Transcript Conditional Probability

“Teach A Level Maths”
Statistics 1
Conditional Probability
© Christine Crisp
Conditional Probability
We talk about conditional probability when the
probability of one event depends on whether or not
another event has occurred.
e.g. There are 2 red and 3 blue counters in a bag
and, without looking, we take out one counter and
do not replace it.
The probability of a 2nd counter taken from the bag
being red depends on whether the 1st was red or
blue.
Conditional probability problems can be solved by
considering the individual possibilities or by using a
table, a tree diagram or a formula.
Harder problems are most easily solved by using a
formula together with a tree diagram.
Conditional Probability
Notation
P(A) means “the probability that event A occurs”
P(A/) means “the probability that event A does not
occur”
P(A B) means
“the probability that event A occurs given that B has
occurred”. This is conditional probability.
Conditional Probability
e.g. 1. The following table gives data on the type of car,
grouped by petrol consumption, owned by 100 people.
Low
Medium
High
Male
12
33
7
Female
23
21
4
Total
100
One person is selected at random.
L is the event “the person owns a low rated car”
Conditional Probability
e.g. 1. The following table gives data on the type of car,
grouped by petrol consumption, owned by 100 people.
Low
Medium
High
Male
12
33
7
Female
23
21
4
Total
100
One person is selected at random.
L is the event “the person owns a low rated car”
F is the event “a female is chosen”.
Conditional Probability
e.g. 1. The following table gives data on the type of car,
grouped by petrol consumption, owned by 100 people.
Low
Medium
High
Male
12
33
7
Female
23
21
4
Total
100
One person is selected at random.
L is the event “the person owns a low rated car”
F is the event “a female is chosen”.
Conditional Probability
e.g. 1. The following table gives data on the type of car,
grouped by petrol consumption, owned by 100 people.
Low
Medium
High
Male
12
33
7
Female
23
21
4
Total
100
One person is selected at random.
L is the event “the person owns a low rated car”
F is the event “a female is chosen”.
Find (i) P(L)
(ii) P(F and L)
(iii) P(F L)
There is no need for a formula to solve this type of
problem.
We just need to be careful which row or column we look at.
Conditional Probability
Solution:
Male
Female
Find (i) P(L)
Low
12
23
35
(ii) P(F and L)
Medium
33
21
High
7
4
Total
100
(iii) P(F L)
(7 I usually leave the answers as fractions
(i) P(L) =
 as they won’t always be exact decimals.
100 20 20
It’s good practice to cancel. )
35 7
Conditional Probability
Solution:
Male
Female
Low
12
23
Medium
33
21
High
7
4
Total
100
Find (i) P(L)
(ii) P(F and L)
(iii) P(F L)
35 7
7
(i) P(L) =

100 20 20
23
(ii) P(F and L) =
100
The probability of selecting a
female with a low rated car.
Conditional Probability
Solution:
Male
Female
Find (i) P(L)
Low
12
23
35
(ii) P(F and L)
Medium
33
21
High
7
4
Total
100
(iii) P(F L)
35 7
7
(i) P(L) =

100 20 20
23
(ii) P(F and L) =
100
(iii) P(F L) 
23
35
We
be careful
with thea
The must
probability
of selecting
denominators
in (ii)
(iii).rated.
Here
female given the
carand
is low
we are given the car is low rated.
We want the total of that column.
Conditional Probability
Solution:
Male
Female
Low
12
23
Medium
33
21
High
7
4
Total
100
Find (i) P(L)
(ii) P(F and L)
(iii) P(F L)
35 7
7
(i) P(L) =

100 20 20
23
(ii) P(F and L) =
100
(iii) P(F L) 
23
35
Notice that
1
7 23
23

P(L)  P(F L)  
20 35 5 100
= P(F and L)
So, P(F and L) = P(F L)  P(L)
Conditional Probability
P(F and L) = P(F L)  P(L)
This result can be used to help solve harder conditional
probability problems.
However, I haven’t proved the formula, just shown
that it works for one particular problem.
Conditional Probability
SUMMARY
The probability that both event A and event B occur is
given by
P(A and B) = P(A B)  P(B)
We often use this in the form
P(A B)  P(A and B)
P(B)
In words, this is “the probability of event A given that B
has occurred, equals the probability of both A and B
occurring divided by the probability of B”.
Reminder:
P(A and B) can also be written as P(A  B)
Conditional Probability
e.g. 3. In November, the probability of a man getting
to work on time if there is fog on the M6 is 2 .
5
If the visibility is good, the probability is 9 .
10
3
The probability of fog at the time he travels is
.
20
(a) Calculate the probability of him arriving on time.
(b) Calculate the probability that there was fog given that
he arrives on time.
There are lots of clues in the question to tell us we are
dealing with conditional probability.
Conditional Probability
e.g. 3. In November, the probability of a man getting
to work on time if there is fog on the M6 is 2 .
5
If the visibility is good, the probability is 9 .
10
3
The probability of fog at the time he travels is
.
20
(a) Calculate the probability of him arriving on time.
(b) Calculate the probability that there was fog given that
he arrives on time.
There are lots of clues in the question to tell us we are
dealing with conditional probability.
Solution:
Let T be the event “ getting to work on time ”
Let F be the event “ fog on the M6 ”
Can you write down the notation for the probabilities
that we want to find in (a) and (b)?
Conditional Probability
(a) Calculate the probability of him arriving on time. P(T)
(b) Calculate the probability that there was fog given that
he arrives on time. P(F T)
Can you also write down the notation for the three
probabilities given in the question?
“ the probability of a man getting to work on time if
2
there is fog is 2 ” P(T F)
Not foggy

5
5
9
“ If the visibility is good, the probability is 9 ”. P(T F/) 
10
“ The probability of fog at the time he travels is
10
3
”.
20
P(F) 
3
20
This is a much harder problem so we draw a tree diagram.
Conditional Probability
P(T F) 
3
20
17
20
2
5
P(T F/) 
9
10
P(F) 
2
5
On
T
time
F
Fog
3
5
/
FNo
Fog
9
10
1
10
3
20
3 2

20 5
Not
on3 3
/
T

time
20 5
On 17  9
T
time 20 10
Not
on17 1
T/

time 20 10
Conditional Probability
P(T F) 
2
5
P(T F/) 
9
10
2
5
3
20
17
20
P(F) 
T
3 2

20 5
T/
3 3

20 5
T
17 9

20 10
T/
17 1

20 10
F
3
5
F/
9
10
1
10
3
20
Because we only reach the 2nd set of branches
after the 1st set has occurred, the 2nd set
must represent conditional probabilities.
Conditional Probability
(a) Calculate the probability of him arriving on time.
2
5
3
20
17
20
T
3 2

20 5
T/
3 3

20 5
T
17 9

20 10
T/
17 1

20 10
F
3
5
F/
9
10
1
10
Conditional Probability
(a) Calculate the probability of him arriving on time.
2
5
3
20
17
20
F
3
5
F/
9
10
1
10
T
6
3 2
 
20 5 100
( foggy
and
he
3
3
/
T
 time )
arrives
on
20 5
T
17 9

20 10
T/
17 1

20 10
Conditional Probability
(a) Calculate the probability of him arriving on time.
2
5
3
20
17
20
T
6
3 2
 
20 5 100
T/
3 3

20 5
F
3
5
9
10
F/
1
10
/
17 9 153
 
T
20 10 200
( not foggy
17 and
1 he
/
T
arrives
on time )
20 10
6
153 16533 33



P ( T )  P (F and T)  P (F and T )
100 200 20040 40
Conditional Probability
(b) Calculate the probability that there was fog given that
he arrives on time.
P (F and T )
P (F T) 
We need P ( F T )
P(T)
Fog on M 6
3
20
Getting to work
2
5
F
P (F and T )
P (F T ) 
P(T)

T
3 2
6
P (F and T )   
20 5 100
33
From part (a), P ( T ) 
40
6
33
4
6 2 40 2
 P (F T) 
P (F T) 



55
100 40 1005 3311
Exercise
Conditional Probability
1. A sample of 100 adults were asked how they travelled
to work. The results are shown in the table.
Walk
Bus
Bike
Car
Men
10
12
6
26
Women
7
18
4
17
Total
100
One person is picked at random.
M is the event “the person is a man”
C is the event “the person travels by car”
Find (i) P(M) (ii) P(M C) (iii) P(M/ and C) (iv) P(C M/)
Conditional Probability
Solution:
Walk
Bus
Bike
Car
Men
10
12
6
26
Women
7
18
4
17
Total
Find (i) P(M)
54
100
(ii) P(M C)
54
27
(i) P(M) 

100 50
(iii) P(M/ and C) (iv) P(C M/)
(ii) P(M C) 
Conditional Probability
Solution:
Walk
Bus
Bike
Car
Men
10
12
6
26
Women
7
18
4
17
Total
Find (i) P(M)
43
(ii) P(M C)
54
27
(i) P(M) 

100 50
(iii) P(M/ and C) 
54
100
(iii) P(M/ and C) (iv) P(C M/)
26
(ii) P(M C) 
43
Conditional Probability
Solution:
Walk
Bus
Bike
Car
Men
10
12
6
26
Women
7
18
4
17
Total
Find (i) P(M)
43
(ii) P(M C)
54
100
(iii) P(M/ and C) (iv) P(C M/)
54
27
(i) P(M) 

100 50
26
(ii) P(M C) 
43
(iii) P(M/ and C)  17
(iv) P(C M/) 
100
Conditional Probability
Solution:
Walk
Bus
Bike
Car
Men
10
12
6
26
54
Women
7
18
4
17
46
43
100
Total
Find (i) P(M)
(ii) P(M C)
(iii) P(M/ and C) (iv) P(C M/)
54
27
(i) P(M) 

100 50
(iii) P(M/ and C)  17
100
26
(ii) P(M C) 
43
17
/
(iv) P(C M ) 
46
Exercise
Conditional Probability
2. The probability of a maximum temperature of 28 or
more on the 1st day of Wimbledon ( tennis competition! )
3
has been estimated as . The probability of a particular
8
British player winning on the 1st day if it is below 28 is
1
3
estimated to be
but otherwise only .
2
4
Draw a tree diagram and use it to help solve the following:
(i) the probability of the player winning,
(ii) the probability that, if the player has won, it was at
least 28.
Solution: Let T be the event “ temperature 28 or more ”
Let W be the event “ player wins ”
Then,
3
P(T) 
8
/
3
P(W T ) 
4
1
P(W T ) 
2
Conditional Probability
Let T be the event “ temperature 28 or more ”
Let W be the event “ player wins ”
Then,
3
P(T) 
8
/
3
P(W T ) 
4
1
2
3
8
5
8
High
T
temp
Lower
/
T
temp
1
2
3
4
1
4
1
P(W T ) 
2
Wins
W
3 1 3
 
8 2 16
W/
Loses
3 1 3
 
8 2 16
W
Wins
5 3 15
 
8 4 32
/
W
Loses
5 1 5
 
8 4 32
Sum
=1
Conditional Probability
1
2
3
8
5
8
T
1
2
3
4
T/
1
4
(i)
W
3 1 3
 
8 2 16
W/
3 1 3
 
8 2 16
W
5 3 15
 
8 4 32
W/
5 1 5
 
8 4 32
P (W)  P ( T and W)  P ( T / and W)
Conditional Probability
1
2
3
8
5
8
T
1
2
3
4
T/
1
4
(i)
P (W)  P ( T and W) 
W
3 1 3
 
8 2 16
W/
3 1 3
 
8 2 16
W
5 3 15
 
8 4 32
W/
5 1 5
 
8 4 32
3 15 6  15 21

P ( T and W)   
16 32
32
32
/
Conditional Probability
1
2
3
8
5
8
T
1
2
3
4
T/
1
4
21
P ( W) 
32
P ( T and W )
(ii) P ( T W ) 
P (W)
W
3 1 3
 
8 2 16
W/
3 1 3
 
8 2 16
W
5 3 15
 
8 4 32
W/
5 1 5
 
8 4 32
Conditional Probability
1
2
3
8
5
8
T
1
2
3
4
T/
1
4
21
P ( W) 
32
W
3 1 3
 
8 2 16
W/
3 1 3
 
8 2 16
W
5 3 15
 
8 4 32
W/
5 1 5
 
8 4 32
3 21
P ( T and W )
(ii) P ( T W ) 
 P ( T W) 

16 32
P (W)
Conditional Probability
1
2
3
8
5
8
T
1
2
3
4
T/
1
4
21
P ( W) 
32
W
3 1 3
 
8 2 16
W/
3 1 3
 
8 2 16
W
5 3 15
 
8 4 32
W/
5 1 5
 
8 4 32
1
2
3 21 3 32 2
P ( T and W )
(ii) P ( T W ) 

 P ( T W) 

 
16 32 161 217 7
P (W)
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Conditional Probability
SUMMARY
P (A) means
P (A / )
“ the probability that event A occurs ”
means “ the probability that event A does not occur ”
P (A B )
means “ the probability that event A occurs given that
B has occurred ”
P (A and B)
P (A B ) 
P (B)
In words, this is “ the probability of event A given that
B has occurred, equals the probability of both A and B
occurring divided by the probability of B ”.
Rearranging:
P (A and B)  P (A B)  P (B)
Reminder: P(A and B) can also be written as P (A  B)
Conditional Probability
e.g. 1. The following table gives data on the type of car,
grouped by petrol consumption, owned by 100 people.
Low
Medium
High
Male
12
33
7
Female
23
21
4
Total
100
One person is selected at random.
L is the event “the person owns a low rated car”
F is the event “a female is chosen”.
Find (i) P(L)
(ii) P(F and L)
(iii) P(F L)
There is no need for a Venn diagram or a formula to
solve this type of problem.
We just need to be careful which row or column we look at.
Conditional Probability
Solution:
Male
Female
Low
12
23
Medium
33
21
High
7
4
Total
100
Find (i) P(L)
(ii) P(F and L)
(iii) P(F L)
35 7
7
(i) P(L) =

100 20 20
23
(ii) P(F and L) =
100
(iii) P(F L) 
23
35
Notice that
7 23
23

P(L)  P(F L)  
20 35 5 100
= P(F and L)
So, P(F and L) = P(F L)  P(L)
Conditional Probability
e.g. 2. I have 2 packets of seeds. One contains 20
seeds and although they look the same, 8 will give red
flowers and 12 blue. The 2nd packet has 25 seeds of
which 15 will be red and 10 blue.
Draw a Venn diagram and use it to illustrate the
conditional probability formula.
Solution: Let R be the event “ red flower ” and
F be the event “ 1st packet ”
8
P(R and F) =
45
P(R F) =
8
20
P(F) =
45
20
45
R
F
12
8
20
8
 P(R F)  P(F) = 

20 45 45
So,
P(R and F) = P(R F)  P(F)
8
15
10
Conditional Probability
e.g. 3. In November, the probability of a man getting
to work on time if there is fog on the M6 is 2 .
5
If the visibility is good, the probability is 9 .
10
3
The probability of fog at the time he travels is
.
20
(a) Calculate the probability of him arriving on time.
(b) Calculate the probability that there was fog given
that he arrives on time.
Solution:
Let T be the event “ getting to work on time ”
Let F be the event “ fog on the M6 ”
Conditional Probability
(a) Calculate the probability of him arriving on time.
2
5
3
20
17
20
T
6
3 2
 
20 5 100
T/
3 3

20 5
T
17 9 153
 
20 10 200
T/
17 1

20 10
F
3
5
9
10
F/
1
10
/
6
153 16533 33



P ( T )  P (F and T)  P (F and T )
100 200 200 40 40
Conditional Probability
(b) Calculate the probability that there was fog given that
he arrives on time.
P (F and T )
P (F T) 
We need P ( F T )
P(T)
Fog on M 6
3
20
Getting to work
2
5
F
P (F and T )
P (F T ) 
P(T)

T
3 2
6
P (F and T )   
20 5 100
33
From part (a), P ( T ) 
40
6
33
4
6 2 40 2
 P (F T) 
P (F T) 



55
100 40 1005 3311
Conditional Probability
If B has no effect on A, then, P(A B) = P(A) and we say
the events are independent.
( The probability of A does not depend on B. )
So, for 2 independent events,
P (A and B)  P (A)  P (B)