Introductory Probability
Conditional Probability: Restricting the Sample Space
Nicholas Nguyen
[email protected]
Department of Mathematics
UK
February 15, 2017
Agenda
I Conditional Probability: Recalculating with Additional
Information
I Bernoulli Trials and Conditional Probability
Announcement: The fth homework is not available yet.
There is no quiz this upcoming Friday.
On A Game Show
You are a contestant on the Who Wants to Be a Millionaire
game show, and are stuck on a question with 4 choices A-D.
P(choose
right answer)
=
1 right answer
4 choices total
= 1/4.
Before 2008, there was a lifeline called 50:50, which could be
used to remove 2 wrong answer choices. Let's say it removed
C and D.
P(choose
right answer)
=
1 right answer
2 choices left
= 1/2.
On A Game Show
You are a contestant on the Who Wants to Be a Millionaire
game show, and are stuck on a question with 4 choices A-D.
P(choose
right answer)
=
1 right answer
4 choices total
= 1/4.
Before 2008, there was a lifeline called 50:50, which could be
used to remove 2 wrong answer choices. Let's say it removed
C and D.
P(choose
right answer)
=
1 right answer
2 choices left
= 1/2.
On A Game Show
You are a contestant on the Who Wants to Be a Millionaire
game show, and are stuck on a question with 4 choices A-D.
P(choose
right answer)
=
1 right answer
4 choices total
= 1/4.
Before 2008, there was a lifeline called 50:50, which could be
used to remove 2 wrong answer choices. Let's say it removed
C and D.
P(choose
right answer)
=
1 right answer
2 choices left
= 1/2.
Millionaire Formalities
The sample space is the 4 choices you can guess:
Ω = {A, B, C , D}.
Let's say A is the correct answer, so that the event
F
given by
you guess correctly is you guess A:
F = {A}.
Then assuming uniform probability,
F
= 1/4.
4 outcomes in Ω
away, then let E be the event
P(F ) =
If C and D are taken
1 outcome in
two remaining choices you can still guess:
E = {A, B}.
with the
Millionaire Formalities
The sample space is the 4 choices you can guess:
Ω = {A, B, C , D}.
Let's say A is the correct answer, so that the event
F
given by
you guess correctly is you guess A:
F = {A}.
Then assuming uniform probability,
F
= 1/4.
4 outcomes in Ω
away, then let E be the event
P(F ) =
If C and D are taken
1 outcome in
two remaining choices you can still guess:
E = {A, B}.
with the
Millionaire Formalities
The sample space is the 4 choices you can guess:
Ω = {A, B, C , D}.
Let's say A is the correct answer, so that the event
F
given by
you guess correctly is you guess A:
F = {A}.
Then assuming uniform probability,
F
= 1/4.
4 outcomes in Ω
away, then let E be the event
P(F ) =
If C and D are taken
1 outcome in
two remaining choices you can still guess:
E = {A, B}.
with the
Let's say A is the correct answer, so that the event
F
given by
you guess correctly is you guess A:
F = {A}.
Then assuming uniform probability,
F
= 1/4.
4 outcomes in Ω
away, then let E be the event
P(F ) =
If C and D are taken
1 outcome in
with the
two remaining choices you can still guess:
E = {A, B}.
Then
P(F ,
It's like
E
given
E) =
F
= 1/2.
in E
1 outcome in
2 outcomes
is a new sample space.
Discrete Conditional Probability
Ω be a discrete sample space, and E be an event with
P(E ) 6= 0. For any event F of Ω, the conditional probability
that F occurs given that E occurs is
Let
P(F |E ) =
P(F ∩ E )
.
P(E )
Discrete Conditional Probability
Ω be a discrete sample space, and E be an event with
P(E ) 6= 0. For any event F of Ω, the conditional probability
that F occurs given that E occurs is
Let
P(F |E ) =
P(F ∩ E )
.
P(E )
We need the intersection: outcomes in
F
that are not in
cannot occur if we are given that an outcome in
E
E
occurred.
Discrete Conditional Probability
Distribution Function
Let
Ω
be a discrete sample space,
m
be a probability
E be an event with P(E ) 6= 0.
m to give the probability that
{ω} occurs given that E occurred. Given m(ω), if ω is not in
E , then if E occurred, then ω did not occur, so
distribution function on
For each
ω ∈ Ω,
Ω,
and
we can modify
m(ω|E ) = 0.
ω is in E , then we can reweigh
of ω occurring relative to E :
If
m(ω|E ) =
Then
E.
m(ω|E )
is the
or rescale the probability
m(ω)
.
P(E )
conditional distribution
given (relative to)
Discrete Conditional Probability
Distribution Function
Then
m(ω|E )
is the
conditional distribution
given (relative to)
E.
To summarize, given a distribution function

ω
0,
m(ω|E ) = m(ω)

, ω
P(E )
m
not in
in
E.
on
E,
Ω,
For any event
F,
which are also in
in
E,
we can split
E (F ∩ E ),
F
into two cases: outcomes in
and outcomes in
so they are also in the complement of
F
E
which are not
instead (F
Remember that
P(F ∩ E ) =
m(ω).
∑
ω in F ∩E
Then
P(F |E ) =
m(ω|E )
∑
ω in F
=
m(ω|E ) +
∑
ω in F ∩E
=
∑
ω in F ∩E
=
1
P(E )
m(ω|E )
∑
ω in F ∩Ẽ
m(ω)
+
P(E )
∑
ω in F ∩E
0
∑
ω in F ∩Ẽ
m(ω) =
F
P(F ∩ E )
.
F (E )
∩ Ẽ ).
Intersection?
We have an 8-sided fair die. Let
F
be the event
E
be the event
{1, 4, 7}
and
{1, 8}.Then
3
P(E ) = ,
8
and
P(F ) =
2
8
1
= .
4
Note that
F ∩ E = {1}.
If
E
occurs, then the die could not have rolled an 8 since 8 is
not in
E.
It could roll a 1 since that is in
P(F |E ) =
E.
Hence
P(roll a 1) P(F ∩ E ) 1/8
=
=
=
P(E )
P(E )
3/8
1/3.
Intersection?
We have an 8-sided fair die. Let
F
be the event
E
be the event
{1, 4, 7}
and
{1, 8}.Then
3
P(E ) = ,
8
and
P(F ) =
2
8
1
= .
4
Note that
F ∩ E = {1}.
If
E
occurs, then the die could not have rolled an 8 since 8 is
not in
E.
It could roll a 1 since that is in
P(F |E ) =
E.
Hence
P(roll a 1) P(F ∩ E ) 1/8
=
=
=
P(E )
P(E )
3/8
1/3.
Order Matters
Note that
F ∩ E = {1}.
If
E
occurs, then the die could not have rolled an 8 since 8 is
not in
E.
It could roll a 1 since that is in
E.
Hence
P(roll a 1) P(F ∩ E ) 1/8
=
=
= 1/3.
P(E )
P(E )
3/8
What is P(E |F )? If F occurs, then the die could not have
rolled a 4 or 7 since they are not in F . It could roll a 1 since
that is in F .
P(F |E ) =
Hence
P(E |F ) =
P(roll a 1) P(E ∩ F ) 1/8
=
=
=
P(F )
P(F )
2/8
This is not the same as
P(F |E ).
1/2.
Order Matters
If
E
occurs, then the die could not have rolled an 8 since 8 is
not in
E.
It could roll a 1 since that is in
E.
Hence
P(F |E ) =
What is
P(roll a 1) P(F ∩ E ) 1/8
=
=
=
P(E )
P(E )
3/8
P(E |F )?
If
F
occurs, then the die could not have
rolled a 4 or 7 since they are not in
that is in
1/3.
F.
It could roll a 1 since
F.
Hence
P(roll a 1) P(E ∩ F ) 1/8
=
=
=
P(F )
P(F )
2/8
same as P(F |E ).
P(E |F ) =
This is not the
1/2.
Bernoulli Trials
We toss a fair coin
heads (p
Let
F
n=3
times and record the number of
= 1/2).
be the event there are exactly two heads. Then
P(F ) = b(3, 1/2, 2) =
3
2
(1/2)2 (1/2)3−2 = 3/8.
What if we knew the rst toss was heads? Let
event. Let us compute
P(F |E )
E
be this
using restricted sample spaces.
If the rst toss was heads, then to get two heads in total, we
need one head among the remaining two tosses.
This is a binomial probability for 2 trials with
success:
p = 1/2
and 1
Bernoulli Trials
We toss a fair coin
heads (p
Let
F
n=3
times and record the number of
= 1/2).
be the event there are exactly two heads. Then
P(F ) = b(3, 1/2, 2) =
3
2
(1/2)2 (1/2)3−2 = 3/8.
What if we knew the rst toss was heads? Let
event. Let us compute
P(F |E )
E
be this
using restricted sample spaces.
If the rst toss was heads, then to get two heads in total, we
need one head among the remaining two tosses.
This is a binomial probability for 2 trials with
success:
p = 1/2
and 1
Bernoulli Trials
We toss a fair coin
heads (p
Let
F
n=3
times and record the number of
= 1/2).
be the event there are exactly two heads. Then
P(F ) = b(3, 1/2, 2) =
3
2
(1/2)2 (1/2)3−2 = 3/8.
What if we knew the rst toss was heads? Let
event. Let us compute
P(F |E )
E
be this
using restricted sample spaces.
If the rst toss was heads, then to get two heads in total, we
need one head among the remaining two tosses.
This is a binomial probability for 2 trials with
success:
p = 1/2
and 1
Bernoulli Trials
We toss a fair coin
heads (p
Let
F
n=3
times and record the number of
= 1/2).
be the event there are exactly two heads. Then
P(F ) = b(3, 1/2, 2) =
3
2
(1/2)2 (1/2)3−2 = 3/8.
What if we knew the rst toss was heads? Let
event. Let us compute
P(F |E )
E
be this
using restricted sample spaces.
If the rst toss was heads, then to get two heads in total, we
need one head among the remaining two tosses.
This is a binomial probability for 2 trials with
success:
p = 1/2
and 1
Bernoulli Trials
We toss a fair coin
heads (p
Let
F
n=3
times and record the number of
= 1/2).
be the event there are exactly two heads. Then
P(F ) = b(3, 1/2, 2) =
3
2
(1/2)2 (1/2)3−2 = 3/8.
What if we knew the rst toss was heads? Let
event. Let us compute
P(F |E )
E
be this
using restricted sample spaces.
If the rst toss was heads, then to get two heads in total, we
need one head among the remaining two tosses.
This is a binomial probability for 2 trials with
success:
p = 1/2
and 1
We toss a fair coin
heads (p
Let
F
n=3
times and record the number of
= 1/2).
be the event there are exactly two heads. Then
P(F ) = b(3, 1/2, 2) =
3
2
(1/2)2 (1/2)3−2 = 3/8.
What if we knew the rst toss was heads? Let
event. Let us compute
P(F |E )
E
be this
using restricted sample spaces.
If the rst toss was heads, then to get two heads in total, we
need one head among the remaining two tosses.
This is a binomial probability for 2 trials with
p = 1/2
and 1
success:
P(F |E ) = b(2, 1/2, 1) =
2
1
(1/2)1 (1/2)2−1 =
1/2
.
The Sequences with Two Heads
There are eight sequences of H's and T's in 3 tosses:
HHH, HHT , HTH, THH,
HTT , THT , TTH, TTT .
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences whose rst toss is heads, we get
E = {HHH, HHT , HTH, HTT },
The Sequences with Two Heads
There are eight sequences of H's and T's in 3 tosses:
HHH, HHT , HTH, THH,
HTT , THT , TTH, TTT .
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences whose rst toss is heads, we get
E = {HHH, HHT , HTH, HTT },
The Sequences with Two Heads
There are eight sequences of H's and T's in 3 tosses:
HHH, HHT , HTH, THH,
HTT , THT , TTH, TTT .
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences whose rst toss is heads, we get
E = {HHH, HHT , HTH, HTT },
The Sequences with Two Heads
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences whose rst toss is heads, we get
E = {HHH, HHT , HTH, HTT },
and 2 out of 4 of them have two H's:
P(exactly
2 heads given rst toss is heads)
= 2/4 =
1/2
.
Bernoulli Trials 2
What if we knew there is at least one head among the three
tosses?
Let
E
be this event. It is a trials process with at least 1
success. Use the complement no successes:
P(E ) = 1 − b(3, 1/2, 0)
3
= 1−
(1/2)0 (1/2)3
0
= 1 − 1/8 = 7/8.
Then
F ∩E
is the event there is at least one head, and there
are two heads total, which is the same as there are two
heads total,
F
itself:
P(F ∩ E ) = P(F ) = 3/8
(in this case).
Bernoulli Trials 2
What if we knew there is at least one head among the three
tosses?
Let
E
be this event. It is a trials process with at least 1
success. Use the complement no successes:
P(E ) = 1 − b(3, 1/2, 0)
3
= 1−
(1/2)0 (1/2)3
0
= 1 − 1/8 = 7/8.
Then
F ∩E
is the event there is at least one head, and there
are two heads total, which is the same as there are two
heads total,
F
itself:
P(F ∩ E ) = P(F ) = 3/8
(in this case).
Bernoulli Trials 2
What if we knew there is at least one head among the three
tosses?
Let
E
be this event. It is a trials process with at least 1
success. Use the complement no successes:
P(E ) = 1 − b(3, 1/2, 0)
3
= 1−
(1/2)0 (1/2)3
0
= 1 − 1/8 = 7/8.
Then
F ∩E
is the event there is at least one head, and there
are two heads total, which is the same as there are two
heads total,
F
itself:
P(F ∩ E ) = P(F ) = 3/8
(in this case).
Let
E
be this event. It is a trials process with at least 1
success. Use the complement no successes:
P(E ) = 1 − b(3, 1/2, 0)
3
= 1−
(1/2)0 (1/2)3
0
= 1 − 1/8 = 7/8.
Then
F ∩E
is the event there is at least one head, and there
are two heads total, which is the same as there are two
heads total,
F
itself:
P(F ∩ E ) = P(F ) = 3/8
Hence
P(F |E ) =
(in this case).
P(F ∩ E ) 3/8
=
=
P(E )
7/8
3/7
.
The Sequences with Two Heads
There are eight sequences of H's and T's in 3 tosses:
HHH, HHT , HTH, THH,
HTT , THT , TTH, TTT .
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences with at least one head, we get
E ={HHH, HHT , HTH, THH,
HTT , THT , TTH}.TTT }.
The Sequences with Two Heads
There are eight sequences of H's and T's in 3 tosses:
HHH, HHT , HTH, THH,
HTT , THT , TTH, TTT .
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences with at least one head, we get
E ={HHH, HHT , HTH, THH,
HTT , THT , TTH}.TTT }.
The Sequences with Two Heads
Three of them have two H's.
Since each sequence is equally likely (probability 1/8 each),
P(exactly
2 heads)
= 3/8.
If we restrict to sequences with at least one head, we get
E ={HHH, HHT , HTH, THH,
HTT , THT , TTH}.TTT }.
and 3 out of 7 of them have two H's:
P(exactly
2 heads given at least 1 head)
=
3/7
.
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