In this explainer, we will learn how to determine whether two events are independent and compute probabilities involving independent events.

Let us start with the definition of independent events.

### Definition: Independent Events

Two events are independent if the occurrence of one has no effect on the probability of the other.

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.

We start with an example that tests our understanding of what it means for two events to be independent.

### Example 1: Determining Independence from a Given Context

In which of the following scenarios are and independent events?

- 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.
- A die is rolled. Event is rolling an even number and event is rolling a prime number.
- 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.
- 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.
- 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

Recall that events and are independent when the occurrence of one has no effect on the probability of the other; otherwise, they are dependent. 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.

- 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 outcome of affects the outcome of , meaning that the two events are dependent.
- Event is rolling an even number ,
while event is rolling a prime number . The probability
of rolling an even number is
because there are 3 even numbers from 6 equally likely outcomes. Similarly, 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. If the event β and β is rolling a number that is both even and prime, then . Therefore, and are dependent events since the number being even means there is only a third of the original chance of it being prime, and vice versa. - Event is rolling a 6 on a die, and event is a coin landing heads up. Suppose 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 an event is still , regardless of what we get from a die roll. Thus, and are independent.
- 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 us 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. On the other hand, 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.

In general, the probability of the second candy taken is always affected by what the child takes first, regardless of which type they take, because the number of crunchy or chewy candies in the bag changes. The way the probability changes will depend on which type of candy they take first, but either way, it will always change. Since the probability of occurring is different when has occurred than when has not occurred, then event is dependent on event . Therefore, they cannot be independent events. - 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. However, if event did not occur, then there would be 5 boys and 4 girls available for the second selection. The two events are dependent.

We conclude that the only scenario where and are independent events is C.

We now introduce an important theorem that enables us to check for independence by doing some simple probability calculations.

### Theorem: Multiplication Rule for Independent Events

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

In general, we can use this multiplication rule to check whether a pair of events are independent.

Note that an alternative way to express the event β and β is to use the notation β,β which means βthe intersection of and .β Therefore, and refer to the same probability.

Let us now look at an example where we apply the multiplication rule.

### Example 2: Determining Independence from a Venn Diagram

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

### Answer

Here, we are given a Venn diagram and must find out if the events shown are independent. Recall that events and are independent if where β and β is the event where events and occur simultaneously. If events and do not satisfy this rule, they must be dependent.

The probability of the event β and ,β which is the intersection of and , is marked below.

This gives us

Next, the probability of is the sum of the two probabilities marked below.

This gives us

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

This gives us

Using the probabilities obtained, we can now check if the condition is satisfied. We have which is not equal to .

Therefore, the answer to the question is no: and are not independent events.

If we know that two events and are independent, we can sometimes work backward from the multiplication rule to find a missing probability. Here is an example of this type.

### Example 3: Finding Probabilities Assuming Independence of Events

Denote by and two independent events. Given that and , determine .

### Answer

Recall that independent events are events that have no effect upon each other. If and are two independent events, they satisfy the multiplication rule where β and β is the event where events and occur simultaneously. An alternative way to express the event β and β is to use the notation β,β which means βthe intersection of and .β Therefore, and refer to the same probability.

Here, we must work backward from the multiplication rule to find . First, note that in the question, denotes the same thing as , which means βthe probability of or or both.β It helps to draw some Venn diagrams to represent the given information.

Notice that by subtracting from , we can obtain .

Therefore, writing for , we can find an expression for .

Next, we can substitute for , , and in the multiplication rule and then rearrange to find the value of :

Subtracting from both sides of the equation gives and dividing both sides by 0.72 gives .

Finally, we can calculate by substituting for in the equation , so

When we have event probabilities presented in Venn diagrams, we can use the fact that the probabilities in a Venn diagram must sum to 1 to help us find a missing probability. Once we have done this, we can work out whether certain events are independent, as demonstrated in the following example.

### Example 4: Finding an Unknown Constant in a Venn Diagram and Determining Independence

The Venn diagram shows the probabilities that students at a college are in a sports club or have had treatment for an injury in the past year. represents the event that the student is in a sports club. represents the event that the student has had treatment for an injury in the past year.

- Find the value of .
- Are and independent events?

### Answer

**Part 1**

The probabilities shown in the Venn diagram are , , and . We must find , which is the missing probability .

Remember that the probabilities in a Venn diagram must sum to 1 because the diagram represents all possible outcomes. Therefore, we can work out the value of by subtracting the three given probabilities from 1, as follows:

**Part 2**

For the second part, we need to work out whether and are independent events. Recall that events and are independent if where β and β is the event where events and occur simultaneously. If events and do not satisfy this rule, they must be dependent.

Thus, before we can use this rule, we must first calculate and . From the Venn diagram, we see that

Substituting from part 1, we get

Similarly, we have

Lastly, we check if the condition is satisfied. We see that which is not equal to . Hence, the answer to part 2 is no: and are not independent events.

As a final note, we have found that events and are dependent, with . In other words, there is a higher probability of the two events occurring together than if they were independent. Given the context, this is perhaps not surprising because students who are members of sports clubs are probably more likely to receive injuries requiring treatment than those who are not.

In our last example, we will use the combined properties of Venn diagrams and independent events to work out two missing probabilities.

### Example 5: Finding Unknown Constants in a Venn Diagram

The independent events and are shown in the Venn diagram below. Work out the two possible values for and .

### Answer

In this question, we are told that and are independent events. Recall that they must therefore satisfy the multiplication rule where β and β is the event where events and occur simultaneously.

The probabilities shown in the Venn diagram are and . The missing probabilities are and .

Our strategy will be to find algebraic expressions involving for and . Since we know that and are independent, we can then substitute into the multiplication rule to form a quadratic equation in . Once we have solved this equation to get the two possible values of , we can then use the fact that the probabilities in a Venn diagram sum to 1 to work out the corresponding values of .

From the Venn diagram, we see that

Similarly, we have

Next, we substitute for , , and in the multiplication rule to get so

Then, subtracting from both sides gives the quadratic equation .

Now, recall the quadratic formula, which gives the solutions to the quadratic equation :

Taking , , and , we get

Therefore, either or .

The last step is to find the corresponding values of . Since the four probabilities in the Venn diagram must sum to 1, we can work out by subtracting the other three probabilities from 1, as follows:

If , we get ; if , we get .

Hence, the two possible values for and are , or , .

Let us finish by recapping some key concepts from this explainer.

### Key Points

- Two events are independent when the occurrence of one has no effect on the probability of the other; otherwise, they are dependent.
- Multiplication rule: events and are independent if and only if
where β and β is the event where events and
occur simultaneously.

An alternative way to express the event β and β is to use the notation β,β which means βthe intersection of and .β Therefore, and refer to the same probability. - Venn diagrams are a convenient way to represent the probabilities of events. We can use the fact that the probabilities in a Venn diagram sum to 1 to work out missing probabilities. We can also combine this property with the multiplication rule to determine whether two given events are independent.