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Year 10 Probability

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**Tree Diagrams Tree diagrams With replacement (independent)**

Without replacement (dependent)

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Tree diagrams Tree diagrams are very useful for showing outcomes for multiple events. A weighted tree diagram has the probabilities on the branches.

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**Independent Probability (Tree Diagrams) red red blue red blue blue**

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. Probability (Tree Diagrams) First Choice Second Choice red red blue red Independent blue blue

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**Characteristics Probability (Tree Diagrams)**

Characteristics of a tree diagram red blue First Choice Second Choice The probabilities for each event are shown along the arm of each branch and they sum to 1. Ends of first and second level branches show the different outcomes. Probabilities are multiplied along each arm. Characteristics

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**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. Probability (Tree Diagrams) black green First Choice Second Choice Q1 beads

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**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. Probability (Tree Diagrams) Q2 Coins head tail First Coin Second Coin P(head and a tail or a tail and a head) = ½ P(2 heads) = ¼

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**Probability (Tree Diagrams)**

Q3 Sports Probability (Tree Diagrams) 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 0.6 0.3 0.7 Peter Win P(Win and Win) for Peter = 0.12 0.4 x 0.3 = 0.12 0.4 x 0.7 = 0.28 0.6 x 0.3 = 0.18 0.6 x 0.7 = 0.42 Peter Win 0.4 Becky Win P(Lose and Win) for Becky = 0.28 0.7 Peter Win Becky Win Becky Win

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**With Replacement Examples:**

A coin is tossed 3 times, show the outcomes using a tree diagram. Each outcome is equally likely. Pr (HHH) = 1 8 Pr ( 2 heads) = Pr(HHT)+Pr(HTH)+Pr(THH) = 3 8

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**What is the probability of: tossing a head then rolling a five? **

A coin is tossed and then a die is rolled. Draw the tree diagram (complete). 1 2 3 4 5 6 H 1 2 3 4 5 6 T What is the probability of: tossing a head then rolling a five? tossing a tail then rolling an even? tossing a tail then rolling a seven? 𝟏 𝟏𝟐 𝟑 𝟏𝟐 = 𝟏 𝟒 𝟎

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**3. Consider the following tree diagram:**

What is the probability of: Drawing two reds? Drawing a red then a blue? Drawing 2 blues? Drawing no blues? Drawing 1 blue? This can be done 2 ways: RB or BR:

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**Dependent Probability (Tree Diagrams) Dependent Events red red blue**

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. Probability (Tree Diagrams) Dependent Events First Choice Second Choice red red blue red Dependent blue blue

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**Probability (Tree Diagrams)**

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. Probability (Tree Diagrams) Dependent Events black green First Choice Second Choice Q4 beads

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**Probability (Tree Diagrams)**

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. Probability (Tree Diagrams) Dependent Events Milk Dark First Pick Second Pick Q5 Chocolates

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**With Replacement Examples:**

A family has three children. What is the probability that: They are all boys? The 1st is a boy and the 2nd and the 3rd are girls? There is one boy and two girls? They are not all boys? a) Pr(BBB) = 𝟏 𝟖 B B B G b) Pr(BGG) = 𝟏 𝟖 B G c) Pr(BGG or GBG or GGB) = 𝟑 𝟖 G B B G G d) Pr(not all boys) = 1 –Pr(all boys) = 1- 𝟏 𝟖 = 𝟕 𝟖 B G G

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Example 2. A mathematics student calculates his chances of passing the next test according to the results on earlier tests. If he passed the last test he thinks his chances are 0.7 of passing the next test. If he failed the last test he estimates that the probability of passing the next test is 0.5. Draw a probability tree diagram to illustrate the possible results obtained on the next two tests, given that he failed the previous test. Find the probability that on the next two tests the student will: (a) pass both (b) pass the first but not the second (c) fail the first and pass the second (d) fail both Pass a) Pr(PP) = 0.5 x 0.7 = 0.35 0.7 Pass 0.5 Fail 0.3 b) Pr(PF) = 0.5 x 0.3 = 0.15 Fail 0.5 Pass c) Pr(FP) = 0.5 x 0.5 = 0.25 0.5 Fail 0.5 d) Pr(FF) = 0.5 x 0.5 = 0.25 Fail

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Example 3. From an urn containing 7 blue and 3 red balls, 2 balls are taken at random (i) with replacement (Independent)and (ii) without replacement (Dependent) Find the probability that: (a) both balls are blue (b) the first ball is red and the second is blue (c) one is red and the other is blue (i) (ii) Pr(BB) 7 10 × 7 10 = b) Pr(RB) 3 10 × 7 10 = c) Pr(RB or BR) (i) (ii) Pr(BB) 7 10 × 7 10 = b) Pr(RB) c) Pr(RB or BR) (i) (ii) Pr(BB) 7 10 × 7 10 = b) Pr(RB) 3 10 × 7 10 = c) Pr(RB or BR) 3 10 × × 3 10 = (i) (ii) Pr(BB) 7 10 × 7 10 = 7 10 × 6 9 = 7 15 b) Pr(RB) 3 10 × 7 10 = c) Pr(RB or BR) 3 10 × × 3 10 = (i) (ii) Pr(BB) 7 10 × 7 10 = 7 10 × 6 9 = 7 15 b) Pr(RB) 3 10 × 7 10 = 3 10 × 7 9 = 7 30 c) Pr(RB or BR) 3 10 × × 3 10 = 3 10 × × 3 9 = (i) (ii) Pr(BB) 7 10 × 7 10 = 7 10 × 6 9 = 7 15 b) Pr(RB) 3 10 × 7 10 = 3 10 × 7 9 = 7 30 c) Pr(RB or BR) 3 10 × × 3 10 = Blue Blue 7 6 With Without Blue Blue 7 3 3 Red 7 Red 7 Blue Blue 7 3 3 Red Red 3 Red Red 2

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**Exercise 8F Proficiency/Enrichment Foundation Standard Advanced**

Understanding 2 — Fluency 3, 5 6 - Problem-solving 9 7*, 10* Reasoning 11, 13 12* Enrichment

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