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Published byRobin Clatterbuck Modified about 1 year ago

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Probability & Tree Diagrams

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

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Attach probabilities H H H T T T HH HT TH TT 2 nd 1 st 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 – 1 st spin has no effect on the 2 nd spin

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Calculate probabilities H H H T T T HH HT TH TT 2 nd 1 st 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? * * *

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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 2 nd 1 st B B B R R R R G G G G RBRB RGRG BRBR BB BGBG GRGR GBGB GG INDEPENDENT EVENTS

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B RR 2 nd 1 st B B B R R R R G G G G RBRB RGRG BRBR BB BGBG GRGR GBGB GG 0.3 0.2 0.5 0.2 0.3 0.5 0.2 0.3 0.5 0.2 0.3 Probabilities P(RR) = 0.3x0.3 = 0.09 P(RB) = 0.3x0.2 = 0.06 P(RG) = 0.3x0.5 = 0.15 P(BR) = 0.2x0.3 = 0.06 P(BB) = 0.2x0.2 = 0.04 P(BG) = 0.2x0.5 = 0.10 P(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

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Choose a meal Main course Salad 0.2 Egg & Chips 0.5 Pizza 0.3 Pudding Ice Cream 0.45 Apple 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

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