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Finding Probability Using Tree Diagrams and Outcome Tables 4.5.

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Presentation on theme: "Finding Probability Using Tree Diagrams and Outcome Tables 4.5."— Presentation transcript:

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)link to movie

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

5 Outcome Table if you flip a coin twice, you can model also model the results with an outcome table Flip 1Flip 2Simple Event HHHH HTHT THTH TTTT

6 Tree Diagrams – For flipping a coin Probability of two or more events 1 st Throw2 nd Throw T HHHH H T TTT 1/2 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)

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. One taken, colour noted, returned to bag, then a second taken. Draw tree diagram for 2 draws. 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 Now add in the probability

9 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 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

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

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 3/9 6/9 7/10 3/10 2/9 7/9 1 st event2 nd event 7 Red 3 Blue. Pick 2, without replacement. a) p(R,R) b) p(B,B) c) p(One of each) OUTCOMESP(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

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


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