2. Probability (Tree Diagrams) Tree diagrams can be used to help solve problems involving both dependent and independent events. The following situation can be represented by a tree diagram. Peter has ten coloured cubes in a bag. Three of the cubes are red and 7 are blue. He removes a cube at random from the bag and notes the colour before replacing it. He then chooses a second cube at random. Record the information in a tree diagram. First Choice Second Choice 3 3 3 9 10 red P(red and red) = x 10 10 100 3 red 10 7 3 7 21 10 blue P(red and blue) = x 10 10 100 3 7 3 21 7 10 red P(blue and red) = x 10 10 100 10 blue Independent 7 10 blue P(blue and blue) = 7 x 7 10 10 100 49
3. Probability (Tree Diagrams) Characteristics of a tree diagram First Choice Second Choice 3 3 3 9 10 red P(red and red) = x 10 10 100 3 red 10 7 3 7 21 10 blue P(red and blue) = x 10 10 100 3 7 3 21 7 10 red P(blue and red) = x 10 10 100 10 blue 7 7 7 49 The blue P(blue and blue) = x 10 10 10 100 probabilities for each event are shown along Ends of first and the arm of Probabilities are second level multiplied along each branch branches show each arm. and they sum to 1. Characteristics the different outcomes.
4. Probability (Tree Diagrams) Question 1 Rebecca has nine coloured beads in a bag. Four of the beads are black and the rest are green. She removes a bead at random from the bag and notes the colour before replacing it. She then chooses a second bead. (a) Draw a tree diagram showing all possible outcomes. (b) Calculate the probability that Rebecca chooses: (i) 2 green beads (ii) A black followed by a green bead. First Choice Second Choice 4 4 4 16 9 black P(black and black) = x 9 9 81 4 black 9 5 4 5 20 9 green P(black and green) = x 9 9 81 4 5 4 20 5 9 black P(green and black) = x 9 9 81 9 green 5 5 5 25 9 green P(green and green) = x 9 9 81 Q1 beads
5. Q2 Coins Probability (Tree Diagrams) Question 2 Peter tosses two coins. (a) Draw a tree diagram to show all possible outcomes. (b) Use your tree diagram to find the probability of getting (i) 2 Heads (ii) A head or a tail in any order. First Coin Second Coin 1 2 head P(head and head) = 1 x 1 1 2 2 4 1 head 2 1 1 1 1 2 tail P(head and tail) = x 2 2 4 1 1 1 1 1 2 head P(tail and head) = x 2 2 4 2 tail 1 1 1 1 2 tail P(tail and tail) = x 2 2 4 P(2 heads) = ¼ P(head and a tail or a tail and a head) = ½
6. Probability (Tree Diagrams) Q3 Sports Question 3 Peter and Becky run a race and play a tennis match. The probability that Peter wins the race is 0.4. The probability that Becky wins the tennis is 0.7. (a) Complete the tree diagram below. (b) Use your tree diagram to calculate (i) the probability that Peter wins both events. (ii) The probability that Becky loses the race but wins at tennis. Race Tennis 0.3 Peter 0.4 x 0.3 = 0.12 Peter Win 0.4 Win Becky 0.7 0.4 x 0.7 = 0.28 Win 0.3 Peter 0.6 x 0.3 = 0.18 0.6 Win Becky Win 0.7 Becky 0.6 x 0.7 = 0.42 Win P(Win and Win) for Peter = 0.12 P(Lose and Win) for Becky = 0.28
7. Probability (Tree Diagrams) Dependent Events The following situation can be represented by a tree diagram. Peter has ten coloured cubes in a bag. Three of the cubes are red and seven are blue. He removes a cube at random from the bag and notes the colour but does not replace it. He then chooses a second cube at random. Record the information in a tree diagram. First Choice Second Choice 2 3 2 6 9 red P(red and red) = x 10 9 90 3 red 10 7 3 7 21 9 blue P(red and blue) = x 10 9 90 3 7 3 21 7 9 red P(blue and red) = x 10 9 90 10 blue Dependent 6 9 blue P(blue and blue) = 7 6 42 x 10 9 90
8. Probability (Tree Diagrams) Dependent Events Question 4 Rebecca has nine coloured beads in a bag. Four of the beads are black and the rest are green. She removes a bead at random from the bag and does not replace it. She then chooses a second bead. (a) Draw a tree diagram showing all possible outcome (b) Calculate the probability that Rebecca chooses: (i) 2 green beads (ii) A black followed by a green bead. First Choice Second Choice 3 8 black P(black and black) = 4 x 3 12 9 8 72 4 black 5 green P(black and green) = 4 x 5 20 9 8 9 8 72 4 5 4 20 5 8 black P(green and black) = x 9 8 72 9 green 4 5 4 20 Q4 beads 8 green P(green and green) = x 9 8 72
9. Probability (Tree Diagrams) Dependent Events Question 5 Lucy has a box of 30 chocolates. 18 are milk chocolate and the rest are dark chocolate. She takes a chocolate at random from the box and eats it. She then chooses a second. (a) Draw a tree diagram to show all the possible outcomes. (b) Calculate the probability that Lucy chooses: (i) 2 milk chocolates. (ii) A dark chocolate followed by a milk chocolate. First Pick Second Pick 17 18 17 306 29 Milk P(milk and milk) = x 30 29 870 18 Milk 30 12 18 12 216 29 Dark P(milk and dark) = x 30 29 870 18 12 18 216 12 29 Milk P(dark and milk) = x 30 29 870 30 Dark 11 12 11 132 Q5 Chocolates 29 Dark P(dark and dark) = x 30 29 870
10. Probability (Tree Diagrams) 3 Independent Events First Choice Second Choice 4 red 20 11 20 blue red 4 5 20 20 yellow 4 red 11 20 11 20 20 blue blue 5 20 yellow 5 4 red 20 20 11 yellow 20 blue 5 20 yellow 3 Ind
11. Probability (Tree Diagrams) 3 Independent Events First Choice Second Choice red red blue 4 20 yellow red 11 20 blue blue yellow 5 red 20 yellow blue 3 Ind/Blank yellow
12. Probability (Tree Diagrams) 3 Independent Events First Choice Second Choice 3 Ind/Blank/2
13. Probability (Tree Diagrams) 3 Dependent Events First Choice Second Choice 3 red 19 11 19 blue red 4 5 20 19 yellow 4 red 11 19 10 20 19 blue blue 5 19 yellow 5 4 red 20 19 11 yellow 19 blue 4 19 yellow 3 Dep
14. Probability (Tree Diagrams) 3 Dependent Events First Choice Second Choice red red blue 4 20 yellow red 11 20 blue blue yellow 5 red 20 yellow blue 3 Dep/Blank yellow
15. Probability (Tree Diagrams) 3 Dependent Events First Choice Second Choice 3 Dep/Blank/2 Dep/Blank
16. Probability (Tree Diagrams) First Choice Second Choice Third Choice 3 red 10 2 Independent Events. 3 Selections 3 blue red 7 10 10 3 3 red red 10 10 7 blue blue 10 7 3 3 10 7 red 10 10 red 10 blue 7 10 3 blue 7 10 10 blue red 3 Ind/3 7 Select 10 blue
17. Probability (Tree Diagrams) First Choice Second Choice Third Choice red 2 Independent Events. 3 Selections blue red 3 red 10 red blue blue 7 red red 10 blue blue blue red 3 Ind/3 Select/Blank blue
18. Probability (Tree Diagrams) First Choice Second Choice Third Choice 2 Independent Events. 3 Selections 3 Ind/3 Select/Blank2
19. Probability (Tree Diagrams) First Choice Second Choice Third Choice 1 red 8 2 Dependent Events. 3 Selections 2 blue red 7 9 8 2 3 red red 8 10 7 blue blue 9 6 2 3 8 7 red 8 red 9 10 blue 6 8 3 blue 6 8 9 blue red 3 Dep/3 Select 5 8 blue
20. Probability (Tree Diagrams) First Choice Second Choice Third Choice red 2 Dependent Events. 3 Selections blue red 3 red 10 red blue blue 7 red red 10 blue blue blue red 3 Dep/3 3 Dep/3 Select/Blank Select blue
21. Probability (Tree Diagrams) First Choice Second Choice Third Choice 2 Dependent Events. 3 Selections 3 Dep/3 Select/Blank2 3 Dep/3 Select
22. Probability (Tree Diagrams) Tree diagrams can be used to help solve problems involving both dependent and independent events. The following situation can be represented by a tree diagram. Peter has ten coloured cubes in a bag. Three of the cubes are red and 7 are blue. He removes a cube at random from the bag and notes the colour before replacing it. He then chooses a second cube at random. Record the information in a tree diagram. Worksheet 1
+1-315-636-4485
+91 98151-74050 
