شارح الدرس: Dependent and Independent Events | نجوى شارح الدرس: Dependent and Independent Events | نجوى

# شارح الدرس: Dependent and Independent Events الرياضيات

In this explainer, we will learn how to calculate probabilities for dependent and independent events and how to check if two events are independent.

We begin by recalling the following definition.

### Definition: Independent and Dependent Events

Events and are independent if the occurrence of does not affect the probability of the occurrence of . That is, where represents the probability of event occurring given that event has occurred.

If the condition above fails, then we say that and are dependent events. For instance, if we are rolling a die twice, rolling an even number in the first roll and rolling a 4 in the second roll are independent events because the fact that we rolled an even number in the first roll neither increases nor decreases the chance of rolling a 4 in the second roll. On the other hand, rolling an even number and rolling a 4 in the same roll are dependent events, since rolling an even number doubles the chance that we have rolled a 4.

Let us consider a few scenarios in the first example and check whether they describe dependent or independent events using the definition.

### Example 1: Identifying Scenarios Representing Independent Events

In which of the following scenarios are and independent events?

1. A student leaves their house on their way to school. Event is them arriving at the bus stop in time to catch the bus and event is them getting to school on time.
2. A die is rolled. Event is rolling an even number and event is rolling a prime number.
3. A die is rolled and a coin is flipped. Event is rolling a 6 on the die, and event is the coin landing heads up.
4. A child takes two candies at random from a bag that contains chewy candies and crunchy candies. Event is them taking a chewy candy first and event is them taking a crunchy candy second.
5. A teacher selects two students at random from a group containing five boys and five girls. Event is the teacher selecting a boy first, and event is the teacher selecting a girl second.

### Answer

We recall that events and are independent if the occurrence of does not affect the probability of the occurrence of . To establish whether and are independent or dependent, we should examine whether the probability of changes when we assume that has already occurred. Let us consider each scenario using the definition.

1. Event is the student arriving at the bus stop in time to catch the bus, and event is the student getting to school on time. We can assume that the probability of getting to school on time greatly increases if the student catches the bus, since the student is more likely to be late to school if he or she misses the bus. So, the occurrence of event of affects the probability of the occurrence of event , meaning that the two events are dependent.
2. Event is rolling an even number , while is rolling a prime number . The probability of rolling a prime number is because there are 3 prime numbers from 6 equally likely outcomes. On the other hand, if we assume that we have already rolled an even number, then the roll has to be a 2 for it to be a prime number. In other words, the event given is the same as getting 2 from . So, . Since , the two events are dependent.
3. Event is rolling a 6 on a die, and event is a coin landing heads up. Say that we have already rolled a 6 on a die. Does the probability of the coin landing heads up change? No, the probability of such event is still regardless of what we get from a die roll. Thus, and are independent.
4. Event is the child taking a chewy candy first, and event is the child taking a crunchy candy second. To determine whether the two events are independent, we need to know whether the probability that event occurs is different depending on whether or not event has already occurred. Let’s say the bag starts with just one chewy candy, and one crunchy candy. If the child takes the chewy candy on their first selection, then we say that event has occurred. Since they will keep that candy out of the bag and make their selection from what remains in the bag, it is certain that they will pick the crunchy candy on their second selection, so . If the child takes the crunchy candy first, then we say that event has not occurred, and in this case, it is impossible for them to take the crunchy candy second because there is only a chewy candy left in the bag, so . Since the probability of occurring is different when has occurred to when has not occurred, then event is dependent on event . and are not independent events.
5. This scenario is similar to scenario D, and the probability of selecting a girl second depends on whether a boy or girl was selected first, since the composition of the group remaining for the second selection will be different depending on whether a boy or girl was selected first. If event occurred, then there will be 4 boys and 5 girls available for the second selection, and . However, if event did not occur, then there will be 5 boys and 4 girls available for the second selection, and . The two events are dependent.

Hence, the correct answer is C.

In the next example, we will use a probability tree diagram to decide whether or not the events are independent.

### Example 2: Using Probability Tree Diagrams to Decide Whether Events Are Independent

A bag contains 5 red candies and 4 blue candies. I take one at random, note its color, and eat it. I then do the same for another candy. The figure below shows the probability tree associated with this problem. Are the events of “getting a blue candy first” and “getting a red candy second” independent?

### Answer

We recall that events and are independent if the occurrence of does not affect the probability of the occurrence of . To establish whether or not and are independent, we should examine whether the probability of changes depending on the outcome of .

If we take a blue candy first, the probability of getting the red candy second is given by the lower branches of the probability tree diagram.

Then we have,

On the other hand, the probability of getting the red candy second, regardless of what was selected first, is the same as the probability of getting the red candy first. This is because the order of the selection does not affect its probability unless we are assuming a condition from the previous choice. The probability of getting the red candy first is noted below.

This is equal to the probability of getting a red candy second, so we have

This gives us

Since the probability of “getting a red candy second” changes when we assume that we get a blue candy first, the two events are not independent.

We recall that the conditional probability is computed by the equation

 𝑃(𝐵∣𝐴)=𝑃(𝐴∩𝐵)𝑃(𝐴). (1)

We also know that for independents events. Substituting this into the equation above and rearranging, we get

This leads to the following theorem.

### Theorem: Multiplication Rule for Independent Events

Events and are independent if and only if where is the event where events and occur simultaneously.

In the next example, we will use this theorem to show that any two mutually exclusive events with nonzero probabilities cannot be independent.

### Example 3: Mutually Exclusive Events and Independent Events

If and and , are and independent?

### Answer

We recall that, if and are independent events, then

We are given that , which means that they are mutually exclusive. In other words, the two events cannot occur at the same time. We recall that the probability of an empty set is equal to zero, so

On the other hand, we can calculate

Since , and are dependent.

The example above demonstrates a more general fact: two mutually exclusive events with nonzero probabilities cannot be independent. This makes sense heuristically because if and are mutually exclusive events, then and cannot occur simultaneously. So, if we assume that event has already occurred, then that eliminates the possibility that could occur. In other words, the outcome of affects the outcome of , making them dependent.

In the next example, we will use probabilities given in a Venn Diagram to decide whether or not two events are independent.

### Example 4: Using Probabilities in a Venn Diagram to Decide Whether Events Are Independent

In a sample space , the probabilities for the combinations of events  and  occurring are shown. Are and independent events?

### Answer

We recall that the events and are independent if

The probability of the intersection, , is marked below.

This gives us

The probability of is given by the sum of the two probabilities marked below.

This give us

Finally, we can calculate the probability of by finding the sum of the two probabilities marked below.

This give us

Using these values, we can check the condition : which is not equal to .

Hence, the answer is no; and are not independent events.

In our next example, we apply the multiplication rule for independent events to compute the probability of the intersection.

### Example 5: Finding the Probability of the Intersection of Independent Events

A jar of marbles contains 4 blue marbles, 5 red marbles, 1 green marble, and 2 black marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. Find the probability that the first is blue and the second is red.

### Answer

We need to find the probability that the first marble is blue and the second marble is red. Let be the event that the first marble is blue, and the event that the second one is red. Using probability notation, we need to identify .

We note that the first marble is replaced before the second marble is chosen. This implies that the condition for choosing the second marble is identical to that of choosing the first marble. The outcome of the first choice does not affect the probability of the second choice, so the events and are independent. Let us draw a tree diagram describing this example.

We note that the second set of branches have the same probabilities as the respective first set of branches. This is because the first selected marble is placed back into the pool before the second one is chosen, restoring the probability space to the original setting.

We recall that, if and are independent events, the multiplication rule states that

Let us find and from the probability tree.

So, we get and . We can apply the multiplication rule to get

Hence, the probability that the first marble is blue and the second marble is red is .

If and are dependent events, then a slightly different version of the multiplication rule applies. We can obtain this version by rearranging equation (1).

### Theorem: General Multiplication Rule

If and are dependent events, then

In our last example, we will apply the general multiplication rule to compute the probability of an intersection.

### Example 6: Determining the Probability of Intersection of Two Dependent Events

A bag contains 18 white balls and 9 black balls. If 2 balls are drawn consecutively without replacement, what is the probability that the second ball is black and the first one is white?

### Answer

We need to find the probability that the first ball is white and the second ball is black. Let be the event that the first ball is white, and the event that the second one is black. Using probability notation, we need to identify .

We note that the second ball is chosen without replacing the first ball. Because of this, the condition for choosing the second ball is different from that of choosing the first one. In other words, the outcome of the first event affects the probability distribution of the second event. So, the events and are dependent.

We recall that, if and are dependent events, the general multiplication rule states that

Let us find and .

In the beginning, there are 18 white balls out of 27 balls in the jar, so the probability of selecting the white ball first is

Next, we consider . If we assume that has occurred, we are assuming that a white ball has been taken out of the bag before the second ball is selected. That leaves 17 white balls and 9 black balls in the bag for the second choice. Then,

We can draw the probability tree diagram with these two values.

Applying the general multiplication rule, the probability of the intersection is

We note that, when we use tree diagrams, the probability of an intersection can be found by multiplying along the branches of the relevant events regardless of whether or not the events are independent.

Hence, the probability that the first ball is white and the second ball is black is .

Let us summarize a few important points from the explainer.

### Key Points

• Events and are independent if the occurrence of does not affect the probability of the occurrence of .
• In a probability notation, events and are independent if
• Events and are independent if and only if
• If and are dependent events, then
• Any two mutually exclusive events with nonzero probabilities cannot be independent.

## حمِّل تطبيق «نجوى كلاسيز»

احضر حصصك، ودردش مع معلمك وزملائك، واطَّلِع على أسئلة متعلقة بفصلك. حمِّل تطبيق «نجوى كلاسيز» اليوم!

#### التحميل على الكمبيوتر

Windows macOS Intel macOS Apple Silicon

تستخدم «نجوى» ملفات تعريف الارتباط لضمان حصولك على أفضل تجربة على موقعنا. اعرف المزيد عن سياسة الخصوصية