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Hi, my name is Mrs. Dennett and in this lesson we're going to be looking at Conditional Probability Word Problems. We'll start with a classic probability problem.

We've got some coloured counters in a bag.

12 pink and eight blue.

We're going to select two counters at random, and calculate some probabilities.

So here goes.

Let's consider what might happen using a tree diagram.

This is a very useful maths tool, which you may have used before.

So we start with our first pick, we can get pink or blue.

Now we've got 12 pink counters out of 20 altogether.

So the probability that we select a pink one at this moment in time is 12 out of 20.

What about blue? We have 8 blue counters so the probability of selecting a blue is eight out of 20.

Now we can simplify these probabilities.

But as we know, we're going to select a second counter, it's probably going to be easier to leave these probabilities as they are for now.

So now, let's think about our second pick.

If I had picked a pink counter on my first go, I would still have the option of picking a pink or blue counter on my second pick, as there are still pink and blue counters left in the bag.

However, the probabilities are going to change slightly, because I no longer have 20 counters in the bag, I now have 19 there is one less pink one.

So the probability of getting a pink is now 11 out of 19.

And for blue, while there's still eight blue counters in the bag, so the numerator is still eight but my denominator is now 19.

So that was a second pick on the condition that I picked pink out first.

So let's now put our pink counter back in the bag, and think about what would have happened to the probabilities if I had picked out a blue counter, first.

Well, there are still pink and blue counters left in the bag.

But now there are only 19 counters all together.

I still have 12 pink counters.

So the probability of selecting pink on my second pick, given that I've selected a blue is now 12 out of 19.

For blue, the probability is now seven out of 19 because there are now seven blue counters in the bag, and 19 counters all together.

Let's put the blue counter back in and we can start to look at the questions considering the possible outcomes.

We firstly want the probability that both counters are blue, which is this outcome.

Pick a blue counter and then another blue counter.

So we multiply these probabilities together to get 56 out of 380.

Next, we consider the probability that both counters are the same colour.

We already found one scenario in part A blue and blue.

So we need to find the probability of selecting pink and pink.

This is 12 out of 20 times 11 out of 19, and gives us a probability of 132 out of 380 for picking two pink counters.

Now, as these are two different outcomes, selecting blue and blue, or pink and pink, we add these together and we get 188 out of 380.

Finally, we want the probability that the counters are two different colours.

There are two possible ways for this to happen.

We could pick pink then blue, or blue, then pink.

So we have to calculate these two probabilities separately using our tree diagram.

And we get a probability of 96 out of 380 for both of them.

So we can just multiply 96 out of 380 by two to get 192 out of 380.

We could of course have just added 96 out of 380 and add 96 out of 380, again, to give us the same total 192 out of 380.

And that's the probability that the counters are two different colours.

Here is a question for you to try.

Pause the video to complete the task and restart when you are finished.

Here are the answers.

Let's calculate the probability that both counters are red, we multiply together seven 12s by six 11s, which gives us 42 over 132.

To find the probability that both counters are the same colour, we can use our answer to part A, and added to the probability that both counters are blue.

So to find the probability that both counters are blue, we multiply together five twelves and four elevens, giving us 20 over 132.

We add together the probability that the counters are blue, to the answer we found in part A, the probability that both counters are red, so we do 20 over 130 to add 42 over 132, giving us 62 over 132.

Finally, to calculate the probability that both counters are different colours, we can use Part B as we know probability add up to one, so we just subtract 62 over 132 from one to give 70 over 132.

Each answer has been simplified.

The next problem is about two sets of traffic lights in a town centre.

The probability that you will have to stop at the first set of traffic lights is 0.

2.

If you had to stop at the first set of traffic lights, the probability that you will have to stop at the second set of traffic lights is 0.

04.

If you didn't have to stop at the first set of traffic lights, the probability that you will have to stop at the second set of traffic lights is 0.

35.

Calculate the probability that you will have to stop for at least one set of traffic lights.

Let's start by drawing a tree diagram, to help us visualise what is happening here, we have two sets of traffic lights and at each set, we either stop or don't stop.

The probability that we stop at the first set of traffic lights is 0.

2.

We're then told if we have to stop here, then the probability we have to stop again at the second set of traffic lights is 0.

04.

So let's put that information in our tree diagram.

And we're finally told that if we don't have to stop at the first set, the probability of being stopped at the second set of traffic lights is 0.

35.

So we can add that to our diagram as well.

We then need to work out the remaining probabilities.

We know that the sum of probabilities is equal to one.

So when we reach the first set of traffic lights, the probabilities for stop and don't stop must add up to one.

So the probability that we don't stop must be 0.

8.

After being stopped at the first set of traffic lights, we then reach the second set.

And again here, there are two outcomes, so their probabilities must add up to one.

So the probability that we don't stop in this scenario must be 0.

96.

Finally, the same is true for this scenario, 0.

35 and 0.

65 is equal to one.

And we've completed our tree diagram.

So we can now answer the question, which was, calculate the probability that you will have to stop for at least one set of traffic lights.

Let's consider which outcomes we need.

We either have to stop once or twice in these scenarios.

There is only one scenario where we don't have to stop at all.

This is useful because we know that the sum of all the probabilities for all four of these outcomes must be one.

We can find the probability of not stopping at either set of traffic lights and subtract from one to find the answer to our question, 0.

52 that's for not stopping at either set of traffic lights.

We can subtract this from one to get 0.

48.

And this gives us the probability we will have to stop at least once.

So the probability that we will have to stop at least once is 0.

48.

Now, if you did want to calculate this using the three outcomes you could do.

So the three outcomes I'm talking about here are stop and stop, stop and don't stop or don't stop and stops.

So each of these scenarios involves stopping at least once.

Here are the probabilities for these scenarios, we would have to add these together and as you can see, we get the same answer of 0.

48 but with a bit extra working out.

Here is a question for you to try.

Pause the video to complete the task and restart when you are finished.

Here are the answers.

I would definitely recommend drawing a probability tree diagram for each of these questions, carefully checking that for each section of the probability branches, that probabilities in each part add up to one.

That's all for this lesson.

Thank you for watching.