Probability - global2methods.global2.vic.edu.au/files/2012/11/Probability-167mhec.pdf · VCE Maths...

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VCE Maths Methods - Unit 2 - Probability Probability • Probability Tree diagrams Laice diagrams Venn diagrams Karnough maps Probability tables Union & intersection rules Conditional probability Markov chains 1

Transcript of Probability - global2methods.global2.vic.edu.au/files/2012/11/Probability-167mhec.pdf · VCE Maths...

VCE Maths Methods - Unit 2 - Probability

Probability

• Probability• Tree diagrams• Laice diagrams• Venn diagrams• Karnough maps• Probability tables• Union & intersection rules• Conditional probability• Markov chains

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VCE Maths Methods - Unit 2 - Probability

Probability

• Probability is the mathematics of chance.• While an outcome of a random event can only be can’t determined in advance, we

can use a prediction based on experimental or theoretical probability.• Probabilities can be expressed as fractions, decimals or percentages.• 0 = no chance at all, 100% = absolutely certain.• The chances of an event happening or not happening add to make 1.

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Pr (E') = 1 - Pr (E)

0 ≤ Pr (E) ≤1

0% - no chance

25% - unlikely

50% - even chance

75% - likely

100% - certain

VCE Maths Methods - Unit 2 - Probability

Probability

• Experimental probability: based on measurement of past outcomes. For example, a 15% chance of a wet grand final day (based on the last 100 years).

• Theoretical probability: based on the number of favourable outcomes from the total set of possible outcomes. For example, a 1 in a 1000 chance of buying the winning ticket in a raffle.

• The long term expected value is found by multiplying the probability & number of trials.

• For example, buying 20 game tickets with a 1 in 10 chance, it is expected that 2 prizes would be won.

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Experimental probability = Number of favourable outcomes observed

Total number of trials

Theoretical probability =

Number of possible favourable outcomesTotal number of possible outcomes

Pr (E) = n (E)n (ε)

Expected value = E(x) = n × Pr(x)

VCE Maths Methods - Unit 2 - Probability

Calculating probability - tree diagrams

• Shows all possible outcomes for sequential events.• Each branch on the tree represents a possible outcome.• For example, tossing three coins:• Each outcome has a 1 in 8 chance (1/2 x 1/2 x 1/2 = 1/8).

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Pr (3 heads) = 1

8

H

T

H

T

H

T

T

T

H

T

H

T

H

T

H

HHH

HHTHTH

HTTTHHTHTTTH

TTT

Pr (2 heads) = 3× 1

8=

38

Pr (1 head) = 3× 1

8=

38

Pr (0 heads) = 1

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VCE Maths Methods - Unit 2 - Probability

Calculating probability - laice diagrams

• Shows the possible outcomes from two sequential events.• Easier to draw than a tree diagram if there is a large number of outcomes.• For example, rolling two dice & finding the product.

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Pr (6) = 4× 1

36 = 1

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1 2 3 4 5 6

1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

The chance of scoring 6:

Pr (36) = 1× 1

36 = 1

36

The chance of scoring 36:

VCE Maths Methods - Unit 2 - Probability

Venn diagrams

• A useful visual tool to show the relationship between two sets or events.

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S '∩A '

A

25 students in the class• 10 competed in the

swimming sports (Set S)• 14 competed in the

athletics sports (Set A)• 8 students competed in

both (Set S∩A)

S∩An = 8 S∩A '

n = 2 S '∩An = 6

n = 9

(Swimming) (Athletics)

• Two events are mutually exclusive if there is no intersection of the two sets.

• (If no students did both sports)

(S∩A) = the intersection of events S & A.

S

VCE Maths Methods - Unit 2 - Probability

Union & intersection rules

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What is the chance that a student that is randomly selected:

a) Is a swimmer?

b) Is an athlete?

c) Is a swimmer and an athlete?

d) Is a swimmer or an athlete?

Pr(S)=10

25=40%

Pr=14

25=56%

Pr= 8

25=32% (Intersection)

Pr=10

25+

1425

−8

25=

1625

=64% (Union)

Pr(A∪B)=Pr(A)+Pr(B)−Pr(A∩B)

Pr(A∪B)=1−Pr(A '∩B ')

VCE Maths Methods - Unit 2 - Probability

Karnough maps

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A A’

S

S’

25 students in the class• 10 competed in the

swimming sports (Set S)• 14 competed in the

athletics sports (Set A)• 8 students competed in

both (Set S∩A)

8 2 10

6 9 15

14 11 25

• Used to show the number of elements of two overlapping sets.

VCE Maths Methods - Unit 2 - Probability

Karnough maps

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B B’

A

A’

n(A∩B) n(A∩B ') n(A)

n(A '∩B) n(A '∩B ') n(A ')

n(B) n(B ') n

VCE Maths Methods - Unit 2 - Probability

Probability tables

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B B’

A

A’

• Used to show the probability of two overlapping events occurring.

Pr(A∩B) Pr(A∩B ') Pr(A)

Pr(A '∩B) Pr(A '∩B ') Pr(A ')

Pr(B) Pr(B ') 1

VCE Maths Methods - Unit 2 - Probability

Conditional probability

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• Two events (A & B) are considered to be independent if the probability of event A occurring has no influence over event B occurring.

• If the events A & B are not independent, then they are said to be conditional.

Pr(A∩B)=Pr(A | B)×Pr(B)

Pr(A | B)= Pr(A∩B)

Pr(B)• For example, a student in the class that competed at the athletics - what

is the chance that they were also at the swimming sports?

Pr(S | A)= Pr(S∩A)

Pr(A)

=

825⎛⎝⎜

⎞⎠⎟

1425⎛⎝⎜

⎞⎠⎟

=8

14 =

47

Independent events: Conditional events:

Pr(A∩B)=Pr(A)×Pr(B)

VCE Maths Methods - Unit 2 - Probability

Markov chains

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• For example, the chance of a team winning is 80% following a win, but only 40% following a loss.

• What is the chance that after winning the first game, the team also wins the third?

• This could be expressed as a tree diagram......

W1

W2

W2’

80%

20%

W3

W3’

80%

20%

Pr(W2∩W3 ) = 80% x80% = 64%

Pr(W2∩W3 ') = 80% x20% = 16%

Pr(W2 '∩W3 ) = 20% x40% = 8%

Pr(W2 '∩W3 ') = 20% x60% = 12%

W3

W3’

40%

60%

Pr(W3 )=Pr(W2∩W3 ) + Pr(W2 '∩W3 ) = 64% + 8% = 72%

Pr(W2∩W3 ) = Pr(W3 | W2 )×Pr(W2 )

Pr(W2∩W3 ') = Pr(W3 ' | W2 )×Pr(W2 )

Pr(W2 '∩W3 ) = Pr(W3 | W2 ')×Pr(W2 ')

Pr(W2 '∩W3 ) = Pr(W3 | W2 ')×Pr(W2 ')

VCE Maths Methods - Unit 2 - Probability

Markov chains

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• Used to find the long term probability of a sequence of a number of repeated events, where the conditional probabilities remain constant.

Pr(B)Pr(B ')

⎣⎢⎢

⎦⎥⎥=

Pr(B | A) Pr(B | A ')Pr(B ' | A) Pr(B ' | A ')

⎣⎢⎢

⎦⎥⎥

nPr(A)Pr(A ')

⎣⎢⎢

⎦⎥⎥

Sn = (Result) State matrix (n states)

T= Transition matrix(probabilities)

S0= State matrix(initial state)

Pr(B) = Pr(B | A)×Pr(A)+Pr(B | A')×Pr(A')

Sn =T n ×S0

VCE Maths Methods - Unit 2 - Probability

Markov chains

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• From the previous example:

0.8 0.4

0.2 0.6⎡

⎣⎢

⎦⎥

210

⎣⎢

⎦⎥

= 0.72

0.28⎡

⎣⎢

⎦⎥

S2 = State (Result) matrix (after 2 more games)T= Transition matrix

(probabilities)S0 = State matrix

(initial state)• As the number of states is increased, the overall probabilities will

converge to defined limits.

n =1,Pr=0.800 n =4,Pr=0.675

n =3,Pr=0.688 n =2,Pr=0.720 n =15,Pr=0.667

Overall chance of winning = 0.4 / (0.4 + 0.2)

= 2/3 n =100,Pr=0.667

Winning

Following a win

Losing

Following a loss

The first game was a win