Lesson Explainer: Dependent and Independent Events Mathematics

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 𝐴 and 𝐡 are independent if the occurrence of 𝐴 does not affect the probability of the occurrence of 𝐡. That is, 𝑃(𝐡∣𝐴)=𝑃(𝐡), where 𝑃(𝐡∣𝐴) represents the probability of event 𝐡 occurring given that event 𝐴 has occurred.

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.

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 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

We recall that events 𝐴 and 𝐡 are independent if the occurrence of 𝐴 does not affect the probability of the occurrence of 𝐡. 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 occurrence of event of 𝐴 affects the probability of the occurrence of event 𝐡, meaning that the two events are dependent.
  2. Event 𝐴 is rolling an even number {2,4,6}, while 𝐡 is rolling a prime number {2,3,5}. 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. In other words, the event 𝐡 given 𝐴 is the same as getting 2 from {2,4,6}. So, 𝑃(𝐡∣𝐴)=13. Since 𝑃(𝐡∣𝐴)≠𝑃(𝐡), the two events are dependent.
  3. Event 𝐴 is rolling a 6 on a die, and event 𝐡 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 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’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 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, so 𝑃(𝐡∣𝐴)=1. 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, so 𝑃(𝐡∣𝐴)=0. Since the probability of 𝐡 occurring is different when 𝐴 has occurred to when 𝐴 has not occurred, then event 𝐡 is dependent on event 𝐴. 𝐴 and 𝐡 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 𝐴 occurred, then there will be 4 boys and 5 girls available for the second selection, and 𝑃(𝐡∣𝐴)=59. However, if event 𝐴 did not occur, then there will be 5 boys and 4 girls available for the second selection, and 𝑃(𝐡∣𝐴)=49. 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 β€œgetting a blue candy first” and β€œgetting a red candy second” independent?

Answer

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

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, 𝑃(∣)=58.gettingaredcandysecondgettingabluecandyfirst

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 𝑃()=59.gettingaredcandysecond

This gives us 𝑃(∣)≠𝑃().gettingaredcandysecondgettingabluecandyfirstgettingaredcandysecond

Since the probability of β€œgetting 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(𝐡∣𝐴)=𝑃(𝐴∩𝐡)𝑃(𝐴).(1)

We also know that 𝑃(𝐡∣𝐴)=𝑃(𝐡) for independents events. Substituting this into the equation above and rearranging, we get 𝑃(𝐴)×𝑃(𝐡)=𝑃(𝐴∩𝐡).

This leads to the following theorem.

Theorem: Multiplication Rule for Independent Events

Events 𝐴 and 𝐡 are independent if and only if 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡), where 𝐴∩𝐡 is the event where events 𝐴 and 𝐡 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 𝑃(𝐴)=0.3 and 𝑃(𝐡)=0.25 and 𝐴∩𝐡=βˆ…, are 𝐴 and 𝐡 independent?

Answer

We recall that, if 𝐴 and 𝐡 are independent events, then 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡).

We are given that 𝐴∩𝐡=βˆ…, 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 𝑃(𝐴∩𝐡)=𝑃(βˆ…)=0.

On the other hand, we can calculate 𝑃(𝐴)×𝑃(𝐡)=0.3Γ—0.25=0.075.

Since 𝑃(𝐴∩𝐡)≠𝑃(𝐴)×𝑃(𝐡), 𝐴 and 𝐡 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 𝐴 and 𝐡 are mutually exclusive events, then 𝐴 and 𝐡 cannot occur simultaneously. So, if we assume that event 𝐴 has already occurred, then that eliminates the possibility that 𝐡 could occur. In other words, the outcome of 𝐴 affects the outcome of 𝐡, 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 𝑆, the probabilities for the combinations of events 𝐴 and 𝐡 occurring are shown. Are 𝐴 and 𝐡 independent events?

Answer

We recall that the events 𝐴 and 𝐡 are independent if 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡).

The probability of the intersection, 𝐴∩𝐡, is marked below.

This gives us 𝑃(𝐴∩𝐡)=519.

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

This give 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 give us 𝑃(𝐡)=519+519=1019.

Using these values, we can check the condition 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡): 𝑃(𝐴)×𝑃(𝐡)=919Γ—1019=90361, which is not equal to 𝑃(𝐴∩𝐡)=519.

Hence, the answer is no; 𝐴 and 𝐡 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 𝐴 be the event that the first marble is blue, and 𝐡 the event that the second one is red. Using probability notation, we need to identify 𝑃(𝐴∩𝐡).

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 𝐴 and 𝐡 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 𝐴 and 𝐡 are independent events, the multiplication rule states that 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡).

Let us find 𝑃(𝐴) and 𝑃(𝐡) from the probability tree.

So, we get 𝑃(𝐴)=412=13 and 𝑃(𝐡)=512. We can apply the multiplication rule to get 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡)=13Γ—512=536.

Hence, the probability that the first marble is blue and the second marble is red is 536.

If 𝐴 and 𝐡 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 𝐴 and 𝐡 are dependent events, then 𝑃(𝐴∩𝐡)=𝑃(𝐡∣𝐴)×𝑃(𝐴).

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 𝐴 be the event that the first ball is white, and 𝐡 the event that the second one is black. Using probability notation, we need to identify 𝑃(𝐴∩𝐡).

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 𝐴 and 𝐡 are dependent.

We recall that, if 𝐴 and 𝐡 are dependent events, the general multiplication rule states that 𝑃(𝐴∩𝐡)=𝑃(𝐡∣𝐴)×𝑃(𝐴).

Let us find 𝑃(𝐴) and 𝑃(𝐡∣𝐴).

In the beginning, there are 18 blue marbles out of 27 marbles in the jar, so the probability of selecting the white ball first is 𝑃(𝐴)=1827=23.

Next, we consider 𝑃(𝐡∣𝐴). If we assume that 𝐴 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, 𝑃(𝐡∣𝐴)=926.

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

Applying the general multiplication rule, the probability of the intersection is 𝑃(𝐴∩𝐡)=𝑃(𝐡∣𝐴)×𝑃(𝐴)=926Γ—23=313.

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 313.

Let us summarize a few important points from the explainer.

Key Points

  • Events 𝐴 and 𝐡 are independent if the occurrence of 𝐴 does not affect the probability of the occurrence of 𝐡.
  • In a probability notation, events 𝐴 and 𝐡 are independent if 𝑃(𝐡∣𝐴)=𝑃(𝐡).
  • Events 𝐴 and 𝐡 are independent if and only if 𝑃(𝐴∩𝐡)=𝑃(𝐴)×𝑃(𝐡).
  • If 𝐴 and 𝐡 are dependent events, then 𝑃(𝐴∩𝐡)=𝑃(𝐡∣𝐴)×𝑃(𝐴).
  • Any two mutually exclusive events with nonzero probabilities cannot be independent.

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