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Probability & Tree Diagrams. Possible Outcomes. For example – a fair coin is spun twice. 1 st. 2 nd. H. HH. H. T. HT. H. TH. T. T. TT. Attach probabilities. 1 st. 2 nd. H. HH. P(H,H)= ½x½=¼. ½. H. ½. ½. T. HT. P(H,T)= ½x½=¼. H. TH. ½. P(T,H)= ½x½=¼. ½. T. ½. T. - PowerPoint PPT Presentation
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Probability & Tree Diagrams
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
For example – a fair coin is spun twice
H
H
H
T
T
T
HH
HT
TH
TT
2nd 1st
Possible Outcomes
Attach probabilities
H
H
H
T
T
T
HH
HT
TH
TT
2nd 1st
½
½
½
½
½
½
P(H,H)=½x½=¼
P(H,T)=½x½=¼
P(T,H)=½x½=¼
P(T,T)=½x½=¼
INDEPENDENT EVENTS – 1st spin has no effect on the 2nd spin
Calculate probabilities
H
H
H
T
T
T
HH
HT
TH
TT
2nd 1st
½
½
½
½
½
½
P(H,H)=½x½=¼
P(H,T)=½x½=¼
P(T,H)=½x½=¼
P(T,T)=½x½=¼
Probability of at least one Head?
*
**
For example – 10 coloured beads in a bag – 3 Red, 2 Blue, 5 Green. One taken, colour noted, returned to bag, then a
second taken.
B
RR
2nd 1st
B
B
BR
R
R
R
G
G
G
G
RB
RGBR
BB
BGGR
GB
GG
INDEPENDENT EVENTS
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
Choose a mealMain courseSalad 0.2Egg & Chips 0.5Pizza 0.3
PuddingIce Cream 0.45Apple Pie 0.55
S
E
P
IC
AP
IC
AP
IC
AP
0.2
0.5
0.3
0.45
0.55
0.45
0.55
0.45
0.55
P(S,IC) = 0.2 x 0.45 = 0.09
P(S,AP) = 0.2 x 0.55 = 0.110
P(E,IC) = 0.5 x 0.45 = 0.225
P(E,AP) = 0.5 x 0.55 = 0.275
P(P,IC) = 0.3 x 0.45 = 0.135
P(P,AP) = 0.3 x 0.55 = 0.165
