Back to AQA Probability (F) Home
5 I) Combined Events – Number of Outcomes
5 I) Combined Events – Number of Outcomes
There are a variety of different ways that we are able to find the outcomes when an event is conducted multiple times or there are two different events occurring. We are also able to work out the probability of the different outcomes occurring, and we will be looking at this in more detail towards the end of this section.
Systematically Listing Events
One way that we can obtain the outcomes for an event that is repeated or two different events is to use systematic listing. This is where by we create a list in an organised way so that we ensure that we do not miss any of the outcomes.
Example 1
What are the possible outcomes when we flip a coin twice?
There are two options when we flip a coin; heads and tails. Let’s start with the outcomes when a coin lands on heads first. The outcomes are:
Let’s now say that a coin lands on tails first and see the possible outcomes that can occur:
Therefore, all of the possible outcomes for the flipping of two coins are as follows:
Example 2
Let’s now have a more complex example. A football team is playing a football match and they can win, lose or draw. The weather throughout the game can be sunny, overcast, rainy or snowy. What are the possible outcomes for the football match and the weather?
Let’s start our list of outcomes with all of the possible weather conditions when the team wins. The possible outcomes are:
Now, let’s do the same for the outcomes when the team loses:
And finally, we’ll list the possible outcomes when the team draws:
Therefore, the list of outcomes for the football game result and the weather conditions are as follows:
It is best to answer questions like this in a methodical way. For the football and weather example, I listed all of the outcomes when the team won, then when the team lost and then when the team drew. By listing the outcomes in a methodical way, we make sure that we do not miss out any of the outcomes.
There are a few more question similar to this in the quiz.
One way that we can obtain the outcomes for an event that is repeated or two different events is to use systematic listing. This is where by we create a list in an organised way so that we ensure that we do not miss any of the outcomes.
Example 1
What are the possible outcomes when we flip a coin twice?
There are two options when we flip a coin; heads and tails. Let’s start with the outcomes when a coin lands on heads first. The outcomes are:
- Heads then heads
- Heads then tails
Let’s now say that a coin lands on tails first and see the possible outcomes that can occur:
- Tails then heads
- Tails then tails
Therefore, all of the possible outcomes for the flipping of two coins are as follows:
- Heads then heads
- Heads then tails
- Tails then heads
- Tails then tails
Example 2
Let’s now have a more complex example. A football team is playing a football match and they can win, lose or draw. The weather throughout the game can be sunny, overcast, rainy or snowy. What are the possible outcomes for the football match and the weather?
Let’s start our list of outcomes with all of the possible weather conditions when the team wins. The possible outcomes are:
- Win and sunny
- Win and overcast
- Win and rainy
- Win and snowy
Now, let’s do the same for the outcomes when the team loses:
- Lose and sunny
- Lose and overcast
- Lose and rainy
- Lose and snowy
And finally, we’ll list the possible outcomes when the team draws:
- Draw and sunny
- Draw and overcast
- Draw and rainy
- Draw and snowy
Therefore, the list of outcomes for the football game result and the weather conditions are as follows:
- Win and sunny
- Win and overcast
- Win and rainy
- Win and snowy
- Lose and sunny
- Lose and overcast
- Lose and rainy
- Lose and snowy
- Draw and sunny
- Draw and overcast
- Draw and rainy
- Draw and snowy
It is best to answer questions like this in a methodical way. For the football and weather example, I listed all of the outcomes when the team won, then when the team lost and then when the team drew. By listing the outcomes in a methodical way, we make sure that we do not miss out any of the outcomes.
There are a few more question similar to this in the quiz.
Product Rule
We are able to quickly find out how many outcomes there are for 2 or more events by multiplying the number of outcomes for each of the individual events.
With the first example, which was flipping the coin twice, there was were 2 outcomes (heads or tails) for each time that the coin was flipped. Therefore, we can multiply the 2 x 2 to see that there are 4 possible outcomes (the outcomes were: HH, HT, TT and TH).
For the second example that was the football match, there were 3 possible football results (win, lose or draw) and 4 possible weather conditions (sunny, overcast, rainy and snowy). We would be able to quickly find out the number of possible outcomes by multiplying the number of outcomes for the football match by the number of outcomes for the weather conditions. We therefore multiply 3 by 4 to see that there are 12 different outcomes (these are the outcomes that are listed in the above section).
We are able to quickly find out how many outcomes there are for 2 or more events by multiplying the number of outcomes for each of the individual events.
With the first example, which was flipping the coin twice, there was were 2 outcomes (heads or tails) for each time that the coin was flipped. Therefore, we can multiply the 2 x 2 to see that there are 4 possible outcomes (the outcomes were: HH, HT, TT and TH).
For the second example that was the football match, there were 3 possible football results (win, lose or draw) and 4 possible weather conditions (sunny, overcast, rainy and snowy). We would be able to quickly find out the number of possible outcomes by multiplying the number of outcomes for the football match by the number of outcomes for the weather conditions. We therefore multiply 3 by 4 to see that there are 12 different outcomes (these are the outcomes that are listed in the above section).
Example 3
Let’s have another example that we have not already looked at. We have two different bags (bag A and bag B) and we pick a ball from each of the bags. Bag A contains red, green and yellow balls. Bag B contains green, orange, red, blue and purple balls. How many possible outcomes are there?
To work out the number of possible outcomes, we multiply the number of outcomes for the first bag by the number of outcomes for the second bag. There are 3 different coloured balls in bag A, which means that there are 3 different outcomes. There are 5 different coloured balls in bag B, which means that there are 5 different outcomes for bag B. Therefore, we multiply the number of outcomes for bag A (3) by the number of outcomes for bag B (5).
Let’s have another example that we have not already looked at. We have two different bags (bag A and bag B) and we pick a ball from each of the bags. Bag A contains red, green and yellow balls. Bag B contains green, orange, red, blue and purple balls. How many possible outcomes are there?
To work out the number of possible outcomes, we multiply the number of outcomes for the first bag by the number of outcomes for the second bag. There are 3 different coloured balls in bag A, which means that there are 3 different outcomes. There are 5 different coloured balls in bag B, which means that there are 5 different outcomes for bag B. Therefore, we multiply the number of outcomes for bag A (3) by the number of outcomes for bag B (5).
This tells us that there are 15 possible outcomes.
Example 4
Here is another example. How many outcomes would there be if we were to roll a normal six-sided dice and spin a spinner that has 5 sections?
The dice is a normal dice that has 6 different outcomes and the spinner has 5 sections, so 5 outcomes. Therefore, to work out the number of possible outcomes when rolling a dice and a spinner we multiply the number of outcomes for each event.
Here is another example. How many outcomes would there be if we were to roll a normal six-sided dice and spin a spinner that has 5 sections?
The dice is a normal dice that has 6 different outcomes and the spinner has 5 sections, so 5 outcomes. Therefore, to work out the number of possible outcomes when rolling a dice and a spinner we multiply the number of outcomes for each event.
Therefore, there are 30 different possible outcomes.
Notation
We can denote all of the possible outcomes for an event by placing them inside curly brackets like the ones that are given below:
We can denote all of the possible outcomes for an event by placing them inside curly brackets like the ones that are given below:
The first event that we looked at was the flipping of two coins. We could denote the outcomes of the flipping of two coins as {HH, HT, TH, TT} where H is head and T is tails.
Example 5
The final example that we are going to look at is slightly different.
A restaurant has three different courses; starters, mains and desserts. There are 4 different starters, 7 different mains and 5 different desserts. An individual is going to choose two out of the three courses. Show that there are 83 different combinations to choose from.
The question tells us that there are three different courses (starter, main and dessert) and our individual chooses two out of the three courses. Therefore, the first step in answering this question is to work out the different combinations of courses that our individual can have. I am going to do this by systematically listing the options and the options are:
We now need to find the number of different combinations from the three different combinations of courses. I am going to do this by using the product rule. After we have found the number of different combinations from the three different combinations of courses, we add the number of outcomes together to find the total number of different combinations.
Let’s start by working out the number of combinations when having a starter and a main. We are told in the question that there are 4 different starters and 7 different mains. Therefore, we can work out the number of combinations of starters and mains by multiplying 4 by 7.
The final example that we are going to look at is slightly different.
A restaurant has three different courses; starters, mains and desserts. There are 4 different starters, 7 different mains and 5 different desserts. An individual is going to choose two out of the three courses. Show that there are 83 different combinations to choose from.
The question tells us that there are three different courses (starter, main and dessert) and our individual chooses two out of the three courses. Therefore, the first step in answering this question is to work out the different combinations of courses that our individual can have. I am going to do this by systematically listing the options and the options are:
- Starter & main
- Starter & dessert
- Main & dessert
We now need to find the number of different combinations from the three different combinations of courses. I am going to do this by using the product rule. After we have found the number of different combinations from the three different combinations of courses, we add the number of outcomes together to find the total number of different combinations.
Let’s start by working out the number of combinations when having a starter and a main. We are told in the question that there are 4 different starters and 7 different mains. Therefore, we can work out the number of combinations of starters and mains by multiplying 4 by 7.
The above calculation tells us that there are 28 different combinations of starters and mains.
We now move onto the second course option, which is to have a starter and a dessert. The question tells us that there are 4 different starters and 5 different desserts. Therefore, we work out the number of combinations of starters and desserts by multiplying 4 by 5.
We now move onto the second course option, which is to have a starter and a dessert. The question tells us that there are 4 different starters and 5 different desserts. Therefore, we work out the number of combinations of starters and desserts by multiplying 4 by 5.
There are 20 different combinations of starters and desserts.
We now move onto the final course option, which is to have a main and a dessert. The question tells us that there are 7 mains and 5 desserts. Therefore, we work out the number of combinations of mains and desserts by multiplying 7 by 5.
We now move onto the final course option, which is to have a main and a dessert. The question tells us that there are 7 mains and 5 desserts. Therefore, we work out the number of combinations of mains and desserts by multiplying 7 by 5.
There are 35 different combinations of mains and desserts.
The final step is to add the number of combinations of outcomes for the three different course options. We add 28 (starter and main; S&M), 20 (starter and dessert; S&D) and 35 (main and dessert; S&D) together.
The final step is to add the number of combinations of outcomes for the three different course options. We add 28 (starter and main; S&M), 20 (starter and dessert; S&D) and 35 (main and dessert; S&D) together.
The above calculation tells us that there are 83 different combinations. This is the same number as the question told us, which means that we have proved that there are 83 different meal combinations when choosing two out of the three courses.