Lesson Video: Conditional Probability: Tree Diagrams | Nagwa Lesson Video: Conditional Probability: Tree Diagrams | Nagwa

# Lesson Video: Conditional Probability: Tree Diagrams Mathematics

In this video, we will learn how to use tree diagrams to calculate conditional probabilities.

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### Video Transcript

In this video, we will learn how to use tree diagrams to calculate conditional probabilities. When working with conditional probabilities, it is helpful to use a tree diagram to illustrate the probability of the different outcomes. To help understand this, letβs first recall the formula for conditional probability.

The probability that an event π΅ occurs given that event π΄ has already occurred is written the probability of π΅ given π΄ is equal to the probability of π΄ intersection π΅ divided by the probability of π΄, where the probability of the intersection is the probability that both π΄ and π΅ occur. By multiplying both sides by the probability of π΄, this can be rewritten as the probability of π΄ intersection π΅ is equal to the probability of π΅ given π΄ multiplied by the probability of π΄.

We will now consider how this can be represented on a tree diagram. We can use the formula to determine the intersection of two events. On a tree diagram, these can be calculated by multiplying across branches, with the first branch representing the probability of π΄ and the second branch representing the probability of π΅ given that π΄ has occurred. Recalling that all probabilities lie in the interval from zero to one such that the probability of π΄ is greater than or equal to zero and less than or equal to one and that the probability of the complement of an event written π΄ prime or π΄ bar is equal to one minus the probability of π΄, we can then complete the top half of the tree diagram.

Extending this to the bottom half, beginning with the probability of π΄ prime, the completed tree diagram is as shown. By multiplying along our branches, weβre able to calculate the probabilities of π΄ intersection π΅, π΄ intersection π΅ prime, π΄ prime intersection π΅, and π΄ prime intersection π΅ prime. Letβs now consider how we can use a tree diagram of this type in context.

A bag contains three blue balls and seven red balls. David selects two balls without replacement and draws the following tree diagram. Given that the first ball is red, find the value of π₯ that represents the probability that the second ball selected is red.

We are told in the question that the two balls are selected without replacement. This means that the color of the second ball is dependent on the color of the first ball. As a result, we need to consider what the outcome of the first event is when calculating the probability of the second event. This is an example of conditional probability. We are trying to find the probability that the second ball is red given that the first one was red.

We are told initially that the bag contains three blue balls and seven red balls. As there are 10 balls in total, the probability that the first ball selected is blue is three out of 10 or three-tenths. The probability that the first ball is red is seven-tenths. In this question, we are told that the first ball is red. This means that there are now six red balls remaining in the bag. There are still three blue balls, giving us a total of nine balls. The probability that the second ball is red given that the first one is red is therefore equal to six-ninths. By dividing the numerator and denominator of this fraction by three, this simplifies to two-thirds. The value of π₯ on the tree diagram is therefore equal to two-thirds.

We can check this answer by considering pairs of branches, as we know these must sum to one. As the probability that the second ball is blue given that the first one is red is one-third and one-third plus two-thirds equals one, we know that our answer is correct. The fraction one-third came from the fact that three of the nine remaining balls are blue. And three-ninths is equivalent to one-third.

Whilst it is not required in this question, it is worth noting that when the first ball selected is blue, there will be seven red and two blue balls left in the bag. This results in probabilities of two-ninths and seven-ninths for the second ball being blue and red, respectively, given that the first ball is blue.

In our next question, we will draw a tree diagram in order to find the probability of a conditional event.

A bag contains 22 red balls and 15 black balls. Two balls are drawn at random. Find the probability that the second ball is black given that the first ball is red. Give your answer to three decimal places.

In this question, we are told that two balls are drawn at random from a bag. One way of representing this is using a tree diagram. We know that the first ball selected can be either red or black. The same is true of the second ball, giving us four possible combinations: red red, red black, black red, or black black. As the two balls are drawn at the same time, we can assume this is done without replacement. This means that we are dealing with dependent events and conditional probability.

Conditional probability can be written using the notation shown: the probability of π΅ given π΄. In this question, we were asked to find the probability that the second ball is black given that the first ball is red. This is the probability corresponding to the branch highlighted in pink. As the events are dependent, we will firstly need to calculate the probability that the first ball drawn is red.

There are 22 red balls and 15 black balls in the bag. This means there are a total of 37 balls. And the probability that the first ball selected is red is 22 out of 37. Whilst it is not required in this question, we can also add to our tree diagram the probability that the first ball is black. This is equal to 15 out of 37. Letβs now consider how many balls are left in the bag if the first one drawn is red. There are now 21 red balls, and there are still 15 black balls. This is a total of 36, and the probability of selecting a black ball now is 15 out of 36. This is the probability we are looking for. The probability that the second ball is black given that the first ball is red is 15 out of 36.

Whilst we would often leave our answer as a fraction, in this case, we are asked to give our answer to three decimal places. 15 divided by 36 or the simplified fraction five divided by 12 is equal to 0.4166 and so on. We can round this to three decimal places, giving us an answer of 0.417. At this stage, it is worth completing the remainder of the tree diagram. If the first ball selected is red, the probability that the second ball is also red is 21 out of 36, as 21 of the remaining balls are red. If we now assume that the first ball drawn was black, there would be 22 red balls left and only 14 black balls. This means that the probability that the second ball is red given that the first ball is black is 22 out of 36. And the probability that the second ball is black given that the first ball is also black is 14 out of 36.

It is also worth checking at this stage that the three pairs of fractions circled sum to one. We can do this before or after simplifying the fractions.

We will now consider one final example.

The probability that it rains on a given day is 0.6. If it rains, the probability that a group of friends play football is 0.2. If it does not rain, the probability that they play football rises to 0.8. Work out the probability that it rains on a given day and the friends play football. Work out the probability that it does not rain on a given day and the friends play football. What is the probability that the friends will play football on a given day?

There are three parts to this question. They all involve conditional probability and the dependent events whether it rains and whether a group of friends play football. One way to represent the information from the question is using a tree diagram. We will now clear some space to do this first.

We will begin by letting π be the event that it rains. We are told that the probability that it rains on any given day is 0.6. We know that the complement of any event π΄, which is written π΄ prime or π΄ bar, has a probability that is equal to one minus the probability of π΄. This means that the probability it does not rain in this question is one minus 0.6. This is equal to 0.4 and can be added to the tree diagram as shown.

If we let the event that the group of friends play football be πΉ, there are four possible scenarios: firstly, that it rains and the friends play football; secondly, that it rains and the friends do not play football; thirdly, that it does not rain and the friends play football; and finally, that it does not rain and the friends do not play football. We are told that if it rains, the probability that the friends play football is 0.2. This is an example of conditional probability, the probability that the friends play football given that it rains. We can then add 0.2 to our tree diagram.

Once again, since the probabilities on each pair of branches sum to one, the probability of the complement of this is 0.8. The probability that the friends do not play football given that it rains is 0.8. We can repeat this for the bottom half of our tree diagram. We are told in the question that if it does not rain, the probability that the friends play football is 0.8. The conditional probability that the friends play football given that it does not rain is 0.8.

Letβs now return to the three specific questions we were asked. Firstly, we were asked to work out the probability that it rains on a given day and the friends play football. As we want both events to occur, this is the intersection of the two events. We recall that given two events π΄ and π΅, the probability of π΄ intersection π΅ is equal to the probability of π΅ given π΄ multiplied by the probability of π΄. In this question, the probability that it rains and the friends play football is equal to the probability that the friends play football given that it rains multiplied by the probability it rains. We need to multiply the probabilities 0.2 and 0.6. This is equal to 0.12.

Letβs now consider the second part of our question. This asked us to work out the probability that it does not rain on a given day and the friends play football. This corresponds to the pink path on our tree diagram. The probability that it does not rain and the friends play football is equal to the probability they play football given that it does not rain multiplied by the probability it does not rain. We need to multiply 0.8 and 0.4. This is equal to 0.32. The probability that it does not rain on a given day and the friends play football is 0.32.

The final part of our question asked us to calculate the probability that the friends play football on a given day. This can occur in one of two ways: either it rains and they play football or it does not rain and they play football. We need to find the union of these two events. From the tree diagram, this involves finding the sum of the probabilities. We need to add 0.12 and 0.32. This is equal to 0.44. We can therefore conclude that the probability that the friends play football on a given day is 0.44.

It is worth noting that the sum of the probabilities for every possible outcome combined is equal to one. In this case, the four probabilities 0.12, 0.48, 0.32, and 0.08 sum to one.

We will now finish this video by summarizing the key points. We saw in this video that when there are a relatively small number of outcomes, a tree diagram is a useful way to illustrate the probability of compound events. We saw that the sum of the probabilities for each set of branches equals one. Likewise, the sum of the probabilities of all the final outcomes equals one. When dealing with conditional probability, we saw that the probability of π΄ intersection π΅ is equal to the probability of π΅ given π΄ multiplied by the probability of π΄.

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