Math 1300: Section 8-3 Conditional Probability, Intersection, and Independence

Conditional Probability
Intersection of Events: Product Rule
Probability Trees
Independent Events
Summary

Math 1300 Finite Mathematics
Section 8-3: Conditional Probability, Intersection, and
Independence

Jason Aubrey

Department of Mathematics
University of Missouri

Jason Aubrey Math 1300 Finite Mathematics

Probability Trees
Independent Events
Summary


Conditional probability is the probability of an event
occuring given that another event has already occured.


Probability Trees
Independent Events
Summary


Conditional probability is the probability of an event
occuring given that another event has already occured.
The conditional probability of event A occuring, given that
event B has occured is denoted P(A|B), and is read as
“probability of A, given B.”


Probability Trees
Independent Events
Summary

Deﬁnition
For events A and B in a arbitrary sample space S, we deﬁne
the conditional probability of A given B by

P(A ∩ B)
P(A|B) = P(B) = 0
P(B)


Probability Trees
Independent Events
Summary

0.25
A B
.35 0.15 .25


Probability Trees
Independent Events
Summary

0.25
A B
.35 0.15 .25

P(A ∩ B) = 0.15


Probability Trees
Independent Events
Summary

0.25
A B
.35 0.15 .25

P(A ∩ B) = 0.15
“out of all outcomes in the
sample space, 15% of
them are in A ∩ B.


Probability Trees
Independent Events
Summary

P(A|B) asks: out of all
0.25
outcomes in B, what
A B proportion of them are also
in A.
.35 0.15 .25

P(A ∩ B) = 0.15
“out of all outcomes in the P(A ∩ B)
P(A|B) =
sample space, 15% of P(B)


Probability Trees
Independent Events
Summary

0.25
outcomes in B, what
in A.
.35 0.15 .25
So, we can think of the
conditional probability as
restricting the sample
space.
P(A ∩ B) = 0.15
P(A|B) =


Probability Trees
Independent Events
Summary

0.25
outcomes in B, what
in A.
.35 0.15 .25
So, we can think of the
conditional probability as
restricting the sample
space.
P(A ∩ B) = 0.15
P(A|B) =
them are in A ∩ B. 0.15
= = 0.375
0.4


Probability Trees
Independent Events
Summary

Example: Suppose events A, B, D and E have probabilities as
given in the table.

A B Totals
D 0.2 0.4 0.6
E 0.28 0.12 0.4
Totals 0.48 0.52 1


Probability Trees
Independent Events
Summary

given in the table.

A B Totals
D 0.2 0.4 0.6
E 0.28 0.12 0.4
Totals 0.48 0.52 1

P(B ∩ D)
P(B|D) =
P(D)


Probability Trees
Independent Events
Summary

given in the table.

A B Totals
D 0.2 0.4 0.6
E 0.28 0.12 0.4
Totals 0.48 0.52 1

P(B ∩ D) .4 2
P(B|D) = = =
P(D) .6 3


Probability Trees
Independent Events
Summary

given in the table.

A B Totals
D 0.2 0.4 0.6
E 0.28 0.12 0.4
Totals 0.48 0.52 1

P(B ∩ D) .4 2
P(B|D) = = =
P(D) .6 3

P(A|E)


Probability Trees
Independent Events
Summary

given in the table.

A B Totals
D 0.2 0.4 0.6
E 0.28 0.12 0.4
Totals 0.48 0.52 1

P(B ∩ D) .4 2
P(B|D) = = =
P(D) .6 3
P(A ∩ E) 0.28
P(A|E) = = = .7
P(E) 0.4


Probability Trees
Independent Events
Summary

Example: In a survey of 255 people, the following data were
obtained relating gender to political orientation:
Republican (R) Democrat (D) Ind. (I) Total
Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255
A person is randomly selected. What is the probability that:


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255
The person is male?


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255
The person is male?
n(M) 118
P(M) = = ≈ .463
n(S) 255


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255
The person is male?
n(M) 118
P(M) = = ≈ .463
n(S) 255

The person is male and a Democrat?


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255
The person is male?
n(M) 118
P(M) = = ≈ .463
n(S) 255

The person is male and a Democrat?
n(M ∩ D) 22
P(M ∩ D) = = ≈ .086
n(S) 255

Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255

P(M) = .46
P(M ∩ D) = 0.086


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255

P(M) = .46
P(M ∩ D) = 0.086
a Democrat?


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255

P(M) = .46
P(M ∩ D) = 0.086
a Democrat?
n(D) 99
P(D) = = ≈ .39
n(S) 255


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255

P(M) = .46
P(M ∩ D) = 0.086
a Democrat?
n(D) 99
P(D) = = ≈ .39
n(S) 255

Male, given that the person is a Democrat?


Probability Trees
Independent Events
Summary

Male (M) 79 22 17 118
Female (F) 40 77 20 137
Total 119 99 37 255

P(M) = .46
P(M ∩ D) = 0.086
a Democrat?
n(D) 99
P(D) = = ≈ .39
n(S) 255

Male, given that the person is a Democrat?
P(M ∩ D) 0.086
P(M|D) = = ≈ .22
P(D) .39


Probability Trees
Independent Events
Summary

Note
Note that for sample spaces with equally likely outcomes

n(A ∩ B)
P(A|B) =
n(B)


Probability Trees
Independent Events
Summary

Note
Note that for sample spaces with equally likely outcomes

n(A ∩ B)
P(A|B) =
n(B)

So, for example, from the previous problem we could have done

n(M ∩ D) 22
P(M|D) = = ≈ .22
n(D) 99


Probability Trees
Independent Events
Summary


In the previous section we sometimes used the addition
principle

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

to ﬁnd P(A ∩ B).


Probability Trees
Independent Events
Summary


In the previous section we sometimes used the addition
principle

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

to ﬁnd P(A ∩ B).
The conditional probability formula gives another method
for ﬁnding P(A ∩ B) when appropriate, called the Product
Rule.


Probability Trees
Independent Events
Summary

P(A ∩ B)
P(A|B) =
P(B)


Probability Trees
Independent Events
Summary

P(A ∩ B)
P(A|B) =
P(B)
P(A ∩ B)
P(B)P(A|B) = P(B)
P(B)


Probability Trees
Independent Events
Summary

P(A ∩ B)
P(A|B) =
P(B)
P(A ∩ B)
P(B)P(A|B) = P(B)
P(B)
P(B)P(A|B) = P(A ∩ B)


Probability Trees
Independent Events
Summary

P(A ∩ B)
P(A|B) =
P(B)
P(A ∩ B)
P(B)P(A|B) = P(B)
P(B)
P(B)P(A|B) = P(A ∩ B)
P(A ∩ B) = P(A|B)P(B)


Probability Trees
Independent Events
Summary

Theorem
For events A and B with nonzero probabilities in a sample
space S,

P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)

and we can use either P(A)P(B|A) or P(B)P(A|B) to compute
P(A ∩ B).


Probability Trees
Independent Events
Summary

Example: A computer chip manufacturer produces 45% of its
chips at Plant A and at and the remainder at Plant B. However,
1% of the chips produced at Plant A are defective, while 0.5%
of the chips produced at Plant B are defective. What is the
probability that a randomly chosen computer chip produced by
this manufactuer is defective and was produced at Plant A?


Probability Trees
Independent Events
Summary

Let A be the event that a chip was produced at plant A. Let D
represent the event that a chip is defective.


Probability Trees
Independent Events
Summary

Let A be the event that a chip was produced at plant A. Let D
represent the event that a chip is defective. Then,

P(A) = 0.45 and P(A ) = 0.55

(Note that A represents the event that a chip was produced at
plant B.)


Probability Trees
Independent Events
Summary

The statement “1% of the chips produced at Plant A are
defective, gives the conditional probability

P(D|A) = 0.01


Probability Trees
Independent Events
Summary


P(D|A) = 0.01

Similarly, the statement ”0.5% of the chips produced at Plant B
are defective“ gives

P(D|A ) = 0.005


Probability Trees
Independent Events
Summary


P(D|A) = 0.01


P(D|A ) = 0.005

The question
What is the probability that a randomly chosen
computer chip produced by this manufactuer is
defective and was produced at Plant A?


Probability Trees
Independent Events
Summary


P(D|A) = 0.01


P(D|A ) = 0.005

The question
What is the probability that a randomly chosen
computer chip produced by this manufactuer is
defective and was produced at Plant A?

is asking for P(D ∩ A).

Probability Trees
Independent Events
Summary

By the product rule

P(D ∩ A) = P(D|A)P(A)


Probability Trees
Independent Events
Summary

By the product rule

P(D ∩ A) = P(D|A)P(A)
P(D ∩ A) = (0.01)(0.45) = 0.0045


Probability Trees
Independent Events
Summary

By the product rule

P(D ∩ A) = P(D|A)P(A)
P(D ∩ A) = (0.01)(0.45) = 0.0045

So, the probability that a randomly chosen computer chip
produced by this manufactuer is defective and was produced at
Plant A is 0.45%.


Probability Trees
Independent Events
Summary

Probability Trees

Probability Trees
We used tree diagrams in the previous chapter to help us
count the number of combined outcomes in a sequence of
experiments.


Probability Trees
Independent Events
Summary

Probability Trees

Probability Trees
experiments.
Similarly, we can use tree diagrams to help us compute the
probabilities of combined outcomes in a sequence of
experiments.


Probability Trees
Independent Events
Summary

Probability Trees

Probability Trees
experiments.
Similarly, we can use tree diagrams to help us compute the
probabilities of combined outcomes in a sequence of
experiments.
A seqence of experiments where the outcome of each
experiment is not certain is called a stochastic process.


Probability Trees
Independent Events
Summary

Constructing Probability Trees
Draw a tree diagram corresponding to all combined
outcomes of the sequence of experiments. Label each
node.
Assign a probability to each tree branch.
Use the results from the ﬁrst two steps to answer various
questions related to the sequence of experiements as a
whole.


Probability Trees
Independent Events
Summary

Example: Suppose that 5% of all adults over 40 have diabetes.
A certain physician correctly diagnoses 90% of all adults over
40 with diabetes as having the disease and incorrectly
diagnoses 3% of all adults over 40 without diabetes as having
the disease.


Probability Trees
Independent Events
Summary

Example: Suppose that 5% of all adults over 40 have diabetes.
A certain physician correctly diagnoses 90% of all adults over
40 with diabetes as having the disease and incorrectly
diagnoses 3% of all adults over 40 without diabetes as having
the disease.
(a) Let

A = the event ”has diabetes", and let
B = the event ”is diagnosed with diabetes."

Draw a tree diagram which models this scenario and label it
with appropriate probabilities.


Probability Trees
Independent Events
Summary

A = ”has diabetes" D = ”is diagnosed with diabetes."

”Suppose that 5% of all
adults over 40 have
diabetes.“

.05 A

.95

A


Probability Trees
Independent Events
Summary


adults over 40 have
.90 D
diabetes.“
”A certain physician .10
correctly diagnoses 90% of .05 A
all adults over 40 with D
diabetes as having the
disease. . . “ .95

A


Probability Trees
Independent Events
Summary


adults over 40 have
.90 D
diabetes.“
”A certain physician .10
correctly diagnoses 90% of .05 A
all adults over 40 with D
diabetes as having the
disease. . . “ .95 .03 D
and incorrectly diagnoses
3% of all adults over 40 A .97
without diabetes as having D
the disease.

Probability Trees
Independent Events
Summary

(b) Find the probability that a randomly
selected adult over 40 does not have .90 D
diabetes, and is diagnosed as having
diabetes. .10
.05 A
D

.95 .03 D

A .97
D


Probability Trees
Independent Events
Summary

diabetes. .10
.05 A
D
“does not have diabetes” (A )
.95 .03 D

A .97
D


Probability Trees
Independent Events
Summary

diabetes. .10
.05 A
D
“is diagnosed as having diabetes” .95 .03 D
(D )
A .97
D


Probability Trees
Independent Events
Summary

diabetes. .10
.05 A
D
“is diagnosed as having diabetes” .95 .03 D
(D )
P(A ∩ D) = P(A )P(D|A )P(A ) = A .97
(0.95)(0.03) = .0285 D


Probability Trees
Independent Events
Summary

(c) Find the probability that a randomly
selected adult over 40 is diagnosed as
not having diabetes.

.90 D

.10
.05 A
D

.95 .03 D

A .97
D


Probability Trees
Independent Events
Summary

not having diabetes. We are looking for
P(D ).

.90 D

.10
.05 A
D

.95 .03 D

A .97
D


Probability Trees
Independent Events
Summary

P(D ).
There are two cases. Such a person
either .90 D
does not have it and is (correctly) .10
diagnosed as not having it, .05 A
D
P(A ∩ D ) = P(A )P(D |A )
.95 .03 D

A .97
D


Probability Trees
Independent Events
Summary

P(D ).
There are two cases. Such a person
either .90 D
does not have it and is (correctly) .10
diagnosed as not having it, .05 A
D
P(A ∩ D ) = P(A )P(D |A )
.95 .03 D
has it, but is (incorrectly) diagnosed
as not having it,
A .97
P(A ∩ D ) = P(A)P(D |A) D


Probability Trees
Independent Events
Summary

Then

P(D ) = P(A ∩ D ) + P(A ∩ D )


Probability Trees
Independent Events
Summary

Then

P(D ) = P(A ∩ D ) + P(A ∩ D )
= P(A )P(D |A ) + P(A)P(D |A)


Probability Trees
Independent Events
Summary

Then

P(D ) = P(A ∩ D ) + P(A ∩ D )
= P(A )P(D |A ) + P(A)P(D |A)
= (0.95)(0.97) + (0.05)(0.1)


Probability Trees
Independent Events
Summary

Then

P(D ) = P(A ∩ D ) + P(A ∩ D )
= P(A )P(D |A ) + P(A)P(D |A)
= (0.95)(0.97) + (0.05)(0.1)
= 0.9215 + 0.005 = 0.9265


Probability Trees
Independent Events
Summary

Then

P(D ) = P(A ∩ D ) + P(A ∩ D )
= P(A )P(D |A ) + P(A)P(D |A)
= (0.95)(0.97) + (0.05)(0.1)
= 0.9215 + 0.005 = 0.9265

So, 92.65% of adults over 40 who take this test will be
diagnosed as not having diabetes.


Probability Trees
Independent Events
Summary

(d) Find the probability that a randomly selected adult over 40
actually has diabetes, given that he/she is diagnosed as not
having diabetes.


Probability Trees
Independent Events
Summary

having diabetes.
P(A ∩ D )
Here we are looking for P(A|D ) = .
P(D )


Probability Trees
Independent Events
Summary

having diabetes.
P(A ∩ D )
P(D )
From the previous part,

P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005


Probability Trees
Independent Events
Summary

having diabetes.
P(A ∩ D )
P(D )

P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005

And P(D ) = 0.9265. So,


Probability Trees
Independent Events
Summary

having diabetes.
P(A ∩ D )
P(D )

P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005

And P(D ) = 0.9265. So,
P(A ∩ D ) 0.005
P(A|D ) = = ≈ 0.0054
P(D ) 0.9265


Probability Trees
Independent Events
Summary

having diabetes.
P(A ∩ D )
P(D )

P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005

And P(D ) = 0.9265. So,
P(A ∩ D ) 0.005
P(A|D ) = = ≈ 0.0054
P(D ) 0.9265
So 0.54% of adults diagnosed as not having diabetes by this
doctor actually do have the disease.

Probability Trees
Independent Events
Summary

Independent Events

In probability, to say that two events are independent
intuitively means that the occurence of one event makes it
neither more nor less probable that the other occurs.


Probability Trees
Independent Events
Summary

Independent Events

For example, if two cards are drawn with replacement from
a deck of cards, the event of drawing a red card on the ﬁrst
trial and that of drawing a red card on the second trial are
independent.


Probability Trees
Independent Events
Summary

Independent Events

For example, if two cards are drawn with replacement from
a deck of cards, the event of drawing a red card on the ﬁrst
trial and that of drawing a red card on the second trial are
independent.
By contrast, if two cards are drawn without replacement
from a deck of cards, the event of drawing a red card on
the ﬁrst trial and that of drawing a red card on the second
trial are not independent.


Probability Trees
Independent Events
Summary

Independent Events

Caution: very often intuition is not a reliable guide to the
notion of independence. Must use mathematics to verify
independence, not intuition.


Probability Trees
Independent Events
Summary

Independent Events

In problems we will either


Probability Trees
Independent Events
Summary

Independent Events

be asked to determine whether or not two events are
independent, or


Probability Trees
Independent Events
Summary

Independent Events

be asked to determine whether or not two events are
independent, or
be asked to assume that two events are independent and to
use that assumption to solve the problem.


Probability Trees
Independent Events
Summary

Deﬁnition (Independence)
If A and B are any events in a sample space S, we say that A
and B are independent if and only if

P(A ∩ B) = P(A)P(B)

Otherwise, A and B are said to be dependent.


Probability Trees
Independent Events
Summary

Example: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 are
the events A and B independent?


Probability Trees
Independent Events
Summary

We are asked to determine whether or not two events are
independent. To do this, we must determine whether or not

P(A ∩ B) = P(A)P(B)


Probability Trees
Independent Events
Summary


P(A ∩ B) = P(A)P(B)

We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B).


Probability Trees
Independent Events
Summary


P(A ∩ B) = P(A)P(B)

We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). This is a
perfect place to use the addition principle:


Probability Trees
Independent Events
Summary


P(A ∩ B) = P(A)P(B)


P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.5 − 0.9 = 0.4


Probability Trees
Independent Events
Summary


P(A ∩ B) = P(A)P(B)


P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.5 − 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4


Probability Trees
Independent Events
Summary


P(A ∩ B) = P(A)P(B)


P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.5 − 0.9 = 0.4

We also have P(A)P(B) = (0.8)(0.5) = 0.4
Since P(A ∩ B) = P(A)P(B) we conclude that A and B are
independent events.

Probability Trees
Independent Events
Summary

Example: On a stormy night, the probability that the electricity
goes out is 7% and the probability that the phone goes out is
3%. Assuming that the two are independent, what is the
probability that neither the electricity nor the phone goes out on
a stormy night?


Probability Trees
Independent Events
Summary

a stormy night?
Here we are asked to assume that two events are independent,
and we will use this assumption to solve the problem


Probability Trees
Independent Events
Summary

a stormy night?
Let E represent the electricity going out. So P(E) = 0.07.


Probability Trees
Independent Events
Summary

a stormy night?
Let F represent the phone going out. So P(F ) = 0.03.


Probability Trees
Independent Events
Summary

a stormy night?
Let F represent the phone going out. So P(F ) = 0.03.
The probability that neither the electricity nor the phone
goes out is given by

P(E ∩ F )


Probability Trees
Independent Events
Summary

If E and F are independent, so are E and F . So

P(E ∩ F ) = P(E )P(F )


Probability Trees
Independent Events
Summary


P(E ∩ F ) = P(E )P(F )

Then to solve the problem:

P(E ∩ F ) = P(E )P(F )


Probability Trees
Independent Events
Summary


P(E ∩ F ) = P(E )P(F )


P(E ∩ F ) = P(E )P(F )
= (1 − P(E))(1 − P(F ))


Probability Trees
Independent Events
Summary


P(E ∩ F ) = P(E )P(F )


P(E ∩ F ) = P(E )P(F )
= (1 − P(E))(1 − P(F ))
= (1 − 0.07)(1 − 0.03) = (0.93)(0.97)


Probability Trees
Independent Events
Summary


P(E ∩ F ) = P(E )P(F )


P(E ∩ F ) = P(E )P(F )
= (1 − P(E))(1 − P(F ))
= (1 − 0.07)(1 − 0.03) = (0.93)(0.97)
= 0.9021


Probability Trees
Independent Events
Summary


P(E ∩ F ) = P(E )P(F )


P(E ∩ F ) = P(E )P(F )
= (1 − P(E))(1 − P(F ))
= (1 − 0.07)(1 − 0.03) = (0.93)(0.97)
= 0.9021

So, the probability that neither the electricity nor the phone
goes out is 90.21%.


Probability Trees
Independent Events
Summary

Theorem
If A and B are independent events with nonzero probabilities in
a sample space S, then

P(A|B) = P(B) and P(B|A) = P(B)

If either equation holds, then A and B are independent.


Probability Trees
Independent Events
Summary

Example: A card is drawn from a standard 52 card deck.
Events M and N are

M = the drawn card is a diamond (♦)
N = the drawn card is even (face cards are not valued)


Probability Trees
Independent Events
Summary

Events M and N are


(a) Find P(N|M).


Probability Trees
Independent Events
Summary

Events M and N are


(a) Find P(N|M).
Notice that N ∩ M is the set of all even diamonds. This is,

N ∩ M = {2♦, 4♦, 6♦, 8♦, 10♦}


Probability Trees
Independent Events
Summary

Events M and N are


(a) Find P(N|M).
Notice that N ∩ M is the set of all even diamonds. This is,

N ∩ M = {2♦, 4♦, 6♦, 8♦, 10♦}

So n(N ∩ M) = 5.


Probability Trees
Independent Events
Summary

M is the set of all diamonds, so n(M) = 13.


Probability Trees
Independent Events
Summary


n(N ∩ M)
P(N|M) =
n(M)


Probability Trees
Independent Events
Summary


n(N ∩ M) 5
P(N|M) = =
n(M) 13


Probability Trees
Independent Events
Summary


n(N ∩ M) 5
P(N|M) = =
n(M) 13

(b) Test M and N for independence.


Probability Trees
Independent Events
Summary


n(N ∩ M) 5
P(N|M) = =
n(M) 13

Here we could check whether P(M ∩ N) = P(M)P(N), but it is
slightly more convenient to check whether P(N|M) = P(N).


Probability Trees
Independent Events
Summary


n(N ∩ M) 5
P(N|M) = =
n(M) 13

N is the set of all even cards in the deck. There are 5 even
cards per suit and 4 suits, so n(N) = 20.


Probability Trees
Independent Events
Summary


n(N ∩ M) 5
P(N|M) = =
n(M) 13

cards per suit and 4 suits, so n(N) = 20. Then

20 5
P(N) = =
52 13


Probability Trees
Independent Events
Summary


n(N ∩ M) 5
P(N|M) = =
n(M) 13

cards per suit and 4 suits, so n(N) = 20. Then

20 5
P(N) = =
52 13
P(N|M) = P(N) so yes, N and M are independent.

Probability Trees
Independent Events
Summary

Deﬁnition (Independent Set of Events)
A set of events is said to be independent if for each ﬁnite
subset {E1 , E2 , . . . , Ek }

P(E1 ∩ E2 ∩ · · · ∩ Ek ) = P(E1 )P(E2 ) · · · P(Ek )


Probability Trees
Independent Events
Summary

Example: One interpretation of a baseball player’s batting
average is as the probability of getting a hit each time the player
goes to bat.


Probability Trees
Independent Events
Summary

goes to bat.
For instance, a player with a .300 average has probability .3 of
getting a hit.


Probability Trees
Independent Events
Summary

goes to bat.
For instance, a player with a .300 average has probability .3 of
getting a hit.
If a player with a .300 batting average bats four times in a game
and each at-bat is an independent event, what is the probability
of the player getting at least one hit in the game?


Probability Trees
Independent Events
Summary

Let Hi be the probability that the player gets a hit at his i th time
at bat. Then


Probability Trees
Independent Events
Summary

at bat. Then

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 )


Probability Trees
Independent Events
Summary

at bat. Then

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 )
= 1 − P(H1 )P(H2 )P(H3 )P(H4 )


Probability Trees
Independent Events
Summary

at bat. Then

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 )
= 1 − P(H1 )P(H2 )P(H3 )P(H4 )
= 1 − (0.7)4 = 0.7599


Probability Trees
Independent Events
Summary

at bat. Then

P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 )
= 1 − P(H1 )P(H2 )P(H3 )P(H4 )
= 1 − (0.7)4 = 0.7599

So if a player with a batting average of .300 bats four times in a
game, then there is about a 76% chance of that player getting a
hit.


Probability Trees
Independent Events
Summary

Summary


P(A ∩ B) P(B ∩ A)
P(A|B) = P(B|A) =
P(B) P(A)

Note: P(A|B) is a probability based on the new sample space
B, while P(A ∩ B) is based on the original sample space S.
Product Rule

P(A ∩ B) = P(B|A)P(A) = P(A|B)P(B)


Probability Trees
Independent Events
Summary

Summary
Independent Events
A and B are independent if and only if
P(A ∩ B) = P(A)P(B)
If A and B are independent events with nonzero
probabilities, then
P(A|B) = P(A) and P(B|A) = P(A)
If A and B are independent events with nonzero
probabilities and either P(A|B) = P(A) or P(B|A) = P(B),
then A and B are independent.
If E1 , E2 , . . . , En are independent, then
P(E1 ∩ E2 ∩ . . . ∩ En ) = P(E1 )P(E2 ) · · · P(En )

Math 1300: Section 8-3 Conditional Probability, Intersection, and Independence

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Math 1300: Section 8-3 Conditional Probability, Intersection, and Independence