) Draw the sample space using points in a Cartesian plane.
(b) List and indicate on your diagram the event K.
(c) List and indicate on your diagram the event L.
(d) List and indicate on your diagram the event M .
Solution. The sample space is drawn in Figure 1.
2. Continuing the above problem, express in words what events are represented by the following
sets of points:
(a) N = {(1, 1), (2, 1), (3, 1), (4, 1)}
(b) O = {(2, 0), (3, 0), (4, 0), (4, 1), (4, 2)}
(c) P = {(0, 0), (1, 0), (1, 1)}
Solution. (a) N is the event that one cab is out on a call.
(b) O is complicated; it’s the event that at least two cabs are operable and none are on a call, or
all four cabs are operable and at most two are on a call.
(c) P is the event that at most one cab is operable.
These events are indicated in Figure 1.
3. Continuing the two previous problems, which of the following events are mutually exclusive?
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2. (a) K and M (d) L and P
(b) K and O (e) O and P
(c) M and N (f ) K and N
Solution. Referring again to Figure 1, we can see which of them are and are not mutually exclusive:
(a) mutually exclusive (d) not mutually exclusive; they share the out-
come (1, 0)
(b) not mutually exclusive; they share the out-
come (4, 2)
(e) mutually exclusive
(c) not mutually exclusive; they share the out-
come (1, 1) (f) mutually exclusive
4. Referring to the sample space of Problems 1 and 2, list the points which constitute the following
events, and express them in words:
(i) K (iv) M ∩ P (vii) O ∩ P
(ii) O (v) K ∪ N (viii) M ∪ P
(iii) L ∪ N (vi) L ∩ M (ix) O ∩ L
Solution. (i) K is the event that at most one cab is out on a call.
K = {(0, 0), (1, 0), (2, 0), (3, 0), (4, 0), (1, 1), (2, 1), (3, 1), (4, 1)}
(ii) As a list,
{ }
O = (0, 0), (1, 0), (1, 1), (2, 1), (3, 1), (2, 2), (3, 2), (3, 3), (4, 3), (4, 4)
O is worse than O. Looking ahead, we could use DeMorgan’s laws and the distributive laws
to get a better description of O in words. Notice
O = ((x ≥ 2) ∩ (y = 0)) ∪ ((x = 4) ∩ (y ≤ 2))
So
O = [((x ≥ 2) ∩ (y = 0)) ∪ ((x = 4) ∩ (y ≤ 2))]
= ((x ≥ 2) ∩ (y = 0)) ∩ ((x = 4) ∩ (y ≤ 2))
( ) ( )
= (x ≥ 2) ∪ (y = 0) ∩ (x = 4) ∪ (y ≤ 2)
= ((x < 2) ∪ (y > 0)) ∩ ((x < 4) ∪ (y > 2))
( ) ( )
= (x < 2) ∩ (x < 4) ∪ (x < 2) ∩ (y > 2)
( ) ( )
∪ (y > 0) ∩ (x < 4) ∪ (y > 0) ∩ (y > 2)
( ) ( )
= (x < 2) ∪ (x < 2) ∩ (y > 2) ∪ (y > 0) ∩ (x < 4) ∪ (y > 2)
( )
= (x < 2) ∪ (y > 0) ∩ (x < 4) ∪ (y > 2)
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3. So O is the event that either
1) fewer than two cabs are operative,
2) more than two cabs are out on a call, or
3) not all cabs are operative and some are out on calls
(iii) L ∪ N is the event that exactly one of the cabs is not out on a call or exactly one of the cabs
is out on a call.
L ∪ N = {(1, 0), (2, 1), (3, 2), (4, 3), (1, 1), (2, 1), (3, 1), (4, 1)}
(iv) M ∩ P is the event that only one of the cabs is operative, and at most one of the cabs is
operable. The first is stricter than the second, so the intersection is M , the event that only
one cab is operative:
M ∩ P = P = {(1, 0), (1, 1)}
(v) K ∪ N is the event that at least two cabs are out on a call, or exactly one cab is out on a call.
To simplify, at least one cab is out on a call:
K ∪ N = {(1, 1), (2, 1), (3, 1), (4, 1), (2, 2), (3, 2), (4, 2), (3, 3), (4, 3), (4, 4)}
(vi) L ∩ M is the event that one cab is operable, and it is not out on a call.
L ∩ M = {(1, 0)}
(vii) By looking at Figure 1, we can see that O ∩ P = O.
(viii) M ∪ P is the event that exactly one cab is operable, or that at least two cabs are operable.
In other words, at least one cab is operable.
{
K ∪ P = (1, 0), (2, 0), (3, 0), (4, 0), (1, 1), (2, 1), (3, 1),
}
(4, 1)(2, 2), (3, 2), (4, 2), (3, 3), (4, 3), (4, 4)
(ix) Since O and L are mutually exclusive, O ∩ L = O.
5. Analyzing business conditions in general, four government officials make the following claims:
The first claims that the probabilities for unemployment to go up, remain unchanged, or go down
are, respectively, 0.51, 0.33, and 0.12, the second claims that the respective probabilities are 0.55,
0.49, and −0.04, the third claims that the respective probabilities are 0.52, 0.38, and 0.10, and the
fourth claims the respective probabilities are 0.48, 0.34, and 0.21. Comment on these claims.
Solution. [Rachel] We can arrange the purported probabilities in a table.
Analyst 1 2 3 4
P (up) 0.51 0.55 0.52 0.48
P (unchg) 0.33 0.49 0.38 0.34
P (down) 0.12 -0.04 0.10 0.21
Total 0.96 1.00 1.00 1.03
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4. Two of the analysts’ probabilities don’t add up to one, as they should: #1 and #4. Analyst #2
has a negative probability, so is also unclear on the concept. Only #3’s probabilities are consistent
with the axioms of probability.
6. If a certain mathematics professor asks a question of one of his students (chosen at random
from a very large class), the probabilities that his student will have received an A, B, C, D, or F in
the last examination are 0.09, 0.23, 0.36, 0.18, and 0.14. What are the probabilities that the student
received
(i) at least a C?
(ii) at least a D?
(iii) a B, C, or D?
(iv) at most a C?
Assume that each student in the class has taken the examination.
Solution. (i) P (A ∪ B ∪ C) = P (A) + P (B) + P (C) = 0.09 + 0.23 + 0.36 = 0.68
(ii) P (A ∪ B ∪ C ∪ D) = P (A ∪ B ∪ C) + P (D) = 0.68 + 0.18 = 0.86
(iii) P (B ∪ C ∪ D) = 0.23 + 0.36 + 0.18 = 0.76
(iv) P (C ∪ D ∪ F ) = 0.36 + 0.18 + 0.14 = 0.68
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