1MA01: Probability
Sinéad Ryan
TCD
November 12, 2013
Definitions and Notation
EVENT: a set possible outcomes of an experiment. Eg
flipping a coin is the experiment, landing on heads is
the event
If an experiment is repeated many times then
number of times event A happens
number of times experiment repeated
is called the Probability of event A, written P(A)
An example:
Roll a die 1000 times and record the frequency with
which each number appears giving something like
VALUE
1
2
3
4
5
6
FREQ
166
176
158
178
153
169
P6
Note that i=1 P(i) = 1.
P(1) =
P(2) =
166
1000
176
1000
..
..
. = .
Note that if P is a probability, 0 ≤ P ≤ 1
= 0.166
= 0.176
Theorem: The Principle of Equally Likely Outcomes
If an experiment is conducted with n possible outcomes,
all equally likely, and A is an event with k possible outcomes then
‚
Œ
k
#favourable outcomes for A
P(A) = =
n
#possible outcomes
Example
Consider a jar containing 7 black balls, 6 yellow balls, 4
green and 3 red balls. The jar is shaken and 1 ball
removed without looking.
What is the probability that
the ball is red?
Theorem: The Principle of Equally Likely Outcomes
If an experiment is conducted with n possible outcomes,
all equally likely, and A is an event with k possible outcomes then
‚
Œ
k
#favourable outcomes for A
P(A) = =
n
#possible outcomes
Example
Consider a jar containing 7 black balls, 6 yellow balls, 4
green and 3 red balls. The jar is shaken and 1 ball
removed without looking.
What is the probability that
the ball is red? P(red) =
3
20
Theorem: The Principle of Equally Likely Outcomes
If an experiment is conducted with n possible outcomes,
all equally likely, and A is an event with k possible outcomes then
‚
Œ
k
#favourable outcomes for A
P(A) = =
n
#possible outcomes
Example
Consider a jar containing 7 black balls, 6 yellow balls, 4
green and 3 red balls. The jar is shaken and 1 ball
removed without looking.
What is the probability that
the ball is red? P(red) =
the ball is white?
3
20
Theorem: The Principle of Equally Likely Outcomes
If an experiment is conducted with n possible outcomes,
all equally likely, and A is an event with k possible outcomes then
‚
Œ
k
#favourable outcomes for A
P(A) = =
n
#possible outcomes
Example
Consider a jar containing 7 black balls, 6 yellow balls, 4
green and 3 red balls. The jar is shaken and 1 ball
removed without looking.
What is the probability that
the ball is red? P(red) =
3
20
the ball is white? P(white) =
0
20
=0
Theorem: The Principle of Equally Likely Outcomes
If an experiment is conducted with n possible outcomes,
all equally likely, and A is an event with k possible outcomes then
‚
Œ
k
#favourable outcomes for A
P(A) = =
n
#possible outcomes
Example
Consider a jar containing 7 black balls, 6 yellow balls, 4
green and 3 red balls. The jar is shaken and 1 ball
removed without looking.
What is the probability that
the ball is red? P(red) =
3
20
the ball is white? P(white) =
0
20
=0
the ball is either black, yellow, green or red?
Theorem: The Principle of Equally Likely Outcomes
If an experiment is conducted with n possible outcomes,
all equally likely, and A is an event with k possible outcomes then
‚
Œ
k
#favourable outcomes for A
P(A) = =
n
#possible outcomes
Example
Consider a jar containing 7 black balls, 6 yellow balls, 4
green and 3 red balls. The jar is shaken and 1 ball
removed without looking.
What is the probability that
the ball is red? P(red) =
3
20
the ball is white? P(white) =
0
20
=0
the ball is either black, yellow, green or red?
20
P(b, y, g, r) = 20
=1
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
drawing an ace?
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
drawing a king?
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
drawing a king? P(king) =
4
52
=
1
13
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
drawing a king? P(king) =
not drawing a king?
4
52
=
1
13
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
drawing a king? P(king) =
4
52
=
1
13
not drawing a king? P(not a king) =
48
52
=
12
13
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
drawing a king? P(king) =
4
52
=
1
13
not drawing a king? P(not a king) =
drawing an ace or a king?
48
52
=
12
13
Addition, Complement and Multiplication
Addition: suppose A and B are disjoint events. Then
P(A or B) = P(A) + P(B)
Two events are disjoint of they cannot occur in the
same experiment. Eg. draw one card from a pack drawing an ace and a king are disjoint events in this
experiment since they cannot both occur.
Let’s look at an example of adding probabilities.
Consider a deck of cards. Shuffle and draw 1 card.
What is the probability of
4
1
drawing an ace? P(ace) = 52
= 13
drawing a king? P(king) =
4
52
=
1
13
not drawing a king? P(not a king) = 48
=
52
drawing an ace or a king?
2
P(ace or king) = P(ace) + P(king) = 13
12
13
The assumption of disjoint events is crucial for addition
of probabilities. So, eg. drawing an ace and drawing a
heart are not disjoint so addition rules don’t hold ie.
P(ace or heart) 6= P(ace) + P(heart)
Note above that P(king) = 1/ 13 and P(not king) = 12/ 13.
These disjoint events are complements and since
exactly one of these must happen then their
probabilities sum to 1
P(king) + P(not king) = 1 → P(not king) = 1 − P(king)
In general then, if A is an event then P(not A) = 1 − P(A).
Multiplication rule
Let A and B be 2 events, then
P(A and B) = P(A)P(B|A)
P(B|A) is called a conditional probability: it is the
probability of event B occuring given event A has
already occurred.
Eg. Consider a deck of cards. Draw 2 cards.
What is the probability the 1st card is a heart?
Multiplication rule
Let A and B be 2 events, then
P(A and B) = P(A)P(B|A)
P(B|A) is called a conditional probability: it is the
probability of event B occuring given event A has
already occurred.
Eg. Consider a deck of cards. Draw 2 cards.
What is the probability the 1st card is a heart?
13
P(A) = 52
= 14
Multiplication rule
Let A and B be 2 events, then
P(A and B) = P(A)P(B|A)
P(B|A) is called a conditional probability: it is the
probability of event B occuring given event A has
already occurred.
Eg. Consider a deck of cards. Draw 2 cards.
What is the probability the 1st card is a heart?
13
P(A) = 52
= 14
Given it’s a heart what is the probability the 2nd
card is a heart?
Multiplication rule
Let A and B be 2 events, then
P(A and B) = P(A)P(B|A)
P(B|A) is called a conditional probability: it is the
probability of event B occuring given event A has
already occurred.
Eg. Consider a deck of cards. Draw 2 cards.
What is the probability the 1st card is a heart?
13
P(A) = 52
= 14
Given it’s a heart what is the probability the 2nd
4
card is a heart?P(B|A) = 12
= 17
52
Multiplication rule
Let A and B be 2 events, then
P(A and B) = P(A)P(B|A)
P(B|A) is called a conditional probability: it is the
probability of event B occuring given event A has
already occurred.
Eg. Consider a deck of cards. Draw 2 cards.
What is the probability the 1st card is a heart?
13
P(A) = 52
= 14
Given it’s a heart what is the probability the 2nd
4
card is a heart?P(B|A) = 12
= 17
52
What is the probability both cards are hearts?
Multiplication rule
Let A and B be 2 events, then
P(A and B) = P(A)P(B|A)
P(B|A) is called a conditional probability: it is the
probability of event B occuring given event A has
already occurred.
Eg. Consider a deck of cards. Draw 2 cards.
What is the probability the 1st card is a heart?
13
P(A) = 52
= 14
Given it’s a heart what is the probability the 2nd
4
card is a heart?P(B|A) = 12
= 17
52
What is the probability both cards are hearts?
1
P(A and B) = P(A)P(B|A) = 17
In the last example you could also use the P.E.L.O. and
list all 2652 ways the 1st and 2nd cards could be dealt.
Clearly using the multiplication rule is easier.
Independence
Events A and B are independent if P(B|A) = P(B) and in
this case P(A and B) = P(A)P(B).
Note that independence is not the same thing as
disjoint.
A simple example of independence: roll a die twice.
What is the probability of rolling 2 sixes?
P(six and six) = P(A)P(B|A) = P(A)P(B) =
11
66
=
1
36
since the result of each roll is independent of the other.