Let us now develop a formula for the conditional probability
P
E
∣
F
size 12{P left (E \lline F right )} {} .
Suppose an experiment consists of
n
size 12{n} {} equally likely events. Further suppose that there are
m
size 12{m} {} elements in
F
size 12{F} {} , and
c
size 12{c} {} elements in
E
∩
F
size 12{E intersection F} {} , as shown in the following Venn diagram.
If the event
F
size 12{F} {} has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset
F
size 12{F} {} . Therefore, we only look at the set
F
size 12{F} {} and at nothing outside of
F
size 12{F} {} . Since
F
size 12{F} {} has
m
size 12{m} {} elements, the denominator in the calculation of
P
E
∣
F
size 12{P left (E \lline F right )} {} is m. We may think that the numerator for our conditional probability is the number of elements in
E
size 12{E} {} . But clearly we cannot consider the elements of
E
size 12{E} {} that are not in
F
size 12{F} {} . We can only count the elements of
E
size 12{E} {} that are in
F
size 12{F} {} , that is, the elements in
E
∩
F
size 12{E intersection F} {} . Therefore,
P
E
∣
F
=
c
m
size 12{P left (E \lline F right )= { {c} over {m} } } {}
Dividing both the numerator and the denominator by
n
size 12{n} {} , we get
P
E
∣
F
=
c
/
n
m
/
n
size 12{P left (E \lline F right )= { {c/n} over {m/n} } } {}
But
c
/
n
=
P
E
∩
F
size 12{c/n=P left (E intersection F right )} {} , and
m
/
n
=
P
F
size 12{m/n=P left (F right )} {} .
Substituting, we derive the following formula for
P
E
∣
F
size 12{P left (E \lline F right )} {} .
For Two Events
E
size 12{E} {} and
F
size 12{F} {} , the Probability of
E
size 12{E} {} Given
F
size 12{F} {} is
P
E
∣
F
=
P
E
∩
F
P
F
size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} Got questions? Get instant answers now!
A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.
Let
E
size 12{E} {} be the event that an even number shows, and
F
size 12{F} {} be the event that a number greater than three shows. We want
P
E
∣
F
size 12{P left (E \lline F right )} {} .
E
=
2,4,6
size 12{E= left lbrace 2,4,6 right rbrace } {} and
F
=
4,5,6
size 12{F= left lbrace 4,5,6 right rbrace } {} . Which implies,
E
∩
F
=
4,6
size 12{E intersection F= left lbrace 4,6 right rbrace } {}
Therefore,
P
F
=
3
/
6
size 12{P left (F right )=3/6} {} , and
P
E
∩
F
=
2
/
6
size 12{P left (E intersection F right )=2/6} {}
P
E
∣
F
=
P
E
∩
F
P
F
=
2
/
6
3
/
6
=
2
3
size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } = { {2/6} over {3/6} } = { {2} over {3} } } {} .
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The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.
Male(M)
Female(F)
Total
Public Transportation(P)
8
13
21
Drive(D)
39
40
79
Total
47
53
100
The events
M
size 12{M} {} ,
F
size 12{F} {} ,
P
size 12{P} {} , and
D
size 12{D} {} are self explanatory. Find the following probabilities.
P
D
∣
M
size 12{P left (D \lline M right )} {}
P
F
∣
D
size 12{P left (F \lline D right )} {}
P
M
∣
P
size 12{P left (M \lline P right )} {}
We use the conditional probability formula
P
E
∣
F
=
P
E
∩
F
P
F
size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} .
P
D
∣
M
=
P
D
∩
M
P
M
=
39
/
100
47
/
100
=
39
47
size 12{P left (D \lline M right )= { {P left (D intersection M right )} over {P left (M right )} } = { {"39"/"100"} over {"47"/"100"} } = { {"39"} over {"47"} } } {} .
P
F
∣
D
=
P
F
∩
D
P
D
=
40
/
100
79
/
100
=
40
79
size 12{P left (F \lline D right )= { {P left (F intersection D right )} over {P left (D right )} } = { {"40"/"100"} over {"79"/"100"} } = { {"40"} over {"79"} } } {} .
P
M
∣
P
=
P
M
∩
P
P
P
=
8
/
100
21
/
100
=
8
21
size 12{P left (M \lline P right )= { {P left (M intersection P right )} over {P left (P right )} } = { {8/"100"} over {"21"/"100"} } = { {8} over {"21"} } } {} . Got questions? Get instant answers now! Got questions? Get instant answers now!
Given
P
E
=
.
5
size 12{P left (E right )= "." 5} {} ,
P
F
=
.7
size 12{P left (F right )=/7} {} , and
P
E
∩
F
=
.3
size 12{P left (E intersection F right )} {} . Find the following.
P
E
∣
F
size 12{P left (E \lline F right )} {}
P
F
∣
E
size 12{P left (F \lline E right )} {} .
We use the conditional probability formula
P
E
∣
F
=
P
E
∩
F
P
F
size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} .
P
E
∣
F
=
.
3
.
7
=
3
7
size 12{P left (E \lline F right )= { { "." 3} over { "." 7} } = { {3} over {7} } } {} .
P
F
∣
E
=
.
3
/
.
5
=
3
/
5
size 12{P left (F \lline E right )= "." 3/ "." 5=3/5} {} . Got questions? Get instant answers now! Got questions? Get instant answers now!
Given two mutually exclusive events
E
size 12{E} {} and
F
size 12{F} {} such that
P
E
=
.
4
size 12{P left (E right )= "." 4} {} ,
P
F
=
.
9
size 12{P left (F right )= "." 9} {} . Find
P
E
∣
F
size 12{P left (E \lline F right )} {} .
Since
E
size 12{E} {} and
F
size 12{F} {} are mutually exclusive,
P
E
∩
F
=
0
size 12{P left (E intersection F right )=0} {} . Therefore,
P
E
|
F
=
0
.9
=
0
size 12{P left (E intersection F right )=0} {} .
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Given
P
F
∣
E
=
.
5
size 12{P left (F \lline E right )= "." 5} {} , and
P
E
∩
F
=
.
3
size 12{P left (E intersection F right )= "." 3} {} . Find
P
E
size 12{P left (E right )} {} .
Using the conditional probability formula
P
E
∣
F
=
P
E
∩
F
P
F
size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} , we get
P
F
∣
E
=
P
E
∩
F
P
E
size 12{P left (F \lline E right )= { {P left (E intersection F right )} over {P left (E right )} } } {}
Substituting,
.
5
=
.
3
P
E
size 12{ "." 5= { { "." 3} over {P left (E right )} } } {} or
P
E
=
3
/
5
size 12{P left (E right )=3/5} {}
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In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.
Let event
E
size 12{E} {} be that the family has two boys and a girl, and let
F
size 12{F} {} be the probability that the family has at least two boys. We want
P
E
∣
F
size 12{P left (E \lline F right )} {} .
We list the sample space along with the events
E
size 12{E} {} and
F
size 12{F} {} .
S
=
BBB
,
BBG
,
BGB
,
BGG
,
GBB
,
GGB
,
GGG
size 12{S= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GGB", ital "GGG" right rbrace } {}
E
=
BBG
,
BGB
,
GBB
size 12{E= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {} and
F
=
BBB
,
BBG
,
BGB
,
GBB
size 12{F= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "GBB" right rbrace } {}
E
∩
F
=
BBG
,
BGB
,
GBB
size 12{E intersection F= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {}
Therefore,
P
F
=
4
/
8
size 12{P left (F right )=4/8} {} , and
P
E
∩
F
=
3
/
8
size 12{P left (E intersection F right )=3/8} {} .
And
P
E
∣
F
−
3
/
8
4
/
8
=
3
4
size 12{P left (E \lline F right ) - { {3/8} over {4/8} } = { {3} over {4} } } {} .
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At a community college 65% of the students use IBM computers, 50% use Macintosh computers, and 20% use both. If a student is chosen at random, find the following probabilities.
The student uses an IBM given that he uses a Macintosh.
The student uses a Macintosh knowing that he uses an IBM.
Let event
I
size 12{I} {} be that the student uses an IBM computer, and
M
size 12{M} {} the probability that he uses a Macintosh.
P
I
∣
M
=
.
20
.
50
=
2
5
size 12{P left (I \lline M right )= { { "." "20"} over { "." "50"} } = { {2} over {5} } } {}
P
M
∣
I
=
.
20
.
65
=
4
13
size 12{P left (M \lline I right )= { { "." "20"} over { "." "65"} } = { {4} over {"13"} } } {} . Got questions? Get instant answers now! Got questions? Get instant answers now!
