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Tree diagrams Tree diagrams are used to display the sample space for the experiment or game. The tree branches out once for every stage of the experiment.

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Presentation on theme: "Tree diagrams Tree diagrams are used to display the sample space for the experiment or game. The tree branches out once for every stage of the experiment."— Presentation transcript:

1 Tree diagrams Tree diagrams are used to display the sample space for the experiment or game. The tree branches out once for every stage of the experiment. Below are the tree diagrams for flipping a coin and rolling a die and flipping three coins. Each branch represents a new event such as each flip of a coin or each child a mother has. Coin 3 Coin 2 Die Coin 1 H T HHH Coin H1 H T 1 2 3 4 5 6 H2 HHT H3 H4 H5 H6 H T O H T O H T HTH HTT THH THT TTH TTT H T 1 2 3 4 5 6 T1 T2 T3 T4 T5 T6 H T H T

2 Probability tree diagrams
With probability tree diagrams you place the probability of each event on each arm of the diagram. If there is no replacement in a question the probabilities on the arms change after each event. 2 D’s and 8 N’s are placed in a hat. 3 “letters” are drawn WITHOUT replacement. Draw a probability tree to show the outcomes. Example 1 Note the probabilities on each set of arms adds to 1 DON’T simplify the probabilities. This means all answers will have the same denominators

3 Example 2 A hat has 5 red disks and 3 black disks.
Two disks are drawn out without replacement. What is the probability of drawing: different coloured disks? 2 red disks? a) P(different) = 15/56 + 15/56 = 30/56 = 15/28 Note the probabilities on each set of arms adds to 1 Disk 2 Disk 1 4/7 20/56 R B RR b) P(RR) = 20/56 R B = 5/14 5/8 3/7 RB 15/56 5/7 R B BR 15/56 3/8 DON’T simplify the probabilities. This means all answers will have the same denominators 2/7 6/56 BB 56/56 Check

4 Example 3 The probability of a Sydney resident catching the flu is 60%. In a sample of 3 Sydney residents, what is the probability that; exactly 2 residents have the flu? at least one resident has the flu? The resident can be either have the flu (F, 3/5) or be healthy (H , 2/5). Resident 3 As the population is so large, the probability does not change. Resident 2 3/5 27/125 3/5 F H FFF F H Resident 1 2/5 FFH 18/125 a) P(2F, 1H) = 18/125 × 3 F H 3/5 3/5 F H FHF 18/125 = 54/125 2/5 2/5 FHH 12/125 b) P(≥1F) = 1 − P(3H) 3/5 3/5 F H HFF 18/125 = 1 − 8/125 = 117/125 F H 2/5 2/5 HFH 12/125 3/5 F H HHF 12/125 2/5 2/5 HHH 8/125 Check 125/125

5 Today’s work Yesterday’s work Exercise 6-07 Page 219 → 220
1 → 4, 7, 9 → 11 & 14 Yesterday’s work Exercise 6-06 Page 214 → 216 Odds


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