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**KS3 Mathematics D4 Probability**

The aim of this unit is to teach pupils to: Use the vocabulary of probability Use the probability scale; find and justify theoretical probabilities Collect and record experimental data, and estimate probabilities based on the data Compare experimental and theoretical probabilities Material in this unit is linked the Framework’s supplement of examples pp276 –285. D4 Probability

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**D4.1 The language of probability**

Contents D4 Probability D4.1 The language of probability D4.2 The probability scale D4.3 Calculating probability D4.4 Probability diagrams D4.5 Experimental probability 1 of 20

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**The language of probability**

Probability is a measurement of the chance or likelihood of an event happening. Words that we might use to describe probabilities include: 50-50 chance likely unlikely poor chance certain very likely possible Ask pupils to give examples of sentences for each phrase. impossible even chance probable

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**Fair games A game is played with marbles in a bag.**

One of the following bags is chosen for the game. The teacher then pulls a marble at random from the chosen bag: bag a bag b bag c Discuss the following questions: a) From which bag are the girls most likely to win a point? Why? b) From which bag are the boys least likely to win a point? Why? c) From which bag is impossible for the girls to win a point? d) From which bag are the boys certain to win a point? e) From which bag is it equally likely for the boys or the girls to win a point? f) Are any of the bags unfair? Why? If a red marble is pulled out of the bag, the girls get a point. If a blue marble is pulled out of the bag, the boys get a point. Which would be the fair bag to use?

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**A game is fair if all the players have an equal chance of winning.**

Fair games A game is fair if all the players have an equal chance of winning. Which of the following games are fair? A dice is thrown. If it lands on a prime number team A get a point, if it doesn’t team B get a point. There are three prime numbers (2, 3 and 5) and three non-prime numbers (1, 4 and 6). Remind pupils that 1 is not a prime number because it does have two factors. Yes, this game is fair.

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Fair games Nine cards numbered 1 to 9 are used and a card is drawn at random. If a multiple of 3 is drawn team A get a point. If a square number is drawn team B get a point. If any other number is drawn team C get a point. There are three multiples of 3 (3, 6 and 9). There are three square numbers (1, 4 and 9). Ask pupils to explain whether or not they think this game is fair. Although there are three cards that would give either team A or team B a point, there are four cards that would give team C a point. If appropriate, stress that the outcome of drawing a multiple of 3 and the outcome of drawing a square number are not mutually exclusive. It is possible to draw a card that is both a multiple of 3 and a square number, that is the card with a 9 on it. Therefore, P(A score a point) + P(B score a point) + P(C score a point) does not equal 1. Only the sum of all mutually exclusive outcomes equals 1. There are four numbers that are neither square nor multiples of 3 (2, 5, 7 and 8). No, this game is not fair. Team C is more likely to win.

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**Fair games A spinner has five equal sectors numbered 1 to 5.**

The spinner is spun many times. If the spinner stops on an even number team A gets 3 points. If the spinner stops on an odd number team B gets 2 points. 1 2 3 4 5 Suppose the spinner is spun 50 times. We would expect the spinner to stop on an even number 20 times and on an odd number 30 times. The game is fair because, although the spinner is less likely to stop on an even number, team A gets proportionally more points when it does. The probability of the spinner stopping on an even number is 0.4. The probability of the spinner stopping on an odd number is 0.6. So, the spinner is 50% more likely to stop on an odd number. The game is fair because team A get 50% more points when the spinner stops on an even number. We can show that the game is fair by considering a theoretical game where the spinner is spun 50 times. Team A would score 20 × 3 points = 60 points Team B would score 30 × 2 points = 60 points Yes, this game is fair.

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**Scratch cards Scratch off a £ sign and win £10! £ £ £ £**

no win no win no win no win Discuss which card is most likely to win. The yellow card and the blue card both have the same number of £ signs. However the blue card has more no win boxes. Conclude that it has a smaller proportion of £ signs and is less likely to win. The red scratch card has five £ signs and so some pupils may believe that this card has a greater probability of winning. Stress that the red card has a smaller proportion of winning squares than the yellow card (5/16 is less than 3/9). The yellow card is therefore more likely to win a prize. By comparing the proportions of winning squares conclude that the yellow card is most likely to win, followed by the red card. The blue card is least likely to win. You are only allowed to scratch off one square and you can’t see what is behind any of the squares. Which of the scratch cards is most likely to win a prize?

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**Bags of counters Choose a blue counter and win a prize!**

bag c bag a bag b bag c Discuss which bag is most likely to win. Stress that is the bag with the largest proportion of blue counters that is the most likely to win. Bag a has 4/12 blue counters, bag b has 3/10 blue counters and in bag c there are 2/5 blue counters. Converting these to decimals we have blue in bag a, 0.3 blue in bag b and 0.4 blue in bag c. Bag c is therefore the most likely to win the prize and then bag a. Bag b is the least likely to win. We can also look at the ratio of blue counters to yellow counters. In bag a there are two yellow counters for each blue counter. In bag b there are 21/3 yellow counters for each blue counter. In bag c there is 11/2 yellow counters for each blue counter. Bag c is the most likely to win because there are fewer yellow counters for each blue counter. Stress the difference between ratio and proportion. Proportion compares the parts to the whole and ratio compares the parts to each other. You are only allowed to choose one counter at random from one of the bags. Which of the bags is most likely to win a prize?

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**Probability statements**

Statements involving probability are often incorrect or misleading. Discuss the following statements: The number 18 is has been drawn the most often in the national lottery so I’m more likely to win if I choose it. I’ve just thrown four heads in a row so I’m much less likely to get a head on my next throw. Discuss each statement in detail. Many misconceptions arise in probability due to a failure to appreciate the random nature of independent events. For example, for the second statement, if we assume that the coin is unbiased then the next throw is just as likely to come up heads than any other throw- the probability is ½- because the coin has no memory and the results are random. It is true that it is very unlikely that five heads in a row would be thrown, however, we are not talking about the probability of getting five heads in a row, we are talking about the probability of the next throw being heads. We could also argue that since it is unlikely that four heads would be thrown in a row the coin must be biased in some way. Based on the coins past history it could therefore be argued that the coin is actually more likely to land heads up. For the last statement, this is only true if the meal served is random and that both curry and pizza are equally likely. If there is a choice involved then the only way to estimate the probability of the next person choosing curry is to carry out a survey to find out which meal people prefer. If 74 out of 100 people surveyed preferred pizza, for example, then we could estimate that the probability of the next person choosing pizza is 0.74 or 74%. There are two choices for lunch, pizza or curry. That means that there is a 50% chance that the next person will choose pizza. I’m so unlucky. If I roll this dice I’ll never get a six.

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**D4.2 The probability scale**

Contents D4 Probability D4.1 The language of probability D4.2 The probability scale D4.3 Calculating probability D4.4 Probability diagrams D4.5 Experimental probability

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The probability scale The chance of an event happening can be shown on a probability scale. Meeting with King Henry VIII A day of the week starting with a T The next baby born being a boy Getting homework this lesson A square having four right angles impossible Discuss the probability scale. The more likely an event is to occur, the further to the right of the line it is placed. The less likely an event is to occur, the further to the left. unlikely even chance likely certain Less likely More likely

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**The probability scale We measure probability on a scale from 0 to 1.**

If an event is impossible or has no probability of occurring then it has a probability of 0. If an event is certain it has a probability of 1. This can be shown on the probability scale as: 1 impossible even chance certain Probabilities are written as fractions, decimal and, less often, as percentages between 0 and 1.

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The probability scale Ask pupils to drag the pointer to the correct position on the scale.

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**D4.3 Calculating probability**

Contents D4 Probability D4.1 The language of probability D4.2 The probability scale D4.3 Calculating probability D4.4 Probability diagrams D4.5 Experimental probability

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Higher or lower Start by revealing the first card and asking the class to predict whether the next card will be higher or lower? How often can pupils correctly predict whether the next card will be higher or lower? When can they be completely sure of their answer? Discuss strategies. Strategies may improve as you play more than once; for example, are pupils taking into account all the cards already turned over, or just the last one turned?

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**Listing possible outcomes**

When you roll a fair dice you are equally likely to get one of six possible outcomes: 1 6 1 6 1 6 1 6 1 6 1 6 Explain that the word ‘fair’ or ‘unbiased’ means that each outcome is equally likely. Some dice are ‘weighted’. That means that the weight of the dice is unevenly distributed and some numbers are more likely to appear than others. The probability of getting any number is 1 (certain) so the probability of getting each different number is 1 ÷ 6 or 1/6. Since each number on the dice is equally likely the probability of getting any one of the numbers is 1 divided by 6 or 1 6

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**Calculating probability**

What is the probability of the following events? 1) A coin landing tails up? 3) Drawing a seven of hearts from a pack of 52 cards? 1 2 P(tails) = 1 52 P(7 of ) = 2) This spinner stopping on the red section? 4) A baby being born on a Friday? For each example ask pupils to tell you the number of equally likely outcomes before revealing the probability. Introduce the notation of P(n) for the probability of an event n. 1 4 1 7 P(red) = P(Friday) =

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**Calculating probability**

If the outcomes of an event are equally likely then we can calculate the probability using the formula: Probability of an event = Number of successful outcomes Total number of possible outcomes For example, a bag contains 1 yellow, 3 green, 4 blue and 2 red marbles. Point out that calculated probabilities are usually given as fractions but that they can also be given as decimals and (less often) as percentages. Ask pupils to give you the probabilities (as decimals, fractions and percentages) of getting A blue marble A red marble A yellow marble A purple marble A blue or a green marble etc. What is the probability of pulling a green marble from the bag without looking? 3 10 P(green) = or 0.3 or 30%

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**Calculating probability**

This spinner has 8 equal divisions: What is the probability of the spinner landing on a red sector? a blue sector? a green sector? 2 8 = 1 4 a) P(red) = 1 8 b) P(blue) = 4 8 = 1 2 c) P(green) =

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**Calculating probability**

A fair dice is thrown. What is the probability of getting a 2? a multiple of 3? an odd number? a prime number? a number bigger than 6? an integer? 1 6 a) P(2) = 2 6 = 1 3 b) P(a multiple of 3) = 3 6 = 1 2 c) P(an odd number) =

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**Calculating probability**

A fair dice is thrown. What is the probability of getting a 2? a multiple of 3? an odd number? a prime number? a number bigger than 6? an integer? 3 6 = 1 2 d) P(a prime number) = Don’t write 6 e) P(a number bigger than 6) = 6 f) P(an integer) = = 1

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**Calculating probability**

The children in a class were asked how many siblings (brothers and sisters) they had. The results are shown in this frequency table: Number of siblings Number of pupils 4 1 8 2 9 3 5 6 7 What is the probability that a pupil chosen at random from the class will have two siblings? There are 30 pupils in the class and 9 of them have two siblings. 9 30 = 3 10 So, P(two siblings) =

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**Calculating probability**

A bag contains 12 blue balls and some red balls. The probability of drawing a blue ball at random from the bag is How many red balls are there in the bag? 3 7 12 balls represent of the total. 3 7 So, 4 balls represent of the total 1 7 Reason that if the probability of drawing a blue ball at random is 3/7, then 3/7 of the balls must be blue. To find the total number of balls we must divide the number of blue balls by 3 and multiply by 7 to get 28. The number of red balls is then found by subtracting 12 from 28 to get 16. Discuss alternative methods to calculate this using proportional reasoning or algebra. and, 28 balls represent of the total. 7 The number of red balls = 28 – 12 = 16

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**The probability of an event not occurring**

The following spinner is spun once: What is the probability of it landing on the yellow sector? 1 4 P(yellow) = What is the probability of it not landing on the yellow sector? Two probabilities that add up to one are sometimes called complementary probabilities (compare with number complements). 3 4 P(not yellow) = If the probability of an event occurring is p then the probability of it not occurring is 1 – p.

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**The probability of an event not occurring**

The probability of a factory component being faulty is What is the probability of a randomly chosen component not being faulty? P(not faulty) = 1 – 0.03 = 0.97 The probability of pulling a picture card out of a full deck of cards is What is the probability not pulling out a picture card? 3 13 P(not a picture card) = 1 – = 3 13 10 13

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**The probability of an event not occurring**

The following table shows the probabilities of 4 events. For each one work out the probability of the event not occurring. Event Probability of the event occurring Probability of the event not occurring 3 5 2 5 A B 0.77 0.23 C 9 20 11 20 D 8% 92%

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**The probability of an event not occurring**

There are 60 sweets in a bag. 10 are cola bottles, 1 4 are fried eggs, 20 are hearts, the rest are teddies. What is the probability that a sweet chosen at random from the bag is: We can work out the number of sweets that are not teddies by finding the sum of 10, 20 and a ¼ of 60 to get 45. Modify the numbers to make this problem more challenging. 5 6 a) Not a cola bottle P(not a cola bottle) = 45 60 = 3 4 b) Not a teddy P(not a teddy) =

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**Mutually exclusive outcomes**

Outcomes are mutually exclusive if they cannot happen at the same time. For example, when you toss a single coin either it will land on heads or it will land on tails. There are two mutually exclusive outcomes. Outcome A: Head Outcome B: Tail When you roll a dice either it will land on an odd number or it will land on an even number. There are two mutually exclusive outcomes. Outcomes are mutually exclusive if we can either have one or the other but not both. Stress that if we can use either … or … when describing two or more outcomes then they are probably mutually exclusive. Outcome A: An odd number Outcome B: An even number

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**Mutually exclusive outcomes**

A pupil is chosen at random from the class. Which of the following pairs of outcomes are mutually exclusive? Outcome A: the pupil has brown eyes. Outcome B: the pupil has blue eyes. These outcomes are mutually exclusive because a pupil can either have brown eyes, blue eyes or another colour of eyes. Outcome C: the pupil has black hair. Outcome D: the pupil has wears glasses. These outcomes are not mutually exclusive because a pupil could have both black hair and wear glasses.

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**Adding mutually exclusive outcomes**

If two outcomes are mutually exclusive then their probabilities can be added together to find their combined probability. For example, a game is played with the following cards: What is the probability that a card is a moon or a sun? 1 3 1 3 P(moon) = and P(sun) = Ask pupils to tell you the probability of getting a crescent card or a star card. Reveal the solution on the board. Stress that only events that are mutually exclusive can be added in this way. For example, If we are drawing a card at random from a pack P(King) = 2/52, P(Club) = 13/52, but P(King or club) 2/ /52 because a card could be both a king and a club. Drawing a moon and drawing a sun are mutually exclusive outcomes so, 1 3 + = 2 3 P(moon or sun) = P(moon) + P(sun) =

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**Adding mutually exclusive outcomes**

If two outcomes are mutually exclusive then their probabilities can be added together to find their combined probability. For example, a game is played with the following cards: What is the probability that a card is yellow or a star? 1 3 1 3 P(yellow card) = and P(star) = Ask pupils to tell you the probability of getting a yellow card or a star card. Stress that this cannot be found by adding. The probability is 5/9 because one of the cards is both yellow and a star. Drawing a yellow card and drawing a star are not mutually exclusive outcomes because a card could be yellow and a star. P (yellow card or star) cannot be found by adding.

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**The sum of all mutually exclusive outcomes**

The sum of all mutually exclusive outcomes is 1. For example, a bag contains red counters, blue counters, yellow counters and green counters. P(blue) = 0.15 P(yellow) = 0.4 P(green) = 0.35 What is the probability of drawing a red counter from the bag? Explain that when we draw a counter from the bag it is either red, blue, yellow or green. These outcomes are therefore mutually exclusive, there are no other possible outcomes and so their combined probabilities must equal 1. Mutually exclusive outcomes can be added together. The decimals on this slide can be changed to make the problem more challenging. P(blue, yellow or green) = = 0.9 P(red) = 1 – 0.9 = 0.1

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**The sum of all mutually exclusive outcomes**

A box contains bags of crisps. The probability of drawing out the following flavours at random are: 2 5 1 3 P(salt and vinegar) = P(ready salted) = The box also contains cheese and onion crisps. What is the probability of drawing a bag of cheese and onion crisps at random from the box? Pupils may need to revise the addition of fractions with different denominators to complete this question. See N6.1 Adding and subtracting fractions. Explain that when we draw a packet of crisps from the box it is either salt and vinegar, cheese and onion or ready salted. These outcomes are therefore mutually exclusive. There are no other possible outcomes and so their combined probabilities must equal 1. 2 5 + 1 3 = 6 + 5 15 = 11 15 P(salt and vinegar or ready salted) = P(cheese and onion) = 1 – 11 15 = 4 15

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**The sum of all mutually exclusive outcomes**

A box contains bags of crisps. The probability of drawing out the following flavours at random are: 2 5 1 3 P(salt and vinegar) = P(ready salted) = The box also contains cheese and onion crisps. There are 30 bags in the box. How many are there of each flavour? 2 5 of 30 = Number of salt and vinegar = Pupils may need to revise finding fractions of amounts to complete this question. See N6.2 Finding a fraction of an amount. Explain that the probability relates to the proportion of each flavour. If we know the probability of getting each flavour and the number of bags altogether, then we can use this information to work out the number of each flavour. 12 packets 1 3 of 30 = Number of ready salted = 10 packets 4 15 of 30 = Number of cheese and onion = 8 packets

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**D4.4 Probability diagrams**

Contents D4 Probability D4.1 The language of probability D4.2 The probability scale D4.3 Calculating probability D4.4 Probability diagrams D4.5 Experimental probability

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**Finding all possible outcomes of two events**

Two coins are thrown. What is the probability of getting two heads? Before we can work out the probability of getting two heads we need to work out the total number of equally likely outcomes. There are three ways to do this: 1) We can list them systematically. Using H for heads and T for tails, the possible outcomes are: Stress that when there is more than one event it is important to list all the possible outcomes systematically. Listing the outcomes systematically means listing them in a logical order to make sure that none are left out. Explain that TH means, ‘a tail on the first coin and a head on the second’ and that HT means, ‘a head on the first coin and a tail on the second’. These are therefore two separate events. TH and HT are separate equally likely outcomes. TT, TH, HT, HH.

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**Finding all possible outcomes of two events**

2) We can use a two-way table. Second coin H T First coin HH HT TH TT From the table we see that there are four possible outcomes one of which is two heads so, 1 4 P(HH) =

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**Finding all possible outcomes of two events**

3) We can use a probability tree diagram. Outcomes Second coin HH H First coin H T HT T H TH T TT Again we see that there are four possible outcomes so, 1 4 P(HH) =

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**Finding the sample space**

A red dice and a blue dice are thrown and their scores are added together. What is the probability of getting a total of 8 from both dice? There are several ways to get a total of 8 by adding the scores from two dice. We could get a 2 and a 6, a 3 and a 5, a 4 and a 4, Stress that the sample space is the set of all possible outcomes. a 5 and a 3, and a 6 and a 2. To find the set of all possible outcomes, the sample space, we can use a two-way table.

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**Finding the sample space**

+ From the sample space we can see that there are 36 possible outcomes when two dice are thrown. 2 3 4 5 6 7 8 3 4 5 6 7 8 8 4 5 6 7 8 9 8 5 6 7 8 9 10 Five of these have a total of 8. As the two-way table is completed ask pupils what patterns they notice. Use the completed table to justify that the probability of the total score on the two dice being 8 is 5/35. Ask pupils to use the table to give the probabilities of other scores such as: P(3) P(a score less than 7) P(an even score) P(a score that is prime) P(a score that is square) Ask pupils to cancel down any fractions if possible. 8 6 7 8 9 10 11 5 36 P(8) = 8 7 8 9 10 11 12

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Scissors, paper, stone In the game scissors, paper, stone two players have to show either scissors, paper, or stone using their hands as follows: scissors paper stone The rules of the game are that: Scissors beats paper (it cuts). Paper beats stone (it wraps). Stone beats scissors (it blunts). If both players show the same hands it is a draw.

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Scissors, paper, stone What is the probability that both players will show the same hands in a game of scissors, paper, stone? We can list all the possible outcomes in a two-way table using S for Scissors, P for Paper and T for sTone. Scissors Paper Stone First player Second player SS SS SP ST PS PP PP PT TS TP TT TT 3 9 = 1 3 P(same hands) =

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Scissors, paper, stone What is the probability that the first player will win a game of scissors, paper, stone? Using the two-way table we can identify all the ways that the first player can win. Scissors Paper Stone First player Second player SS SP ST PS PP PT TS TP TT SP PT Remember the rules of the game: Scissors beats paper; paper beats stone; stone beats scissors. TS 3 9 = 1 3 P(first player wins) =

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Scissors, paper, stone What is the probability that the second player will win a game of scissors, paper, stone? Using the two-way table we can identify all the ways that the second player can win. Scissors Paper Stone First player Second player SS SP ST PS PP PT TS TP TT ST PS Remember the rules of the game: Scissors beats paper; paper beats stone; stone beats scissors. TP 3 9 = 1 3 P(second player wins) =

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**Scissors, paper, stone Is scissors, paper, stone a fair game? 1**

P(first player wins) = 1 3 P(second player wins) = 1 3 P(a draw) = 1 3 Both player are equally likely to win so, yes, it is a fair game. Review what is meant by a fair game. In a fair game all players are equally likely to win. Allow pupils to play the game in pairs and to record their results. Discuss the fact that in 30 games we would expect to get 10 wins for the first player, 10 wins for the second player and 10 draws. Discuss why this does not happen in reality. Play scissors paper stone 30 times with a partner. Record the number of wins for each player and the number of draws. Are the results as you expected?

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**D4.5 Experimental probability**

Contents D4 Probability D4.1 The language of probability D4.2 The probability scale D4.3 Calculating probability D4.4 Probability diagrams D4.5 Experimental probability

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**Estimating probabilities based on data**

What is the probability a person chosen at random being left-handed? Although there are two possible outcomes, right-handed and left-handed, the probability of someone being left-handed is not ½, why? The two outcomes, being left-handed and being right-handed, are not equally likely. There are more right-handed people than left-handed. To work out the probability of being left-handed we could carry out a survey on a large group of people.

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**Estimating probabilities based on data**

Suppose 1000 people were ask whether they were left- or right-handed. Of the 1000 people asked 87 said that they were left-handed. From this we can estimate the probability of someone being left-handed as or 87 1000 If we repeated the survey with a different sample the results would probably be slightly different. The more people we asked, however, the more accurate our estimate of the probability would be.

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Relative frequency The probability of an event based on data from an experiment or survey is called the relative frequency. Relative frequency is calculated using the formula: Relative frequency = Number of successful trials Total number of trials For example, Ben wants to estimate the probability that a piece of toast will land butter-side-down. He drops a piece of toast 100 times and observes that it lands butter-side-down 65 times. Relative frequency = 65 100 = 13 20

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Relative frequency Sita wants to know if her dice is fair. She throws it 200 times and records her results in a table: Number Frequency Relative frequency 1 31 2 27 3 38 4 30 5 42 6 32 31 200 = 0.155 27 200 = 0.145 38 200 = 0.190 30 200 = 0.150 If the dice were fair we would expect to get each outcome an equal number of times. Stress that in an experiment the results are random and unpredictable. If we repeated this experiment we would get a different set of results. Conclude that this dice seems to be fair because the relative frequencies are all close to 1/6 or 42 200 = 0.210 32 200 = 0.160 Is the dice fair?

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Expected frequency The theoretical probability of an event is its calculated probability based on equally likely outcomes. If the theoretical probability of an event can be calculated, then when we do an experiment we can work out the expected frequency. Expected frequency = theoretical probability × number of trials For example, if you rolled a dice 300 times, how many times would you expect to get a 5? The theoretical probability of getting a 5 is . 1 6 So, expected frequency = × 300 = 1 6 50

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Expected frequency If you tossed a coin 250 times how many times would you expect to get a tail? 1 2 Expected frequency = × 250 = 125 If you rolled a fair dice 150 times how many times would you expect to a number greater than 2? Stress that the greater the number of trials the closer the experimental frequency will be to the expected frequency. 2 3 Expected frequency = × 150 = 100

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Spinners experiment Use the spinners experiment to compare the theoretical probability with the relative frequency for each spinner. Notice that the more times the spinner is spun the closer the relative frequency gets to the theoretical probability.

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Random results Remember that when an experiment is carried out the results will be random and unpredictable. Each time the experiment is repeated the results will be different. The more times an experiment is repeated the more accurate the estimated probability will be. Jenny throws a dice 12 times and doesn’t get a six. She concludes that the dice must be biased. Discuss the unpredictability of random processes. Although you would expect to get two sixes in twelve throws it is possible that you won’t. You would have to thrown the dice many more times to find out if it is biased.

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