Ch2 Probability 02(Conditional Probability).ppt fileConditional probabilityConditional probability...
Transcript of Ch2 Probability 02(Conditional Probability).ppt fileConditional probabilityConditional probability...
Conditional probability
Conditional probabilityConditional probability
So far, all (unconditional) probabilities were calculated with respect to the sample space S.
However, in many situations, we can obtain some new information such that we can update the sample space.
In such cases, we are able to update the probability calculations or
More specifically, we can restrict to a smaller sample space, instead of S.
to calculate conditional probabilities.
Conditional probabilityConditional probability
This problem is posed to Ben Campbell (played by Jim Sturgess) by Professor Micky Rosa (played by Kevin Spacey) in the movie "21“. Without hesitation Ben answers this correctly, which convinces Professor Rosa that Ben would be a good addition to their "card counting team".
“ The movie is inspired by the true story of a group of very smart M.I.T. students who use their math skills to count cards in Las Vegas and they
d ki t f “end up making a ton of money. “
http://www.youtube.com/watch?v=-EVsGaUVaFg&feature=related
Let’s see how Ben solves the problem.
ExamplesExamples Question 1: Fli f i i th ti Li t ll th ibl t iQuestion 1: Flip a fair coin three times. List all the possible outcomes, i.e. find the sample space of this experiment.
S = { HHH, HHT, HTH, THH, TTH, THT, HTT, TTT }.
Denote by A the event that “heads” occurs at the first time. Find A.
A = {HHH, HHT, HTH, HTT}.
.21
84
)()()( ===
SnAnAP
28)(Sn
New information: exactly two heads are obtained.New information: exactly two heads are obtained.
ExamplesExamples Question 1: Fli f i i th ti Li t ll th ibl t iQuestion 1: Flip a fair coin three times. List all the possible outcomes, i.e. find the sample space of this experiment.
S = { HHH, HHT, HTH, THH, TTH, THT, HTT, TTT }.
S” = {HHT, HTH, THH}.Given the new information find the updated P(A) or P(A|S”)Given the new information, find the updated P(A) or P(A|S )
A = {HHH, HHT, HTH, HTT}.
="SAI {HHT, HTH}
New information: exactly two heads are obtained..32
)"()"()"|( ==
SnSAnSAP I
New information: exactly two heads are obtained.3)"(Sn
ExamplesExamples There are 50 students, who can be classified in the following table.
Female (F) Male (M) TotalFemale (F) Male (M) Total
Major in 16 4 20Major in Engineering (E)
16 4 20
Not major in 10 20 30Not major in Engineering (EC)
10 20 30
T l 26 24 50Total 26 24 50
ExamplesExamples
Now we randomly select one of these 50 students, then what is the probability that the student is major in Engineering (E)?
All possible selections are equally likely.
ExamplesExamples There are 50 students, who can be classified in the following table.
Female (F) Male (M) TotalFemale (F) Male (M) Total
Major in 16 4 20Major in Engineering (E)
16 4 20
Not major in 10 20 30Not major in Engineering (EC)
10 20 30
Total 26 24 50
ExamplesExamples
Now we randomly select one of these 50 students, then what is the probability that the student is major in Engineering (E)?
n(S) = 50 n(E) = 20
All possible selections are equally likely.
n(S) 50 n(E) 20
220)(E .52
5020
)()()( ===
SnEnEP
550)(SnNew information: the student is femaleNew information: the student is female.
ExamplesExamples
Now we randomly select one of these 50 students, then what is the probability that the student is major in Engineering (E),
given that the student is female?given that the student is female?Based on this information, now we can update the sample space S to a smaller
F { ll f l t d t }one, i.e. F = {all female students}.
ExamplesExamples There are 50 students, who can be classified in the following table.
Female (F) Male (M) TotalFemale (F) Male (M) Total
Major in 16 4 20Major in Engineering (E)
16 4 20
Not major in 10 20 30Not major in Engineering (EC)
10 20 30
Total 26 24 50
ExamplesExamples
Now we randomly select one of these 50 students, then what is the probability that the student is major in Engineering (E),
given that the student is female?given that the student is female?Based on this information, now we can update the sample space S to a smaller
F { ll f l t d t }one, i.e. F = {all female students}.
Thus, the probability that the student is major in Engineering given that the student is a female is
ExamplesExamples There are 50 students, who can be classified in the following table.
Female (F) Male (M) TotalFemale (F) Male (M) Total
Major in 16 4 20Major in Engineering (E)
16 4 20
Not major in 10 20 30Not major in Engineering (EC)
10 20 30
Total 26 24 50
ExamplesExamples
Now we randomly select one of these 50 students, then what is the probability that the student is major in Engineering (E),
given that the student is female?given that the student is female?Based on this information, now we can update the sample space S to a smaller
F { ll f l t d t }one, i.e. F = {all female students}.
Thus, the probability that the student is major in Engineering given that
816)()|( FEnFEP I
the student is a female is
.1326)(
)()|( ===Fn
FEP
ExamplesExamples
)()(/)()( FEPSnFEnFEn III .)(
)()(/)(
)(/)()(
)()|(FP
FEPSnFn
SnFEnFn
FEnFEP III===
Conditional probabilityConditional probability
Definition (Conditional probability of event)
If A and B are any two events in a sample space SIf A and B are any two events in a sample space Sand P(B) > 0, then the conditional probability of A given B is
)( BAP I
then the conditional probability of A given B is
.)(
)()|(BP
BAPBAP I=
Thi f l b d f t f th f
)(BPThis formula can be used for any cases, not necessary for the case of equally likelihood outcomes.
Conditional probabilityConditional probability
Multiplication ruleMultiplication rule
)()|()( I
For any two events A and B with P(B) > 0,
.)()|()( BPBAPBAP =I
,0)( >CBP IFor any three events A, B and C with
.)()|()|()( CPCBPCBAPCBAP III =
,0)( CIy
)()|()|()(
ExampleExampleA b f f t i 20 f f hi h 5 d f tiA box of fuses contains 20 fuses, of which 5 are defective. If 3 of the fuses are selected at random and removed from the box in succession without replacement, what is the probability that all three fuses are defecti e?defective?
Let A be the event that the first fuse is defective,B be the event that the 2nd fuse is defective andB be the event that the 2nd fuse is defective andC be the event that the 3rd fuse is defective.Then 3)|(4)|(5)( BACPdABPAP I
Substitution into the formula yields
183)|(
194)|(,
205)( === BACPandABPAP I
1345)|()|()()( = BACPABPAPCBAP III
Substitution into the formula yields
.1141
183
194
205
=××=
Tree diagramTree diagramExample 10:Suppose that we are asked for the probability of exactly ONE ace in the draw of three cards from a bridge deck.
⎟⎞
⎜⎛⎟⎞
⎜⎛ 484 4’s aces
Choose 1
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛21
=)(An
Choose 2)(Sn 48’s non-aces
Choose 2
⎟⎟⎞
⎜⎜⎛52
⎟⎟⎠
⎜⎜⎝ 3
Tree diagramTree diagramExample 10:Suppose that we are asked for the probability of exactly ONE ace in the draw of three cards from a bridge deck.
We can imagine that we draw the cards sequentially, and the possibilities on each draw are ace or non-ace. Thenthe possibilities on each draw are ace or non ace. Then we can draw a tree to show all the possible outcomes.
Tree Diagram
513
AceAce
48
502
Case 1
51
48
AceNon-Ace50
48
Ace50352
4 Case 2
Case 3
5148 Non-Ace
Non-Ace
50
5047
3Case 4
514
Ace
N A48
Ace
Non-Ace
503
47
Case 5
Case 6
Non-Ace
Non-Ace5247
Non Ace50
Ace504
Case 6
Case 7
51 Non-Ace5046
Case 8
Tree Diagram
513
AceAce
48
502
Case 1
51
48Ace
Non-AceAce 50
48
50352
4 Case 2
Case 3
5148
Non-Ace
Non-Ace50
5047
3Case 4
514 Ace
Non-AceAce
N A48
503
47
Case 5
Case 6
Ace
Non Ace
Non-Ace
Non-Ace5247 50
450Case 6
Case 7
Non-Ace515046
Case 8
Tree Diagram
513
AceAce
48
502
Case 1
51
48Ace
Non-AceAce 50
48
50352
4 Case 2
Case 3
5148
Non-Ace
Non-Ace50
5047
3Case 4
514 Ace
Non-AceAce
N A48
503
47
Case 5
Case 6
Ace
Non Ace
Non-Ace
Non-Ace5247 50
450Case 6
Case 7
Non-Ace515046
Case 8
the probability of exactly ONE ace in the draw of three cards from a bridge deck is
47484××
the probability of exactly ONE ace in the draw of three cards from a bridge deck is
505152××
Tree Diagram
513
AceAce
48
502
Case 1
51
48Ace
Non-AceAce 50
48
50352
4 Case 2
Case 3
5148
Non-Ace
Non-Ace50
5047
3Case 4
514 Ace
Non-AceAce
N A48
503
47
Case 5
Case 6
Ace
Non Ace
Non-Ace
Non-Ace5247 50
450Case 6
Case 7
Non-Ace515046
Case 8
the probability of exactly ONE ace in the draw of three cards from a bridge deck is
47484×× 47448
××
the probability of exactly ONE ace in the draw of three cards from a bridge deck is
505152××
505152××
Tree Diagram
513
AceAce
48
502
Case 1
51
48Ace
Non-AceAce 50
48
50352
4 Case 2
Case 3
5148
Non-Ace
Non-Ace50
5047
3Case 4
514 Ace
Non-AceAce
N A48
503
47
Case 5
Case 6
Ace
Non Ace
Non-Ace
Non-Ace5247 50
450Case 6
Case 7
Non-Ace515046
Case 8
the probability of exactly ONE ace in the draw of three cards from a bridge deck is
47484×× 47448
×× 44748××
the probability of exactly ONE ace in the draw of three cards from a bridge deck is
505152××
505152××
505152××
⎞⎛⎞⎛= 0.2042⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛248
14
⎠⎝⎠⎝=
⎟⎞
⎜⎛52
⎟⎟⎠
⎞⎜⎜⎝
⎛3