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Conditional Probability. P (A B) means. “the probability that event A occurs given that B has occurred”. Find ( i ) P (L) (ii) P (F ∩ L) (iii) P (F L). Ex 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people. Low. Medium. High. - PowerPoint PPT Presentation
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Conditional Probability
Conditional Probability
Conditional Probability
“the probability that event A occurs given that B has occurred”.
P(A B) means
Conditional Probability
Female
LowMale
Ex 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people.
42123
73312
HighMedium
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 ∩ L) (iii) P(F L)
100
Total
Conditional Probability
2312
100100
(i) P(L) =
Solution:
Find (i) P(L) (ii) P(F ∩ L) (iii) P(F L)
35421Female733Male
TotalHighMediumLow
(leave the answer as fraction20 20
7
2312
735100
Low
Conditional Probability
(i) P(L) =
Solution:
Find (i) P(L) (ii) P(F and L) (iii) P(F L)
10042123Female73312Male
HighMediumLow
20 207
735
100
(ii) P(F ∩ L) = The probability of selecting a female with a low rated car.
23
23
100
Total
100
Conditional Probability
(i) P(L) =
Solution:
Find (i) P(L) (ii) P(F and L) (iii) P(F L)
1003542123Female73312Male
HighMediumLow
20 207
735
100
(ii) P(F ∩ L) =23
Total
100
(iii) P(F L)
The probability of selecting a female given the car is low rated.
23
23
35
12
Conditional Probability
(i) P(L) =
Solution:
Find (i) P(L) (ii) P(F and L) (iii) P(F L)
10042123Female73312Male
HighMediumLow
20 207
735
100
(ii) P(F ∩ L) =23
Total
100
(iii) P(F L)
2335
Conditional Probability
• If you don’t have a table, you can use this formula
( )( )( )
P A BP B AP A
Conditional Probability
Conditional ProbabilityResearchers asked people who exercise regularly whether they jog or
walk. Fifty-eight percent of the respondents were male. Twenty percent of all respondents were males who said they jog. Find the
probability that a male respondent jogs.
P( male ) = 58%P( male and jogs ) = 20%Let A = male.Let B = jogs.
The probability that a male respondent jogs is about 34%.
Write: P( A | B ) =P( A and B )P( A )
= Substitute 0.2 for P(A and B) and 0.58 for P(A). 0.344 Simplify.
0.2 0.58
Conditional Probability
Using Tree Diagrams Jim created the tree diagram
after examining years of weatherobservations in his hometown. Thediagram shows the probability ofwhether a day will begin clear orcloudy, and then the probabilityof rain on days that begin clear
and cloudy.a. Find the probability that a day will start out clear, and then will rain.
The path containing clear and rain represents days that start out clear and then will rain.
P(clear and rain) = P(rain | clear) • P(clear)= 0.04 • 0.28
= 0.011The probability that a day will start out clear and then rain is about 1%.
Conditional Probability
Conditional Probability(continued)
b. Find the probability that it will not rain on any given day.
The paths containing clear and no rain and cloudy and no rain both represent a day when it will not rain. Find the probability for both paths and add them.
P(clear and no rain) + P(cloudy and no rain) =P(clear) • P(no rain | clear) + P(cloudy) • P(no rain | cloudy)
= 0.28(.96) + .72(.69)= 0.7656
The probability that it will not rain on any given day is about 77%.
Conditional Probability
Ex 3 In November, the probability of a man getting to work on time if there is fog on 285 is .
52
109If the visibility is good, the probability is .
203The probability of fog at the time he travels is .
(a)Calculate the probability of him arriving on time. (b)Calculate the probability that there was fog given that he arrives on time.
Conditional Probability
Not on time
52
2017
203
109
101
52
203
53
203
109
2017
101
2017
Fog
No Fog
On time
On timeNot on time
52
P(T F) 10
9P(T F/)
203
P(F)
53
F
F/
T
T/
T
T/
Conditional Probability
52
P(T F) 10
9P(T F/)
203
P(F)
52
2017
203
109
101
52
203
53
203
109
2017
101
2017
53
F
F/
T
T/
T
T/
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.
52
2017
203
109
101
52
203
53
203
109
2017
101
2017
53
F
F/
T
T/
T
T/
Conditional Probability
52
2017
203
109
101
53
203
109
2017
101
2017
53
F
F/
T
T/
T
T/
52
203
(a)Calculate the probability of him arriving on time.
1006
( foggy and he arrives on time )
Conditional Probability
52
2017
203
109
101
53
203
53
F
F/
T
T/
T
T/
109
2017
101
2017
52
203
1006
(a)Calculate the probability of him arriving on time.
200153
200153
1006
200165)(()( Tand/F Tand FT ) PPP
33
40 4033
( not foggy and he arrives on time )
Conditional Probability
Fog on 285 Getting to work
F
T
203
52
4033)( TPFrom part
(a),
)( Tand FP1006
52
203
(b)Calculate the probability that there was fog given that he arrives on time. We need
)( TFP
)()()(
T Tand F TF
PP
P
554( )P TF
)()((
T Tand F
T)FP
PP
4033
1006( ) TFP
3340
1006
5
22
11
Conditional ProbabilityExampl
e 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
Find (i) (ii) (iii) (iv)P(M)
P(M/ and C)
One person is picked at random.M is the event “the person is a man”C is the event “the person travels by car” P(M C) P(C M/)
Conditional Probability
100100
54
Total
174187Women
2661210Men
CarBikeBusWalk
Find (i) (ii) (iii) (iv)P(M)
P(M/ and C)
P(M C) P(C M/)
Solution:
(i) P(M)
(i)10054
5027
(ii) P(M C)
Conditional Probability
10043Total
174187Women
542661210Men
CarBikeBusWalk
Find (i) (ii) (iii) (iv)P(M)
P(M/ and C)
P(M C) P(C M/)
Solution:
(i) P(M) 10054
5027
(ii) P(M C)
(iii) P(M/ and C)
4326
Conditional Probability
10010043Total
174187Women
542661210Men
CarBikeBusWalk
Find (i) (ii) (iii) (iv)P(M)
P(M/ and C)
P(M C) P(C M/)
Solution:
(i) P(M) 10054
5027
(ii) P(M C)
(iii) P(M/ and C)
4326
10017 P(C M/) (iv)
Conditional Probability
10043Total
46174187Women
542661210Men
CarBikeBusWalk
Find (i) (ii) (iii) (iv)P(M)
P(M/ and C)
P(M C) P(C M/)
Solution:
(i) P(M) 10054
5027
(ii) P(M C)
(iii) P(M/ and C)
4326
10017 P(C M/) (iv)
4617
Conditional Probability
Independent Events
If the probability of the occurrence of event A is the same regardless of whether or not an outcome B occurs, then the outcomes A and B are said to be independent of one another. Symbolically, if
then A and B are independent events.
)()|( APBAP