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Elementary Probability
by James Bernhard
Suppose a coin is equally likely to land heads or tails
If we flip this coin twice, there are four possible outcomes: HH,
HT, TH, and TT
This can be depicted graphically as follows. . .
Possible Outcomes for 2 Coin Tosses
HT
HH
TH
TT
The pie is divided into 4 slices (in no particular order) since there
are 4 possible outcomes
The slices are equal sizes since each outcome is equally likely
Sample Space for 2 Coin Tosses
HT
HH
TH
TT
The whole pie can be thought of as depicting the collection of all
possible outcomes, which is called the sample space
Outcomes and collections are outcomes are called events, and can
be depicted similarly
The Event HH or TH
The Event TT
HT
HH
HT
HH
TH
TT
TH
TT
The Event HT, TH or TT
The Event HH, HT, TH, or TT
HT
HH
HT
HH
TH
TT
TH
TT
The probability of an event can be thought of as the fraction of
the time that the event would occur in the long run over many
repetitions of the process
Graphically, the probability of an event can be depicted as the
fraction of the pie which the event occupies
From either point of view, probability is a number between 0 and 1
In addition to being shown by the areas, probabilities of individual
outcomes can also be written into the picture explicitly as well. . .
Sample Space for 2 Coin Tosses
with Probabilities
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
Since the event that is the whole sample space takes up the whole
pie, its probability must always be 1
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
In this case, .25 + .25 + .25 + .25 = 1
The complement of an event is the collection of all outcomes not
included in that event
The Event HH, HT, or TH
The Complement of HH, HT, or TH
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
The complement of an event is the collection of all outcomes not
included in that event
The Event HH, HT, or TH
The Complement of HH, HT, or TH
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
The probability of an event’s complement equals 1 minus the
probability of the event
Two events are disjoint if they do not overlap in the sample space
Disjoint:
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
Not disjoint:
For two disjoint events, the probability of one or the other is the
sum of the probabilities of the individual events
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
In this case, .5 + .25 = .75, as combining them shows:
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
In general, if a random process has n equally likely outcomes, each
one will have a probability of 1/n (shown here for n = 8, with
1/8 = .125):
Event 3
(0.125)
Event 2
(0.125)
Event 4
(0.125)
Event 1
(0.125)
Event 5
(0.125)
Event 8
(0.125)
Event 6
(0.125)
Event 7
(0.125)
In this case, an event made up of several outcomes will have
probability equal to
number of outcomes in the event
number of outcomes in the sample space
Event 3
(0.125)
Event 2
(0.125)
Event 4
(0.125)
Event 1
(0.125)
Event 5
(0.125)
Event 8
(0.125)
Event 6
(0.125)
Event 7
(0.125)
To keep things simple, let’s return to the example of tossing 2 coins
Sample Space for 2 Coin Tosses
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
Two events in this sample space are:
Event A The 2 tosses come up different from each other.
Event B The second toss comes up heads.
Event A
Event B
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
Event A
Event B
HT
(0.25)
HH
(0.25)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
TH
(0.25)
TT
(0.25)
What is the probability that Events A and B both occur?
This can be seen graphically by looking at their intersection
Events A and B
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
In order to compute the probability of both Events A and B
occurring, it is useful to understand conditional probability
The probability of Event B, given the information that Event A has
occurred, is called the conditional probability of B given A, written
P(B|A)
In the picture, P(B|A) is given by the area of B that lies within A,
expressed as a fraction of the area of A, so
P(B|A) =
P(A and B)
Area of the intersection of A and B
=
Area of A
P(A)
HT
(0.25)
HH
(0.25)
TH
(0.25)
TT
(0.25)
In this case,
P(B|A) =
.25
= .5
.5
In general, if we want to know the probability that both Events A
and B will occur, we have:
P(A and B) = Area of the intersection of A and B
= (Area of A) · P(B|A)
= P(A) P(B|A)
That is,
P(A and B) = P(A) P(B|A)
Geometrically, this says that the area of the intersection of A and B
equals the area of A times the fraction of B that is in A
Now assume that Events A and B are any two events with nonzero
probabilities
Events A and B are called independent if P(B|A) = P(B), which
means geometrically that the fraction of B that is in A equals the
fraction of B in the sample space
(This turns out to be the same as the condition P(A|B) = P(A))
In other words, events are independent if knowing that one of them
is true tells you nothing new about the probability of the other one
For independent Events A and B, the formula for the probability
that both will happen simplifies to
P(A and B) = P(A) P(B)
Geometrically, this simplification is due to the fact that the
fraction of B that is in A equals the fraction of B that is in the
sample space