Here is the first branch of a probability tree diagram.

Here is the first branch of a different probability tree diagram. What should the missing probability be?

This tree diagram shows Amir taking red and blue counters from a jar. He replaces the counter after he has picked one up.

Draw a tree diagram for independent events

﻿Hi, my name is Mr Chan. And in this lesson, we're going to learn how to calculate probabilities of independent events. Let's have a look at this example. A bag contains four pink and three blue counters. A counter is chosen at random and then replaced, then another counter is chosen at random, we've got to calculate the probabilities that two pink counters are chosen. So here's a representation of our of probability of counters. We've got four pink ones, and three blue ones. So we're going to create our probability tree diagram. So we've been going to pick out the counters twice, first pick and second pick, and we can only pick out pink or blue counters. So my tree diagram will begin with two branches to represent picking out a pink counter or a blue counter. So we have four pink counters out of a total Seven, so that would be my probability of picking out a pink counter for sevens. And similarly with the blue counters, we have three out of seven. So that would be my probability of three sevens for picking out a blue counter. For the second branch for the second pick of the counters, I need another branch there. Again, because we are replacing the counters once we picked the counters out, so if we picked a pink one, we putting it back into the bag, the probability for picking out the pink counter will be the same for seventh because we've still got four pink counters are the blue three sevens because we still have three blue counters out of a total seven. Similarly, if our first pick we picked out a blue counter. We are replacing that blue counter. So in our second branch, should we have picked a blue counter first of all, we would still have the same probabilities for sevens and three sevens for pink and blue respectively. So these are our outcomes. So in our first pick, we could have picked up pink and then another pink. And then our second outcome would be pink and then blue, and then blue and then pink, and then blue and then blue. So the probabilities, we can work out by multiplying these probabilities across the branches. So if I want to work out the probability of picking out a pink and then a pink counter, I would multiply four sevens with four sevens, which equals 1649. So I've gone across the branches, first pick for pink, second pick for pink and multiply those probabilities together. For the outcome, pink and then blue, I would multiply four sevens with three sevens to get the answer 1249. Similarly, for the probability of picking out a blue and then a pink counter, three sevenths multiplied by four sevenths. It's 1249, and the probability of picking up the blue counter first and then the blue counter second would be three sevens multiplied by three sevens, which gives me an answer 949. So those are all my probability values. Let's go back to the question, which one are we looking at? We're looking at the probability of two pink counters being chosen. And I can see from my calculations and travelling across the branch to the tree, that's the probability I'm looking for the probability of pink, then the pink counter would be 1649. Continuing on from the previous example, what's changed is we're asked a different question this time. So in this question, we've got to calculate the probability that one of each colour counter is chosen. So what that means is I'm picking out a pink and then a blue and then a blue and then a pink. So I would be looking at the probabilities for one of each colour. So I've got a pink and a blue probability. Then an outcome of pink and blue, and also an outcome of blue or pink. So I've got one of each colour that now I can see that there are two outcomes here. Now when we have a situation where there are two outcomes that I've got to consider what we do with that, put those two probabilities that we would add those together. So 1249, add 1249 gives me a total answer of 2449. Let's look at this example. We're told the probability of raining on Saturday is 0.7. And the probability of it raining on Sunday is 0.4. And I need to calculate the probability that it will rain on both days. So I'm going to use a probability tree diagram to help me answer this question. So we've got the probabilities of it raining on Saturday or Sunday. And these are the two outcomes on Saturday it could rain or it could not rain. So let's draw the branches for those two outcomes. I know from the question that it tells me that the probability rating on Saturday 0.7. So let's fill that in. Now, whether it rains or doesn't rain on Saturday, it could rain or not rain on Sunday. So I need branches on Sunday to represent that event. So on Sunday, the probability of raining would be 0.4. I can put that in. And similarly, if it doesn't rain on Saturday, it could still rain on Sunday, or it could not rain on Sunday. So again, I've got two branches on Sunday there. And I know the probability of raining on Sunday 0.4. There are some properties missing from the branches. But I do know along one branch, the probabilities must add up to one. So I can fill that interval. 0.3 and 0.7. Add 0.3 equals one. Similarly on Sunday, if it rains, the property of 0.4, the property must add up to one for it not raining. So that must be 0.6. And again, 0.6 on the branch at the bottom there. Now I can use this probability tree to calculate the probabilities that what happens on Saturday and Sunday. So we've got four outcomes there, the probability, they would rain and then rain probability it rains and then not rains, and then not rains and rains and not rains and not rains. So I've got all the four outcomes there. Now to calculate the probabilities, I've multiplied the values across the branches. So I follow the branch for Saturday raining 0.7. And for Sunday raining, 0.4, multiply those two together, I get a probability of 0.28. If I do the same with the all the other branches, I get these values here. And I get the probability well for outcomes there. Now those of us that what you will notice is that this should add up to one. Let's just check, yes, they do. So let's answer the question now calculate the probability they will run on both days. So the probability I'm looking for is the probability that will rain on Saturday, and the probability it will rain on Sunday. That's both days, this would be this answer here. So my final answer is the probability that will rain and then rain will be 0.28. Following on from the last example, the question changes slightly, and we've got to calculate the probability now that it will rain on at least one day. So raining on at least one day would mean that it rains on Saturday, or it rains on Sunday. So that's at least one day and the probability is we're looking for Ahh it raining on Saturday. On Sunday, so does read at least one day, rains on Saturday, it doesn't rain on Sunday. So it does rain at least one day. And again, if it rains, if it doesn't rain on Saturday, but it does rain on Sunday, that's raining at least one day. So what we have here is we've got to consider three outcomes there. And when we're we've got three outcomes to consider. When we calculated the probabilities, what we do with these probabilities now is just add them together. So we add 0.28 to 0.42, and 0.12 to get a final answer of 0.82. Here's a question for you to try. Pause the video to complete the task, resumed the video once you finished. Here's the answer. In this question, we've got to calculate the probability that Tim picks two blue counters. So using the tree diagram, we look for the probability of the first pick being in blue, and the second pick also being blue. And those two probabilities we will multiply together. So we can see that we would multiply one quarter with one quarter, so that would equal 116. Here's another question for you to try. Pause the video to complete the task, resume the video once you're finished. Here are the answers. In Part B, you've got to calculate the probability of flipping exactly one tail. Now there are two outcomes where that could happen. So you could flip ahead and then a tail or a tail and then head. So when you've worked out the probability of those two outcomes happening, make sure you add those together to get your answer. Here's a question for you to try. Pause the video to complete the task, resume the video once you're finished. Here are the answers. You could have used the diagram drawn of the spinner to help you work out the probabilities for the tree diagram. However, you could also realise that the probabilities long each branch must add up to one. So you're given the first probability of the probability of the first spin of red being one fifth. That means the probability of blue would be forfeits because those two probabilities add up to one. In this question, the tree diagram isn't drawn for you. So you have to do that in part A. Pause the video to have a go resume the video once you're finished. Here are the answers. I hope that your tree diagram looks very similar to the one shown in the answer here. Now in Part B, you have to calculate the probability that it will rain on at least one of the days. So that means if it rains on either Friday or Saturday, you've got to include that in your answer by adding those probabilities together once you've calculated them. That's all for this lesson. Thank you for watching.

For the probability tree shown, which calculation will find the probability of red followed by blue happening?

The probability that Amy and Joe win a game of chess is shown in the tree diagram. Work out the probability that over 2 games Amy wins twice.

James draws this probability tree diagram to help him calculate probabilities of taking sweets from a bag. What assumptions has James made when completing this probability tree?

Calculate probabilities of independent events

In this lesson, we will learn how to calculate probabilities of outcomes for independent events from tree diagrams

Maths

Y11 Maths (Higher)

Key Stage 4

Year 11

Y11 Maths (Core)

Y11 Maths (Foundation)

Probability 3 (Tree diagrams)

Handling Data/Probability (D)

THEME

UNIT

Draw a tree diagram for dependent events

Calculate probabilities of dependent events

Amy is going to pick 2 biscuits from a box. Some are chocolate and some are plain. The probability tree shows the probability information.

James is also going to pick 2 biscuits from a box. Some are chocolate and some are plain. The probability tree shows the probability information.

Here is a probability tree showing Amir taking red and blue sweets from a bag.

Probability of a dependent event is also called?

Simon has drawn this probability tree for a question on dependent probability. What mistake has he made?

This tree diagram shows Amir taking red and blue sweets from a bag. He does not replace the sweet after he has taken one.

Lesson overview:Calculate probabilities of independent events

Core Content

Probability 3 (Tree diagrams):