The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Independent Events Mathematics • 10th Grade

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 𝐴 and 𝐡 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 𝐴 and 𝐡 independent events?

  1. A student leaves their house on their way to school. Event 𝐴 is them arriving at the bus stop in time to catch the bus and event 𝐡 is them getting to school on time.
  2. A die is rolled. Event 𝐴 is rolling an even number and event 𝐡 is rolling a prime number.
  3. A die is rolled and a coin is flipped. Event 𝐴 is rolling a 6 on the die, and event 𝐡 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 𝐴 is them taking a chewy candy first and event 𝐡 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 𝐴 is the teacher selecting a boy first, and event 𝐡 is the teacher selecting a girl second.

Answer

Recall that events 𝐴 and 𝐡 are independent when the occurrence of one has no effect on the probability of the other; otherwise, they are dependent. To establish whether 𝐴 and 𝐡 are independent or dependent, we should examine whether the probability of 𝐡 changes when we assume that 𝐴 has already occurred. Let us consider each scenario using the definition.

  1. Event 𝐴 is the student arriving at the bus stop in time to catch the bus, and event 𝐡 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 𝐴 affects the outcome of 𝐡, meaning that the two events are dependent.
  2. Event 𝐴 is rolling an even number {2,4,6}, while event 𝐡 is rolling a prime number {2,3,5}. The probability of rolling an even number is 𝑃(𝐴)=12 because there are 3 even numbers from 6 equally likely outcomes. Similarly, the probability of rolling a prime number is 𝑃(𝐡)=12 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€œπ΄ and 𝐡” is rolling a number that is both even and prime, then 𝑃(𝐴𝐡)=16and. Therefore, 𝐴 and 𝐡 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.
  3. Event 𝐴 is rolling a 6 on a die, and event 𝐡 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 12, regardless of what we get from a die roll. Thus, 𝐴 and 𝐡 are independent.
  4. Event 𝐴 is the child taking a chewy candy first, and event 𝐡 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 𝐡 occurs is different depending on whether or not event 𝐴 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 𝐴 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 𝐴 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 𝐡 occurring is different when 𝐴 has occurred than when 𝐴 has not occurred, then event 𝐡 is dependent on event 𝐴. Therefore, they cannot be 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 𝐴 occurred, then there will be 4 boys and 5 girls available for the second selection. However, if event 𝐴 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 𝐴 and 𝐡 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 𝐴 and 𝐡 are independent if and only if 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡),and where β€œπ΄ and 𝐡” is the event where events 𝐴 and 𝐡 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€œπ΄ and 𝐡” is to use the notation β€œπ΄βˆ©π΅,” which means β€œthe intersection of 𝐴 and 𝐡.” Therefore, 𝑃(𝐴∩𝐡) and 𝑃(𝐴𝐡)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 𝑆, the probabilities are shown for the combinations of events 𝐴 and 𝐡 occurring. Are 𝐴 and 𝐡 independent events?

Answer

Here, we are given a Venn diagram and must find out if the events shown are independent. Recall that events 𝐴 and 𝐡 are independent if 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡),and where β€œπ΄ and 𝐡” is the event where events 𝐴 and 𝐡 occur simultaneously. If events 𝐴 and 𝐡 do not satisfy this rule, they must be dependent.

The probability of the event β€œπ΄ and 𝐡,” which is the intersection of 𝐴 and 𝐡, is marked below.

This gives us 𝑃(𝐴𝐡)=519.and

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

This gives us 𝑃(𝐴)=419+519=919.

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

This gives us 𝑃(𝐡)=519+519=1019.

Using the probabilities obtained, we can now check if the condition 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡)and is satisfied. We have 𝑃(𝐴)×𝑃(𝐡)=919Γ—1019=90361, which is not equal to 𝑃(𝐴𝐡)=519and.

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

If we know that two events 𝐴 and 𝐡 are independent, we can sometimes work backward from the multiplication rule 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡)and to find a missing probability. Here is an example of this type.

Example 3: Finding Probabilities Assuming Independence of Events

Denote by 𝐴 and 𝐡 two independent events. Given that 𝑃(𝐡)=0.28 and 𝑃(𝐴βˆͺ𝐡)=0.46, determine 𝑃(𝐴).

Answer

Recall that independent events are events that have no effect upon each other. If 𝐴 and 𝐡 are two independent events, they satisfy the multiplication rule 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡),and where β€œπ΄ and 𝐡” is the event where events 𝐴 and 𝐡 occur simultaneously. An alternative way to express the event β€œπ΄ and 𝐡” is to use the notation β€œπ΄βˆ©π΅,” which means β€œthe intersection of 𝐴 and 𝐡.” Therefore, 𝑃(𝐴∩𝐡) and 𝑃(𝐴𝐡)and refer to the same probability.

Here, we must work backward from the multiplication rule to find 𝑃(𝐴). First, note that in the question, 𝑃(𝐴βˆͺ𝐡) denotes the same thing as 𝑃(𝐴𝐡)or, which means β€œthe probability of 𝐴 or 𝐡 or both.” It helps to draw some Venn diagrams to represent the given information.

Notice that by subtracting 𝑃(𝐡) from 𝑃(𝐴𝐡)or, we can obtain 𝑃(𝐴𝐡)butnot.

Therefore, writing π‘₯ for 𝑃(𝐴𝐡)and, we can find an expression for 𝑃(𝐴).

Next, we can substitute for 𝑃(𝐴𝐡)and, 𝑃(𝐴), and 𝑃(𝐡) in the multiplication rule 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡)and and then rearrange to find the value of π‘₯: π‘₯=(π‘₯+0.18)Γ—0.28=π‘₯Γ—0.28+0.18Γ—0.28=0.28π‘₯+0.0504.

Subtracting 0.28π‘₯ from both sides of the equation π‘₯=0.28π‘₯+0.0504 gives 0.72π‘₯=0.0504, and dividing both sides by 0.72 gives π‘₯=0.07.

Finally, we can calculate 𝑃(𝐴) by substituting for π‘₯ in the equation 𝑃(𝐴)=π‘₯+0.18, so 𝑃(𝐴)=0.07+0.18=0.25.

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. 𝐢 represents the event that the student is in a sports club. 𝑇 represents the event that the student has had treatment for an injury in the past year.

  1. Find the value of π‘₯.
  2. Are 𝐢 and 𝑇 independent events?

Answer

Part 1

The probabilities shown in the Venn diagram are 𝑃(𝐢𝑇)=0.52butnot, 𝑃(𝑇𝐢)=0.04butnot, and 𝑃(𝐢𝑇)=0.31neithernor. We must find π‘₯, which is the missing probability 𝑃(𝐢𝑇)and.

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 π‘₯ by subtracting the three given probabilities from 1, as follows: π‘₯=1βˆ’0.52βˆ’0.04βˆ’0.31=1βˆ’0.87=0.13.

Part 2

For the second part, we need to work out whether 𝐢 and 𝑇 are independent events. Recall that events 𝐢 and 𝑇 are independent if 𝑃(𝐢𝑇)=𝑃(𝐢)×𝑃(𝑇),and where β€œπΆ and 𝑇” is the event where events 𝐢 and 𝑇 occur simultaneously. If events 𝐢 and 𝑇 do not satisfy this rule, they must be dependent.

Thus, before we can use this rule, we must first calculate 𝑃(𝐢) and 𝑃(𝑇). From the Venn diagram, we see that 𝑃(𝐢)=𝑃(𝐢𝑇)+𝑃(𝐢𝑇)=π‘₯+0.52.andbutnot

Substituting π‘₯=0.13 from part 1, we get 𝑃(𝐢)=0.13+0.52=0.65.

Similarly, we have 𝑃(𝑇)=𝑃(𝑇𝐢)+𝑃(𝑇𝐢)=π‘₯+0.04=0.13+0.04=0.17.andbutnot

Lastly, we check if the condition 𝑃(𝐢𝑇)=𝑃(𝐢)×𝑃(𝑇)and is satisfied. We see that 𝑃(𝐢)×𝑃(𝑇)=0.65Γ—0.17=0.1105, which is not equal to 𝑃(𝐢𝑇)=0.13and. Hence, the answer to part 2 is no: 𝐢 and 𝑇 are not independent events.

As a final note, we have found that events 𝐢 and 𝑇 are dependent, with 𝑃(𝐢𝑇)>𝑃(𝐢)×𝑃(𝑇)and. 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 𝐴 and 𝐡 are shown in the Venn diagram below. Work out the two possible values for π‘₯ and 𝑦.

Answer

In this question, we are told that 𝐴 and 𝐡 are independent events. Recall that they must therefore satisfy the multiplication rule 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡),and where β€œπ΄ and 𝐡” is the event where events 𝐴 and 𝐡 occur simultaneously.

The probabilities shown in the Venn diagram are 𝑃(𝐴𝐡)=0.05butnot and 𝑃(𝐡𝐴)=0.6butnot. The missing probabilities are 𝑃(𝐴𝐡)=π‘₯and and 𝑃(𝐴𝐡)=𝑦neithernor.

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

From the Venn diagram, we see that 𝑃(𝐴)=𝑃(𝐴𝐡)+𝑃(𝐴𝐡)=π‘₯+0.05.andbutnot

Similarly, we have 𝑃(𝐡)=𝑃(𝐡𝐴)+𝑃(𝐡𝐴)=π‘₯+0.6.andbutnot

Next, we substitute for 𝑃(𝐴𝐡)and, 𝑃(𝐴), and 𝑃(𝐡) in the multiplication rule 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡)and to get π‘₯=(π‘₯+0.05)Γ—(π‘₯+0.6)=π‘₯+0.6π‘₯+0.05π‘₯+0.05Γ—0.6, so π‘₯=π‘₯+0.65π‘₯+0.03.

Then, subtracting π‘₯ from both sides gives the quadratic equation π‘₯βˆ’0.35π‘₯+0.03=0.

Now, recall the quadratic formula, which gives the solutions to the quadratic equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0: π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Taking π‘Ž=1, 𝑏=βˆ’0.35, and 𝑐=0.03, we get π‘₯=βˆ’(βˆ’0.35)±(βˆ’0.35)βˆ’(4Γ—1Γ—0.03)2Γ—1=0.35±√0.1225βˆ’0.122=0.35±√0.00252=0.35Β±0.052.

Therefore, either π‘₯=0.35+0.052=0.42=0.2 or π‘₯=0.35βˆ’0.052=0.32=0.15.

The last step is to find the corresponding values of 𝑦. Since the four probabilities in the Venn diagram must sum to 1, we can work out 𝑦 by subtracting the other three probabilities from 1, as follows: 𝑦=1βˆ’π‘ƒ(𝐴𝐡)βˆ’π‘ƒ(𝐡𝐴)βˆ’π‘ƒ(𝐴𝐡)=1βˆ’0.05βˆ’0.6βˆ’π‘₯=0.35βˆ’π‘₯.butnotbutnotand

If π‘₯=0.2, we get 𝑦=0.35βˆ’0.2=0.15; if π‘₯=0.15, we get 𝑦=0.35βˆ’0.15=0.2.

Hence, the two possible values for π‘₯ and 𝑦 are π‘₯=0.2, 𝑦=0.15 or π‘₯=0.15, 𝑦=0.2.

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 𝐴 and 𝐡 are independent if and only if 𝑃(𝐴𝐡)=𝑃(𝐴)×𝑃(𝐡),and where β€œπ΄ and 𝐡” is the event where events 𝐴 and 𝐡 occur simultaneously.
    An alternative way to express the event β€œπ΄ and 𝐡” is to use the notation β€œπ΄βˆ©π΅,” which means β€œthe intersection of 𝐴 and 𝐡.” Therefore, 𝑃(𝐴∩𝐡) and 𝑃(𝐴𝐡)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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.