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Topic outline
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Video transcript
This unit of work is on the probability of combined events
Students often struggle with combined event problems although calculating probabilities for these is similar process to that of single events in that it amounts to counting up the number of equally likely outcomes that fit a particular situation.
In this unit we will use possibility space diagrams and explore how probabilities can be calculated using fraction arithmetic.
We will explore the different ways of representing combined probability situations diagrammatically, introducing tree diagrams as a way of finding the probabilities of combined events.
A separate lesson covers the language associated with conditional probability.
This unit of work is just one of several approaches that you could take when teaching this topic and you should aim to adapt the resources to match ability level of your learners as well as your school context.
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Lesson 5 video
Video transcript
Life is full of uncertainties. Take for example, the likelihood that when playing a game you correctly guess the first number drawn from 10 tokens consecutively numbered from 1 to 10 in a bag.
We can calculate this probability easily enough because there are only 10 outcomes, one of which matches our guess. We therefore have a 1 out of 10 chance of guessing correctly or a probability of one tenth
However, selecting a single token would be a very short and dull game. It is more interesting to continue to draw tokens from the bag and to keep guessing. This makes listing the likely outcomes and calculating the required probabilities much more complicated.
When we are calculating probabilities based on two more events needing to be matched we call this calculating combined probabilities.
Let us consider another example.
At a fairground, visitors play a game that involves using a spinner.
They spin the spinner twice and win a prize if the spinner lands on the same colour for both spins.
What is the probability of winning a prize?
To calculate the probability of winning we need to consider the number of equally likely outcomes.
We can do this by methodically listing all of the possible outcomes.
Suppose on the first spin the spinner lands at the blue shown.
Then on the second spin the spinner could land on either the same blue section
Or the first green section
Or the second blue section
And so on
Altogether there are 8 possible outcomes for both spins with the first spin landing on the top blue sector.
Now lets consider how many possible outcomes are there for both spins if the first spin lands on the first green sector instead?
There are 8 of these as well.
With this information we should now be able to work out how many outcomes are there altogether for both spins.
We can illustrate all the possible outcomes by putting the results into a table.
A table like this is called a possibility space diagram
We can see from the diagram that there are 64 equally likely outcomes
And we look to see how many of these have the same colour on both spins
Altogether there are 24 of these.
This means that the probability of getting the same colour when the spinner is spun twice is 24 out of 64.
Or a probability of twenty four, sixty fourths. Which simplifies to three eighths.
Possibility spaces are great way of visualising simple combined probability problems where the probability of the second event does not depend on the first.
For example a student plays a game where they draw a ticket from the bag which contains six tickets numbered from 1 to 6.
If they pick a number less than 3, they eat a chocolate.
If they put the ticket back in the bag every time, the probability of picking a winning ticket from the bag each time remains the same.
This makes it easy to find the probability they win 2 chocolates if they play the game twice.
Using the possibility space diagram it is still relatively easy to see that the probability of winning in both games is 4 out of 36 or one ninth.
It is also possible to calculate this probability, without the need to draw a possibility space diagram by considering the probability of the student winning each of the games separately and multiplying them together.
The probability that he draws a winning ticket in the first game is 2/6 and if he replaces the ticket the probability of him drawing a winning ticket in the second game is still 2/6, meaning that the probability that he wins two consecutive games is the same as 1/3 times 1/3, which is equal to 1/9.
If the student fails to return the first ticket to the bag after the first game, the possibility space diagram becomes difficult to draw, as the results of the second game would depend on the first result.
However, we can still calculate the probability that the student wins both games by multiplying together the probability of him winning each game.
In this case, the probability that he wins the first game is 1/3 as before but the probability that he wins the second game is reduced to 1/5, as there are only 5 tickets left in the bag and only one of them is a winning ticket. This means that the probability that he wins in both games is 1/3 multiplied by 1/5 which is 1/15. We can illustrate this method using a tree diagram. The first part of the tree represents the outcomes for the first game. And we can extend the branches to reflect the outcome of the second game If the student wins the first game, there will still only be five tickets left, but only 1 winning ticket. If you put these together you can find the probability of winning both games by multiplying along the branches. From the tree diagram it is possible to calculate the probability of any outcome from the two games by multiplying along the branches and adding up the probabilities of the outcomes required. For example you can find the probability of getting exactly 1 winning ticket by multiplying along two sets of branches and adding the probabilities together. The probability that when you play the game you win the first time and lose the second is 8/30 and the probability you lose the first time, but win the second is also 8/30. This means that the probability of winning exactly one chocolate is 16/30 which is the same as 8/15
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