Lesson video

Good day to you all.

I am Mr. Gratton, and thank you so much for joining me for this probability lesson where we will look at and compare different diagrams that we can use to show and calculate probabilities into different ways.

The first to show probabilities of a two-stage trial, and secondly, to show probabilities of two distinct sets of events.

Pause here to have a quick look at some of the probability representations that we'll be using throughout this lesson.

The first of the two uses for probability diagrams is to show the outcomes of a two-stage trial.

Let's see which ones we can use.

Jun is correct.

We can use an outcome table to show the probabilities of a two-stage trial, where one header shows the outcome from one stage.

In this case, the outcomes of spinner one and the other header shows the outcomes from the other stage, the outcomes of spinner two for this trial.

The inner part of this outcome table is the sample space of outcomes for the entire trial.

The probability of, for example, spinning an X and an A can be found by looking at the number of outcomes in the sample space that are XA, which is 10, over the number of outcomes in the whole sample space, which is 24.

And so the probability of getting both, and X and an A is 10 out of 24.

Furthermore, we can also find the probability of the event X or A.

That is to say the outcomes contain either an X or an A or both, by looking at these three regions of the sample space.

The total number of outcomes in these three regions is 21.

And again, over the total number of outcomes in the sample space of 24.

Therefore, the probability of getting an X or an A or both is 21 out of 24.

I agree with Jun here.

The sample space is huge, and contains so many repeated outcomes.

I mean, we wrote down the outcome XA 10 times.

Is this really efficient? Is there a different diagram that shows the probabilities of a two-stage trial more efficiently? Well, yes.

A probability tree can show the same two-stage trial.

Remember, each layer of branches represents one stage of the trial.

The outcomes of spinner one can go on this first layer of branches and the outcomes, of spinner two go on this second twice.

Once at the end of each outcome from spinner one, and the probability of each outcome in the sample space at the end of the tree, the one which shows the outcomes of the entire trial, is the product of the probabilities of the two connected branches that lead to that outcome.

And so for example, the probability of getting both an A and an X is five eighths times by two thirds, which equals 10 over 24.

Notice how this is the exact same probability as the one that we got from the outcome table.

Similarly, we can also find the probability of the event X or A.

We do this by considering each outcome in the sample space with either an X, an A, or both, and adding together the probabilities of each outcome in that event.

As with the outcome table, the outcomes with either an X or an A in them are AXE, AY, and BX.

So we can add together their probabilities to get 21 out of 24.

Again, the exact same probability as the outcome table.

This helps show that a trial, even shown across different diagrams, will still show the same probabilities for each outcome or each event.

Both diagrams have their benefits.

As Jun says, the tree is far more efficient at showing the probabilities of each stage of the trial, as well as the trial itself in the sample space at the end of the tree.

Whereas the table is definitely more long-winded, but it does help show where these probabilities came from.

As you can see, the breakdown of the number of times each individual outcome may happen in the sample space that leads to the probability.

Okay, let's compare an event across both diagrams at the same time.

Where on each sample space is the event either X or B, but not both shown? And how do we find the probability of this event happening? So first, X or B but not both, means the outcome must contain only one of X or B.

Therefore, the outcomes AXE and BY fit that description for this event, but not BX, as this outcome does include both the B and the X.

We can see that these two outcomes along with their frequencies are also shown on the outcome table.

The probability of AXE or XA is 10 over 24.

This is either the probability explicitly seen on the probability tree, or by considering the 10 XA outcomes in the table out of a total of 24 outcomes.

Similarly, the probability of BY is three over 24.

The probability of this total event is the sum of these two outcomes at 13 over 24, which can be seen by both of these diagrams. Okay, onto a check across these two diagrams that show the same trial.

Using the outcome table, what two probabilities go on the two branches of this probability tree? Pause now to consider where you'd see the outcomes four and five on the outcome table, and what probabilities they would result in.

Both the four and the five are outcomes from the individual stages of the trial.

Out of the three outcomes in stage one, two of them are fours, resulting in the probability two out of three.

Similarly, out of the seven outcomes in stage two of the trial, three of them are fives, resulting in the probability three out of seven.

The probability tree has now been completed.

Pause here to answer which two fractions must be multiplied together to find the probability of getting a nine.

And hence, what is the probability of getting a nine? The two branches leading up to the nine are four and five, which have the probabilities of two thirds and three sevenths.

We multiply these branches together to get a probability of six over 21.

The probability of getting a nine is six over 21.

Linked to the last check, pause here to consider, how many outcomes in the outcome table are nine? Six out of the 21 outcomes are nine.

Wait a second.

Six out of 21 is the exact same probability that we gained from the probability tree.

And finally, let's have a look at two heavily linked check questions.

For your answer to A, how many outcomes are prime, and which diagram can you find that information on? For B, what is the probability of the result being a prime number? And can you come to the same conclusion using both diagrams? Pause now for time to look through or to discuss these two questions.

There are 15 prime outcomes.

We can tell that by looking at the outcome table.

The probability of the result being a prime number is 15 out of 21.

We can gain that information from both the probability tree and from the outcome table.

Jun acknowledges that a probability tree is straightforward to draw and use if there are only a few distinct outcomes that may repeat themselves, such as in spinner one, where outcome A is repeated five times.

However, a tree is a lot less straightforward if there are many unique or distinct outcomes, as each different outcome needs to be represented by its own branch.

For this new spinner one, that means four branches.

And for spinner two, that's three branches for each of the four outcomes from the first spinner.

That's a total of 12 branches.

That is a lot of probabilities to calculate and honestly, quite a lot to show on a very busy diagram.

If there are a lot of distinct, unique outcomes, then the outcome table probably is more neat and simple to calculate the probabilities from.

And for this check, let's look at whether a probability tree or outcome table is more suitable for this two-stage trial.

The dice has integers, one, two, three and four written on it once each.

Whereas the spinner has integers three written three times, and four written twice.

Pause now to look through both of these statements and choose which one you think is a suitable reason for using a tree over a table or vice versa.

Both answers are correct.

There are some arguments for and against using an outcome table or a probability tree for this specific trial.

Okay, onto some practise tasks.

For question one, the trial is, the spinner is spun twice.

Complete both, the probability tree and outcome table for this trial.

When complete, circle the probabilities and outcomes across both diagrams that show the event even number, and find the probability of the event multiple of seven.

Pause now to do question one.

And for question two, here is a correctly completed outcome table.

Pause now to use this table to complete the probability tree and answer the questions, how many prime outcomes are there, and what is the probability of the event six or eight? And here are the answers.

Pause now to compare your probability tree and outcome table to the ones on screen.

Pay particular attention to the outcomes and probabilities that have been circled to show the event even number.

And for question two, pause here to compare your answers with the ones on screen.

So those were the two diagrams that we could use to show a two-stage trial with its probabilities.

But what about a one-stage trial, where we need to look into two combined events happening? Will the diagrams we use change? And if not, will we use them any differently? Let's have a look.

When comparing the probabilities across different events, we can use outcome tables and probability trees that are similar to, but not identical to the ones that we saw in the first cycle.

But in addition to that, we can also use different types of Venn diagram.

For example, in this trial, we can consider two distinct groups of events.

Group one is whether Sam does or does not win.

And group two, whether Alex does or does not win.

To show these events on a probability tree, we must first categorise the outcomes for each event into two groups, similar to how we can group outcomes on an outcome tree.

So for this spinner outcomes, zero, two, three and four, we can create one layer of branches which represents the events of Sam winning, and its opposite event of Sam not winning.

Each individual outcome can be split into the two events like so; the outcomes for the event that Sam wins, and the outcomes for the event that Sam does not win.

Our second layer of branches will show the second group of events, Alex winning or not winning.

We can then take the outcomes from each of the two previous events and distribute them further across the events in layer two.

And so the twos and fours from the Sam winning event, can be further distributed to Alex winning and Alex not winning.

Similarly, the zeros and threes from the Sam doesn't win event can be further distributed to Alex winning and Alex not winning.

This spinner has a total of 16 outcomes.

We can see that five of the outcomes, all of the fours, lie on the path of both sound winning and Alex winning.

And so the probability of the events Sam and Alex winning is five out of 16.

Notice the importance of the words both and and here to show the single path, where both Sam and Alex win.

Now for the events Sam or Alex winning, we need to consider multiple paths.

The first being where both Sam and Alex win the fours.

The second being where only Sam wins the twos.

And last being where only Alex wins, which are the threes.

In total, there are 10 outcomes across these three paths.

And so the probability of the events, either Sam or Alex winning is 10 over 16.

Again, notice the importance of the words either and or describing the multiple paths, where one or both of them win.

Okay, what set of events do you think these two paths of branches represent? Well, these two paths show the outcomes of the event where exactly one person wins at five out of 16.

And which events do you think this path of branches represent? Well, the first branch shows Sam winning, and the second shows Alex not winning.

So this is the set of events where only Sam wins at two out of 16.

The same events we looked at on the probability tree can also be represented on a Venn diagram.

We could show each individual outcome from the spinner categorised into different events.

The outcomes inside the left circle show the outcomes of the event that Sam wins.

All outcomes outside this left circle represent the event that Sam does not win.

Similarly, the outcomes in the right circle show the outcomes of the event that Alex wins.

All outcomes outside the right circle represent the event that Alex does not win.

Okay, let's use some of the important words from earlier.

Both Sam and Alex win if the outcome is four.

Only Sam wins if the outcome is two.

And only Alex wins if the outcome is three.

And so the remaining outcomes, the zeros, go outside both the two circles.

All the outcomes inside the two circles show the event of either Sam winning or Alex winning.

This means that there are 10 outcomes where either Sam or Alex wins.

The six outcomes outside the circle show the event that neither Sam nor Alex win.

If however, there are way too many individual outcomes to sort into events on a Venn diagram, then these same events can be represented on a different type of Venn diagram.

We could show the number of outcomes in each event rather than listing each individual one.

So for example, there are two outcomes in the event that only Sam wins and two plus five equals seven outcomes in the event that Sam wins, regardless of what happens to Alex, we can also see there are two plus five plus three equals 10 outcomes where either Sam or Alex wins.

But only five outcomes where both Sam and Alex win.

That is the middle of the Venn diagram.

Okay, let's compare both the probability tree and Venn diagram together.

Where can we find the probability of the events Sam or Alex wins? Let's have a look.

On the tree, we have both Sam and Alex winning with five out of 16 outcomes.

Sam, not Alex winning at two out of the 16 outcomes.

And Alex, but not Sam winning at three out of the 16 outcomes.

So the probability of Sam or Alex winning is the sum of the probabilities at the end of each of these three branches.

And so the probability of Sam or Alex winning is 10 over 16.

And now onto the Venn diagram.

We know that the outcomes for either Sam or Alex or both winning is the total of all of the outcomes inside both circles.

So two plus three plus five equals 10, are the 10 outcomes across either event.

And there are 16 outcomes in total, or two plus three plus five plus six equals 16, giving once again, 10 out of 16 as the probability of these two events occurring.

For this check, this Venn diagram shows the individual outcomes from a spinner, and notice how some of the outcomes are repeated such as the two tens.

Pause here to consider how many individual outcomes there are in this Venn diagram.

There are 14 outcomes because there are 14 numbers across all of the regions of the diagram.

We need to consider the events prime number and multiple of five.

A probability tree is constructed to show these events.

Pause now to answer what probability goes to the end of this pair of branches.

There is one outcome that is both prime and a multiple of five out of the 14 outcomes total.

And so the probability is one out of 14.

The probability tree is completed.

Pause here to use the tree or Venn diagram to answer what is the probability of the event prime number.

This event does not give any condition on whether or not the outcome is also a multiple of five.

Meaning we only need to consider any outcome in the left-hand circle, the one showing the event prime.

Or the two pairs of branches, both on the path showing prime numbers.

The probability is then six out of 14, either because there are six outcomes in the left circle out of a total of 14 outcomes, or because the two branches on the probability tree denoting prime numbers are one out of 14 plus five out of 14.

And lastly, pause here to use the diagrams to find the probability of prime or a multiple of five.

Prime or a multiple of five is shown by all of the outcomes inside the two circles and by these three paths of branches.

There are then a total of eight out of 15 outcomes that satisfy the events, either prime or a multiple of five.

And lastly, let's compare a two-way table with a Venn diagram.

For this trial, we have a toy that will be chosen at random.

Let's categorise these toys into the event that the toy chosen is a bear and it is wearing a hat.

The outcome table correctly shows the individual outcomes of this trial.

How can the outcome table be used to complete a Venn diagram that shows the number of outcomes in each event? This part of the sample space shows the bears that have a hat.

Therefore, these toys satisfy both the events bear and hat.

Therefore, I can say, in the middle region of the Venn diagram, there are four bears with hats.

However, this part of the outcome table shows toys that are bears, but these bears do not wear hats.

Therefore, we can say in the leftmost region, that there are five bears that do not wear hats.

What do you think this part of the outcome table represents? What part of the Venn diagram does this represent? This part of the outcome table shows toys that are neither bears, nor do they wear hats.

This satisfies neither of the events bear or hat.

And so we can say that there is one outcome outside the two circles showing the one toy that is neither a bear nor does it wear a hat.

And lastly, this part of the outcome table has three toys, which will go in the rightmost region of the Venn diagram.

But what does this event even show? That's right, this shows the toys that wear hats but are not bears.

This makes sense as we put the three toys in the rightmost region where that region belongs to the event hat, but not the event bear.

Okay, let's compare the two diagrams. What part of the outcome table and Venn diagram shows the event of choosing a toy that is either a bear or it wears a hat? We know that either a bear or a hat in the Venn diagram is the combination of all three regions that are inside the circles, and it is also represented by these three parts of the outcome table.

So the probability of bear or hat is four plus five plus three equals 12 outcomes over a 13 outcome total from this trial.

For this final set of checks, the trial is to randomly select a card with a number on it.

And the events we must consider are the card shows an even number, and the card shows a multiple of three.

Pause here to use the correctly completed Venn diagram to identify which part of the outcome table has a missing card, and suggest a possible value for that card.

The Venn diagram states that there are four multiples of three that are not even.

However, the table only shows three outcomes that match these events.

Therefore, any odd multiple of three can be included in the table, so that it accurately represents the Venn diagram.

And up next, pause here to find the probability of each of these three events using either or both of these diagrams. And the probabilities are seven over 14, four over 14, and three over 14.

And once more, pause here to find the probability of each of these three events occurring.

And the probabilities are eight over 14, 12 over 14, and two over 14.

Okay, onto the final few practise questions.

For question one, use the outcome table to complete the Venn diagram.

Pause now to do this and use both diagrams to find these probabilities.

And for question two, use the Venn diagram showing the number of outcomes to complete this probability tree.

Pause now to do this and use both diagrams to find the probabilities of these events.

And finally, question three, use the outcome table to complete the probability tree.

And use the completed probability tree to identify which event in the outcome table has a missing outcome.

Pause now to do this, and use both diagrams to find the probabilities of these three events.

Okay, everyone, great job, here are the answers.

Pause now to match all of the information on screen to all of the information that you've acquired from question one.

And for question two, pause here to check if your probability tree and probabilities of events match those that are on screen.

And finally, pause here again for question three to see if your Venn diagram matches the one on screen.

The missing outcome on the outcome table was an outcome that was odd but not square such as 11 or 17.

And pause here to check if your probabilities match or are equivalent to the ones on screen.

Note that the probabilities on the probability tree are simplified versions of the probabilities out of 28 that you might have found if you used the outcome table.

Okay, take a breather, and a very well done.

That was an intense lesson, where we focused on comparing different representations to find probabilities, either by looking at outcomes in the sample space of an outcome table or the probabilities and outcomes in the sample space at the end of a probability tree.

And by looking at the outcomes of two events shown on either an outcome table, a probability tree, or different types of Venn diagram.

Thank you all once again for joining me for this lesson on probability.

I'll see you again soon for some more maths, but until that time arrives, stay safe, and have an amazing rest of your day.