Download Section 1: Basic Probability Concepts

Section 1: Basic Probability Concepts
August 28th, 2014
Lesson 2
We start with some definitions.
A probability space (or sample space) is the collection
of outcomes of a given experiment.
The outcome of an experiment is called a sample point.
For example, our experiment could be the roll of a six-sided
die. The probability space is the set S = {1, 2, 3, 4, 5, 6}, and
1, 2, 3, etc. are sample points.
Mutually exclusive (or disjoint) outcomes cannot
occur simultaneously
Outcomes are exhaustive if they make up the entire
probability space.
Lesson 2
An event is a collection of sample points, or equivalently
a subset of a probability space.
The event “roll an even number” is the set E = {2, 4, 6}. We
say an event has occurred if the outcome falls in the event set.
The union/intersection/complement of events is just the
union/intersection/complement of the sets.
Events are mutually exclusive if the sets are disjoint.
Events are exhaustive if their union is the entire
probability space.
Lesson 2
Events C1 , C2 , . . . , Cn partition an event A if
S
A = ni=1 Ci and the Ci ’s are mutually exclusive.
Since events are just sets, the usual operations and statements
for sets hold. For example, DeMorgan’s Law is true for events.
Let S = {1, 2, 3, 4, 5, 6} be the probability space for rolling a
six-sided die. Let
A = {1, 3, 5} = “an odd number is rolled”
B = {2, 4, 6} = “an even number is rolled”
C = {1, 2, 3} = “a number less than 4 is rolled”
Then A and B partition S, C 0 = {4, 5, 6}, A ∩ C = {1, 3}, etc.
Lesson 2
Here we point out some important relations for set/events:
For any events A, B1 , . . . , Bn we have
A ∩ (B1 ∪ . . . ∪ Bn ) = (A ∩ B1 ) ∪ . . . ∪ (A ∩ Bn ) and
A ∪ (B1 ∩ . . . ∩ Bn ) = (A ∪ B1 ) ∩ . . . ∩ (A ∪ Bn ).
If B1 , . . . , Bn are exhaustive, then
A = A ∩ (B1 ∪ . . . ∪ Bn ) = (A ∩ B1 ) ∪ . . . ∪ (A ∩ Bn ).
For any events A and B,
A = A ∩ (B ∪ B 0 ) = (A ∩ B) ∪ (A ∩ B 0 ).
Lesson 2
Now we start the actual probability.
A discrete probability space S is a finite or countably infinite
set. For a point ai ∈ S, P[ai ] denotes the probability that the
outcome ai occurs.
The probability function P for a discrete space must satisfy
two properties:
(i) 0 ≤ P[ai ] ≤ 1 for each ai ∈ S.
X
(ii) P[a1 ] + P[a2 ] + · · · =
P[ai ] = 1
all i
If A is an event, the probability of A is
P[A] =
X
P[ai ].
ai ∈A
Lesson 2
If S has finitely many points, say S = {a1 , a2 , . . . , ak }, then a
probability function P is uniform if P[ai ] = k1 .
This says that each outcome is equally (or uniformly) likely to
occur.
When rolling a fair six-sided die, each side is equally likely to
come up. In fact, the probability of getting any one side is 16 .
Thus we have a uniform probability function P defined by
P[j] = 61 for j ∈ {1, 2, 3, 4, 5, 6}.
Let A be the event “roll an odd number”. Then
P[A] = P[1] + P[3] + P[5] =
Lesson 2
1
6
+ 16 +
1
6
= 21 .
A continuous probability space is an uncountable set, for
example an interval in R.
Imagine spinning a pointer on a wheel. If we measure the
position of the spinner in radians, measuring clockwise, we
have a continuous probability space S = (0, 2π]. If this is a fair
spinner (so that each outcome is equally likely), we measure
the probability of landing in a given interval as the length of
that interval divided by the length of the total interval.
For example, the probability of landing between 12:00 and
3:00 (or 0 radians and π2 radians) is 14 , since that section is
one-fourth of the entire wheel.
Lesson 2
Before working some examples, we point out some useful facts
about probability functions.
For mutually exclusive events A1 , . . . , An ,
S
P
P[ ni=1 Ai ] = ni=1 P[Ai ].
P[A ∪ B] = P[A] + P[B] − P[A ∩ B]
P[A] = P[A ∩ B] + P[A ∩ B 0 ]
P[A0 ] = 1 − P[A]
If B1 , . . . , Bn form a partition of the probability space,
P
then P[A] = ni=1 P[A ∩ Bi ]
Notice the similarities between these statements and those
about the number of elements in a set.
Lesson 2
Example (1.1)
Suppose that P[A ∩ B] = 0.3, P[A] = 0.5, and P[B] = 0.6.
Find
P[A0 ∪ B 0 ],
P[A0 ∩ B 0 ],
P[A0 ∪ B].
Lesson 2
Example (1.2)
Suppose that P[A] = 0.7, P[A ∩ B 0 ] = 0.1, and
P[A0 ∩ B 0 ] = 0.15. Find
P[A0 ∪ B],
P[A ∪ B],
P[A ∩ B].
Lesson 2
Example (1.3)
A survey is made to determine the number of households
having electric appliances in a certain city. It is found that
70% have radios (R), 55% have electric irons (I), 60% have
electric toasters (T), 30% have (IR), 45% have (RT), 25%
have (IT) and 15% have all three. Find the probability that a
household has at least one of these appliances.
Lesson 2
Example (1.4)
A survey is made to determine the number of households
having video game systems in a certain city. It is found that
40% have XBox Ones (X), 60% have Playstation 4’s (P), 30%
have Wii U’s (W), 10% have (XP), 15% have (PW), 15%
have (XW) and 5% have all three. Find the probability that a
household has at least two of these systems.
Lesson 2
Example (1.5)
In a survey of 130 students, the following data was obtained.
52 took English, 48 took Math, 56 took Chemistry, 17 took
English and Math, 13 took Math and Chemistry, 19 took
English and Chemistry, 7 took all three subjects.
Find the number of students who took
None of the subjects,
Math, but not English or Chemistry,
English and Math, but not Chemistry.
Lesson 2