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Chapter 2: Outcomes, events,
and probability
CIS 3033
2.1 Sample spaces
• See observed phenomena as outcomes of some
repeated experiment
• Sample space: set of elements describing the
outcomes of the experiment
• Example: the tossing of a coin, Ω = {H, T}
• Example: the month of a day, Ω = {Jan, ..., Dec}
• Example: the weight of an object, Ω = (0,∞)
• Assumption: the scope of possible outcomes are
known in advance, though its size can be infinite
2.1 Sample spaces
Example: If we have three envelopes and number
them 1, 2, and 3, the following sample space consists
of every different permutation we can make using all
three envelopes:
Ω = {123, 132, 213, 231, 312, 321}
There are n · (n − 1) · · · · 3 · 2 · 1 = n! possible
permutations of n objects, so here |Ω| = n factorial
2.2 Events
An event is a subset of the sample space.
Or, an event is a description of the outcome.
Example: If Ω is the sample space of months, then
event L can be "the months with 31 days", and event
R can be "the months whose full name contain the
letter 'r' ".
Since events are sets, they can be manipulated by
the operations defined in set theory.
2.2 Events
• The intersection of A and B, A∩B, occurs if both
events A and B occur
• The union of A and B, A∪B, occurs if at least one of
the two events occurs
• The complement of A, Ac, occurs if and only if A does
not occur
2.2 Events
• Ωc = ∅, the impossible event
• A and B disjoint or mutually exclusive if they have
no outcome in common, i.e., A∩B = ∅
• Event A implies event B if A is a subset of B, i.e., A
⊂B
2.2 Events
DeMorgan’s Laws: For any two events A and B
(A ∪ B)c = Ac ∩ Bc
(A ∩ B)c = Ac ∪ Bc
Example: “It is certainly not true that neither John nor
Mary is to blame.” is equivalent to: “John or Mary is to
blame, or both.”
2.3 Probability
The probability of an event: how likely it will occur.
Example: tossing a fair coin, we know
P({H}) = P({T}) = 1/2
which can be written as
P(H) = P(T) = 1/2
2.3 Probability
If the sample space contains n equally probable
elements, then
Ω = {e1, e2, ... , en}
and
P(e1) = P(e2) = ... = P(en)
which leads to
P(e1) = P(e2) = ... = P(en) = 1/n
If event A corresponds to m of the n elements, then
P(A) = m/n
2.3 Probability
Example: What is the probability for people's birthday
to be in January?
What is the sample space?
1.12 months
2.365 days
3.how about leap years?
Can we assume equal probability in each case?
2.3 Probability
Assumption: an event has only one probability.
Often we only assign probabilities to the individual
outcomes of the experiments, then use the additivity
property to calculate the probabilities of events.
In general, additivity of P implies that the probability
of an event is obtained by summing the probabilities
of the outcomes belonging to the event.
2.3 Probability
If events A and B that are not disjoint:
A = (A∩B) ∪ (A∩Bc), which is a disjoint union; hence
P(A) = P(A ∩ B) + P(A ∩ Bc) .
Split A ∪ B in the same way:
P(A ∪ B) = P((A ∪ B) ∩ B) + P((A ∪ B) ∩ Bc)
= P(B) + P(A ∩ Bc)
= P(A) + P(B) − P(A ∩ B)
Also, since A ∪ Ac = Ω, P(Ac) = 1 − P(A)
2.4 Products of sample spaces
If an event consists of several steps, the overall
sample space is the product of the individual
sample spaces, such as Ω = Ω1 × Ω2
If Ω1 has r elements and Ω2 has s elements, then
Ω1 × Ω2 has r × s elements.
Example: If a fair die is thrown twice, what is the
probability for you to get (1) double 6, (2) no 6, (3)
at least one 6, and (4) exactly one 6?
2.4 Products of sample spaces
In general, if Ω = Ω1 × Ω2 × ... × Ωn (where the
sample spaces in each step can either be the same
or be different), then the probability of the sequence
of outcomes (w1, w2, ..., wn) is the product of the
probabilities of the individual outcomes, that is,
P((w1, w2, ..., wn)) = P(w1) × P(w2) × ... × P(wn)
Later this result will be extended from outcomes to
events.
2.5 An infinite sample space
Definition. A probability function on an infinite (or
finite) sample space Ω assigns to each event A in Ω a
number P(A) in [0, 1] such that
(i) P(Ω) = 1
(ii) P(A1 ∪ A2 ∪ A3 ∪ · · ·)
= P(A1) + P(A2) + P(A3) + · · ·
if A1,A2,A3, . . . are disjoint events.
2.5 An infinite sample space
Example: If a coin is tossed repeatedly until the first
head turns up, what is the probability for each
number of tossing?
Ω = {1, 2, 3, ... }, assume the probability of a head is
p, then P(1) = p, P(2) = (1 − p)p, ..., or in general,
P(n) = (1 − p)n−1p, n = 1, 2, 3, ...
P(Ω) = P(1) + P(2) + ... + P(n) + ...
= p + (1 − p)p + ... (1 − p)n−1p + ... = p × 1/p = 1
since when x < 1, 1 + x + ... + xn + ... = 1/(1-x)
and the sequence is called a geometric series.