Finding Probability Using Tree Diagrams and Outcome Tables 4.5.

Finding Probability Using Tree Diagrams and Outcome Tables

4.5

What are Tree Diagrams

A way of showing the possibilities of two or more events

Simple diagram we use to calculate the probabilities of two or more events

Example: link to movie (only till first pause)

For example – a fair coin is flipped twice

H

H

H

T

T

T

HH

HT

TH

TT

2nd 1st

Possible Outcomes

Outcome Tableif you flip a coin twice, you can model also model the results with an outcome table

Flip 1 Flip 2 Simple

Event

H H HH

H T HT

T H TH

T T TT

Tree Diagrams – For flipping a coin

Probability of two or more events

1st Throw 2nd Throw

THHHHH TTTT 1/21/21/21/21/21/21/2

OUTCOMES

H,H

H,T

T,H

T,T

P(Outcome)

P(H,H) =1/4=1/2x1/2

P(H,T) =1/4=1/2x1/2

P(T,H) =1/4=1/2x1/2

P(T,T) =1/4=1/2x1/2

Total P(all outcomes) = 1

Total=4 (2x2)Total=4 (2x2)

Multiplicative Principle for Probability of Independent Events

if two events are independent the probability of both occurring is…

P(A and B) = P(A) · P(B)

or P(A ∩ B) = P(A) · P(B) INDEPENDENT EVENTS

two events are independent of each other if an occurrence in one event does not change the probability of an occurrence in the other

if this is not true, then the events are dependent

Example – 10 coloured beads in a bag – 3 Red, 2 Blue, 5 Green. One taken, colour noted, returned to bag, then a second taken. Draw tree diagram for 2 draws.

B

RR

2nd 1st

B

B

BR

R

R

R

G

G

G

G

RB

RGBR

BB

BGGR

GB

GG

Now add in the Now add in the probabilityprobability

B

RR

2nd 1st

B

B

BR

R

R

R

G

G

G

G

RB

RGBR

BB

BGGR

GB

GG

0.3

0.2

0.5

0.5

0.20.3

0.5

0.20.3

0.5

0.20.3

Probabilities

P(RR) = 0.3x0.3 = 0.09

P(RB) = 0.3x0.2 = 0.06

P(RG) = 0.3x0.5 = 0.15P(BR) = 0.2x0.3 = 0.06

P(BB) = 0.2x0.2 = 0.04

P(BG) = 0.2x0.5 = 0.10P(GR) = 0.5x0.3 = 0.15

P(GB) = 0.5x0.2 = 0.10

P(GG) = 0.5x0.5 = 0.25

All ADD UP to 1.0

Multiplicative Principle for Counting

The total number of outcomes is the product of the possible outcomes at each step in the sequence

if a is selected from A, and b selected from Bn (a,b) = n(A) x n(B)

– (this assumes that each outcome has no influence on the next outcome)

Problems

How many possible three letter combinations are there? – you can choose 26 letters for each of the three

positions, so there are 26 x 26 x 26 = 17576

how many possible license plates are there in Ontario (4 L and 3#)?– 26 x 26 x 26 x 26 x 10 x 10 x 10

Problemif you rolled 1 die and then flipped a coin you have how many possible outcomes

n(d,c) = n(d) x n(c) = 6 x 2 =12

HT

HTHT

HT

HT

HT

1

2

3

4

5

6

(2,H)

(1,H)

(3,H)

(4,H)

(5,H)

(6,H)

(2,T)

(1,T)

(3,T)

(4,T)

(5,T)

(6,T)

Revisit - Sample Spacethe sample space for the last example would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin in which there are 12 possible outcomes

P (even, head)– there are 3 possible outcomes for an even die

and a head – P(odd roll, head) = 3/12 =¼

P(heads | even) – these are independent events, so knowing the outcome

of the second does not change the probability of the first

)()|(,2

1)(

2

1

6

312

3

)(

)()|(

headsPevenheadsPsaycanweheadsPas

evenP

evenheadsPevenheadsP

Problem: Conditional Probability

Conditional Probability for Independent Events

if A and B are independent events, then…

– P(B | A) = P(B)

if this is not true, then the events are dependent

Another way to prove multiplicative principle

( )( | )

( )

( ) ( ) ( | )

( | ) ( )

( ) ( ) ( )

P B AP B A

P A

P B A P A P B A

but

P B A P B

therefore

P B A P A P B

Dependent Events

two or more events are said to be dependent if the occurrence or non-occurrence of one of the events affects the probabilities of occurrence of any of the others.

3/9

6/9

7/10

3/10

2/9

7/9

1st event 2nd event

7 Red 3 Blue. Pick 2, without replacement. a) p(R,R) b) p(B,B) c) p(One of each)

OUTCOMES P(Outcome)

R,R

R,B

B,R

B,B

P(R,R)=42/90

P(R,B)=21/90

P(B,R)=21/90

P(B,B)=6/90

Total P(all outcomes) = 1

Exampleif you draw a card, replace it and draw another, what is the probability of two aces?– P(1stA and 2ndA)=4/52 x 4/52 =1/169– independent events

if you draw a card and then draw a second card (no replacement), what is the probability of two aces?– P(1stA and 2ndA)=4/52 x 3/51=1/221– second event depends on first event– the sample space is reduced by the first event

Exercises

read the examples on pages 242-244

page 245# 1-11– make sure that you are understanding these

concepts