1. Finding Probability Using Tree Diagrams and Outcome Tables
4.5

2. 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

3. Example: link to movie (only till first pause)

4. For example – a fair coin is flipped twice
1st
2nd H HH
Possible Outcomes H T HT H TH T T TT

5. Outcome Table
if you flip a coin twice, you can model also model the results with an outcome table
Flip 1
Flip 2
Simple
Event H HH T HT TH TT

6. Tree Diagrams – For flipping a coin
Probability of two or more events
Tree Diagrams – For flipping a coin
1st Throw
2nd Throw
OUTCOMES
P(Outcome)
H,H
P(H,H) =1/4=1/2x1/2
H,T
P(H,T) =1/4=1/2x1/2
T,H
P(T,H) =1/4=1/2x1/2
P(T,T) =1/4=1/2x1/2
T,T H H H H H T T T T T
1/2
1/2
1/2
1/2
1/2
1/2
1/2
Total=4 (2x2)
Total P(all outcomes) = 1

7. 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

8. Example – 10 coloured beads in a bag – 3 Red, 2 Blue, 5 Green
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.
1st
2nd R RR B RB R G RG R BR
Now add in the probability B B BB G BG R GR G GB B G GG

9. Probabilities 1st 2nd R RR B RB R G RG R BR B B BB G BG R GR G GB B G
P(RR) = 0.3x0.3 = 0.09
0.3
0.2 B RB
P(RB) = 0.3x0.2 = 0.06 R
0.3
0.5 G RG
P(RG) = 0.3x0.5 = 0.15 R BR
P(BR) = 0.2x0.3 = 0.06
0.3
0.2
0.2 B B BB
P(BB) = 0.2x0.2 = 0.04
0.5 G BG
P(BG) = 0.2x0.5 = 0.10 R GR
0.3
P(GR) = 0.5x0.3 = 0.15
0.5 G
0.2 GB B
P(GB) = 0.5x0.2 = 0.10 G GG
0.5
P(GG) = 0.5x0.5 = 0.25
All ADD UP to 1.0

10. 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 B
n (a,b) = n(A) x n(B)
(this assumes that each outcome has no influence on the next outcome)

11. 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

12. Problem
if 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 H T 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)

13. Revisit - Sample Space
the 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

14. Problem: Conditional Probability
P(heads | even)
these are independent events, so knowing the outcome of the second does not change the probability of the first

15. 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

16. Another way to prove multiplicative principle

17. 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.

18. 7 Red 3 Blue. Pick 2, without replacement.
a) p(R,R) b) p(B,B) c) p(One of each)
OUTCOMES
P(Outcome)
1st event
2nd event
6/9
R,R
P(R,R)=42/90
7/10
3/10
3/9
R,B
P(R,B)=21/90
7/9
B,R
P(B,R)=21/90
2/9
B,B
P(B,B)=6/90
Total P(all outcomes) = 1

19. Example
if 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

20. Exercises read the examples on pages 242-244 page 245# 1-11
make sure that you are understanding these concepts