In this unit we investigate ways of systematically counting all the possible outcomes of an event. One approach is to use tree diagrams.
- systematically find all possible outcomes of an event using tree diagrams and organised lists
A quick glance at newspapers shows the extent to which the language of probability has become important. Individuals need a knowledge of probability to understand consumer reports, surveys and samples. Nearly all activity in the working world requires making decisions in uncertain conditions.
The goal is to help students develop the critical thinking skills needed to reach sound conclusions based on appropriate data samples. Dice and spinner games which may be fair or unfair help develop the essential skills for understanding probability. These skills include: methods of organised counting comparing results of experiments to theoretical probabilities correctly using the language of probability In this unit we focus on the use of tree diagrams as a systematic way of counting all possible outcomes for a simple event.
Let’s consider the outcomes of tossing three coins. Step 1: List the possible outcomes for the first of the three coins Step 2: Draw branches from each of these outcomes. The number of branches will be the number of possible outcomes for the second coin. Step 3: Draw branches for each of the possible outcomes of tossing the third coin. Step 4: Read down the chart from the top to identify the 8 different combinations. Students may need practice in spacing the outcomes especially in the first column. Tree diagrams can also be drawn horizontally (pointing to the left or to the right) or vertically (pointing up or down). As you will see, tree diagrams are quite useful ways to systematically produce all possible outcomes of a situation. However, if a large number of possibilities can occur at each stage, tree diagrams become somewhat messy. If there are too many outcomes the number of branches at a point grows and there can then be too many branches to keep track of. From the tree diagram above we see that the eight possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. This enables us to determine the probabilities of events related to the tossing of three coins. As there is only one outcome with three tails, the probability of getting three tails is 1/8. As there are three outcomes with one head, the probability of getting one head is 3/8. You can probably see that the probability of getting at least one head is 7/8. One thing that needs to be pointed out here is that in all these examples we are using situations where a multiplicative rule of counting applies. This means that we can get the answer by multiplying two lots of numbers together. For instance to get the result of tossing two coins we multiply 2 by 2 as we can get two outcomes (T and H) on each toss. (To get the 8 outcomes from three tosses we multiply 2 by 2 by 2.) This is because coin tossing in this way involves independent events. This is because even if we have one head on the first toss, we can still get a head on the second toss. The second toss does not rely on the first one in any way. The two events are independent.
You can see another example of this under the Problem Solving section of this site. Look for Statistics Level 3, Coin Shake Up. All of the problems in this unit involve independent events. We can use tree diagrams because each branch represents the choice at each stage and in each branch we have events which are independent of the previous branch. This allows us to multiply the possibilities together to get the number of outcomes. Compared to independent events we have dependent ones. For instance, choosing two students from a group of four involves dependent choices. Once the first student is chosen that student can’t be chosen again. So the second choice is, to some extent, dependent on the first choice. So when we choose two students from four, there are not 4 by 4 = 16 possibilities as you might at first suspect. (In fact there are just 6.) Counting dependent events requires different techniques to counting independent events. You can see an example of this under the Problem Solving section of this site. Look for Statistics Level 3, Grabbing CDs and Dressing in the Dark. As a rough and ready rule to decide between independent and dependent events, try a tree diagram. If you can’t seem to make the tree diagram work, then you’re almost certainly involved with dependent choices. But as we said above, you only have to worry about independent events in this unit.
odd, even, fairness, outcomes, probability, likelihood, prediction, choice, chance, divisible, theoretical probability, tree diagram
- We begin the unit by playing Odds and Evens using a spinner divided into thirds with the segments labelled 1, 2 or 3.
Odds and Evens
Players (in two teams or pairs) take turns spinning the spinner twice.
One player (or team) wins if the sum is even.
The other player or team wins if the sum is odd.
Before the students start playing the game ask:
Do you think this is a fair game?
Why or why not?
Let the teams or pairs play the game keeping track of the numbers of wins. Discuss the results and the students’ ideas about fairness.
How did you decide if the game was fair?
Why do some think it is fair and others think it isn’t fair? Can it be both?
- Deciding if the game is fair or not should lead into the idea that to predict fairness you need to know all possible outcomes, that is, all possible totals of numbers on the two spinners. From this you can see how many are odd and how many are even.
How could we find all possible outcomes?
- If no one suggests any version of a tree diagram introduce it as a useful technique when dealing with 2 or 3 variables.
- First list the possible outcomes of the first spin, (1, 2, 3).
- Next decide how many outcomes are possible for the second spin and draw that number of branches from each of the outcomes of the first spin.
- Next calculate the sum at the end of each branch.
- The sums can now be separated into even and odd numbers giving 5 even and 4 odd. Therefore the game is not fair. Over a long series of trials (games) you would expect to get an even sum more often than an odd one.
- Conclude by discussing the probability of getting an odd and an even total.
- Probability of an even total = 5/9
- Probability of an odd total = 4/9
Over the next 2-3 days give the students the following problems to solve. Encourage them to think of ways to justify their solutions and their selection of a counting method. Ask them also to write the probabilities associated with each of the problems.
Problem 1: Sam’s Sandwich
Every morning Sam makes his own sandwich. He can use brown or white bread and can choose between honey, cheese, vegemite, lettuce or tomato.
How many different sandwiches can Sam make with one filling?If Sam makes each possible sandwich equally often, what is the probability that he made a white sandwich with honey for lunch today?
Problem 2: Pete’s Pizza
At Pete’s Pizza there is a special on. You can choose two toppings on your pizza. One choice has to be made from the ham, mushroom, pepperoni and bacon bin, while the other choice has to come from the pineapple, tomato and extra cheese bin.
How many different pizzas are possible with 2 toppings?
If people choose the toppings at random, what’s the probability that Pete will sell a bacon and tomato pizza next?
Problem 3: Coin Tossing
Georgia is captain of her netball team and always selects heads at the start of the game. What is the probability that she will win the toss for the next four games?
Problem 4: Licence Plates
In Botutuland, car licence plates have two letters and four digits. If all of the 10 digits can occur in any one of the four spots how many different licence plates are possible?
If all the licence plates are being used, what’s the probability of a Botutulandian seeing a licence plate with 0000?
Problem 5: Travel
Laura can travel to the city by train, ferry or car. She can then walk, taxi or take a bus to her office.
How many different ways can she travel to her office?
If she is just as likely to select any of the transport types what is the probability that she will travel by ferry and then walk to her office?
Problem 6: Sports
At Tim’s school, they can play one of cricket, hockey, tennis or touch rugby in the summer and one of soccer, rugby or basketball in the winter. What is the probability that he played two games at school last year with a round ball? What event has probability ¼ (or 3/12)?
Problem 7: Spinners II
Mala has a spinner that is divided into quarters with the numbers 1, 2, 3, 4 put on to each quarter. She plays the odd and even game with her friend Erik. Is this a fair game? What event(s) has probability 10/16?
Problem 8: Spinners III
Mala has a spinner that is divided into quarters with 1s in two of the opposite quarters and 2 and 3 in the other two quarters. She plays the odd and even game with her friend Erik. Is this a fair game?
Problem 9: Spinners IV
Mala has a spinner that is divided into quarters with 1s in two of the opposite quarters and 2 and 3 in the other two quarters. She and her friend Erik spin the spinner twice and add the scores of the two spins. Here you win if you get a number divisible by 3. Is this a fair game?
Problem 10: Spinners V
Mala has a spinner that is divided into quarters with 1s in two of the opposite quarters and 2 and 3 in the other two quarters. She plays the following game with her friends Erik and Nelio. The scores of two spins are added. Mala wins if the sum is divisible by 3; Erik wins if the sum has a remainder of 1 when divided by 3; and Nelio wins if the remainder is 2 when the sum is divided by 3. Is this a fair game?
Problem 1 Answer: Sam has the choice of 2 breads and 5 fillings. So he has the choice of 2 x 5 = 10 sandwiches.
This can be done using a tree diagram that first has 2 branches (one for each of the bread types) and then 5 branches at the end of the first branches (one for each of the fillings). This will give 10 ends to the tree.
Now a white sandwich with honey is just one of the ten possible sandwiches. So the chances of making that particular sandwich is 1/10.
Problem 2 Answer: For the pizza you get a choice of 4 from one bin and 3 from the other. Hence there are 12 possible pizzas with two toppings, one topping coming from each bin.
Bacon and tomato is just one of the 12 possible pizzas, so the chances of that pizza are 1/12.
Problem 3 Answer: If you construct the tree here you start off with two branches and you add two branches at the end of each branch until four lots of branch levels have been added. This gives 16 possible outcomes. Only one of these is all heads. So the probability that she wins four calls in a row is 1/16.
Problem 4 Answer: It should be pointed out that this problem is a little hard to do by a tree diagram. First of all there are 10,000 possible numbers that can be used (going from 0 to 9999). Now there are 26 letters that can go in the first letter position and 26 for the second. Altogether there are 26 x 26 x 10,000 = 6,760,000 number plates.
Now there are 26 x 26 = 676 number plates with all zeros. So the probability if seeing one of these is 676/6,760,000 = 1/10,000. (You probably wouldn’t see one of these very often in Botutuland.)
Problem 5 Answer: Laura has 3 ways of getting to the city and then 3 ways of getting to the office. So she has 3 x 3 = 9 ways of getting to work.
The probability of going by ferry and then walking is 1/9.
Problem 6 Answer: Tim has 4 x 3 = 12 choices of sports. He has a choice of 3 round-ball games in summer and 2 in winter for a total of 6. Hence his chances of playing a round-ball game is 6/12 = ½.
There are many ways of getting a probability of 1/4. One way is to play a round-ball game in summer and a non round-ball game in winter.
Problem 7 Answer: There are 16 outcomes here and 8 of them are even. The game is fair.
One way of getting a probability of 10/16 is to take the event ‘the sum is less than 6’. But ‘more than 3’ works equally well. But there are other possibilities.
Problem 8 Answer: Here 10 of the 16 outcomes are even. So it is best to be even. This is definitely not a fair game.
Problem 9 Answer: There are only 5 outcomes here out of 16 that have a number divisible by 3. Hence this is not a fair game.
Problem 10 Answer: This is not a fair game either. There are 5 sums divisible by 3; 5 sums with a remainder of 1 when divided by 3; and 6 with a remainder of 2 when divided by 3. So it is better to be Nelio.
Incidentally, since there are 16 outcomes here and 16 is not divisible by 3, then there is no way that this game could be fair.
Problem Homelink 1: There are 6 outcomes for the dice and 2 for the coin so there are 6 x 2 = 12 altogether.
There are 1 x 3 = 3 ways of getting a head and an even number. So the probability of this is 3/12 = ¼.
Problem Homelink 2: There are 18 different outcomes here. Since 12 of these are even, the probability of getting an even number is 12/18 = 2/3. The probability of getting an odd number is 6/18 = 1/3.
The probability of getting a number less than 4 is 5/18. (There are other possibilities here too. For instance, getting a number in double figures.)
In today’s session the students work in pairs to make up their own counting problems for other students to solve.
- Introduce the session by discussing the variety of T-shirts worn (or owned) by class members. List the different ways that T-shirts can vary, for example, colour, size, logo (or not), neck style.
- Ask the students to work in pairs to make up a T-shirt problem using 3 of the variables listed.
- Tell them to write the problem on one side of a piece of paper (or on the outside cover of a folded piece) and the solution on the other side.
- Then get them to trade problems with other students. When they have solved the problem compare their solution with that of the problem writers.
- Leave the problems on display