Explainer: Compound Events

In this explainer, we will learn how to find and interpret the probability of compound events.

In probability terms, when we speak of a simple event, we are looking for the probability that a single event occurs, for example, getting “heads” with a single toss of a coin. There are two sides to a coin, only one of which is a heads. The probability of getting “heads” is, therefore,

Recall also that probabilities can be written as either fractions, decimals (between 0 and 1), or percentages (between 1% and 100%). So, with a single toss of a coin, the probability of getting “heads” is 12=0.5=0.5×100%=50%.

Compound probability is concerned with finding the probability of more than one event occurring. Let us look at an example.

Example 1: Compound Probability: Throwing a Coin Twice

If we throw a coin twice, what is the probability of getting “heads” both times?

Answer

We are considering a compound event since there are two events involved here: throwing a heads on the first throw and throwing a heads on the second throw of the coin. We can illustrate all the possible outcomes on a tree diagram to help us find the probability of getting “heads” both times. The first set of branches represents the first throw, with the two possible outcomes “heads” and “tails.”

The probabilities for “heads” and “tails” are each 0.5. Now if we attach branches to each of these outcomes, covering all possible outcomes for the second throw, we can work out the compound probability of getting “heads” both times.

Using the same unbiased coin, the probability of getting “heads” on our second throw has not changed from 12 or 0.5. And we obtain the compound probability of getting one “heads” after another by multiplying the probability of “heads” on the first throw with the probability of “heads” on the second throw. We can write this as 𝑃()=0.5×0.5=0.25.HeadsHeads

Before looking at another example, let us remind ourselves of one or two probability concepts and rules that will be useful to us.

Some Probability Rules and Definitions

For any event 𝐴, if 𝑃(𝐴) is the probability of event 𝐴 occurring, then we have the following:

Rule 10𝑃(𝐴)1

Rule 2 Total probability: the sum of the probabilities of all possible outcomes is equal to 1 (or 100%).

The compliment of event 𝐴, written as 𝐴, refers to everything that is not  𝐴.

Rule 3𝑃𝐴=1𝑃(𝐴)

Note: The compliment of event 𝐴 is sometimes also written as 𝐴.

If two events 𝐴 and 𝐵 cannot occur at the same time, we say that they are mutually exclusive or disjoint events. In this case, the joint probability of 𝐴 and 𝐵 occurring at the same time is zero. We write this as 𝑃(𝐴𝐵)=0. If events 𝐴 and 𝐵 are mutually exclusive, the probability that 𝐴 or 𝐵 occurs is the sum of their probabilities. That is,

Rule 4𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)

If events 𝐴 and 𝐵 are not mutually exclusive, the probability that either 𝐴 or 𝐵 or both occur is

Rule 5𝑃(𝐴𝐵)=𝑃(𝐴)+𝑃(𝐵)𝑃(𝐴𝐵)

Let us look at an example of compound probabilities for nonmutually exclusive events.

Example 2: Compound Probability: Probabilities for Nonmutually Exclusive Events

In a sample of 55 people, 28 have brown hair and 22 have blue eyes. 5 of them have neither brown hair nor blue eyes. What is the probability that a random person from the sample has at least one of these features?

Answer

Our compound events here are “brown hair”, and “blue eyes.” To find the probability that a person chosen at random from the sample of 55 people has at least one of these features, we can simply note that since 5 of the 55 have neither feature, all the rest must have at least one, that is, that 555=50 of the 55 have either brown hair or blue eyes or both.

The probability that a person chosen at random has either brown hair, blue eyes, or both is, therefore, 𝑃()=5055=1011.BrownhairBlueeyes

In essence, we have used our total probability rule to calculate this. Remembering that, for event 𝐴, 𝑃(𝐴)=1𝑃𝐴 and letting brownhairBr= and blueeyesBl=, then BrBl means neither brown hair nor blue eyes. And 𝑃()=1𝑃=1555=5055=1011.BrBlBrBl

Hence, as noted, the probability that a person chosen at random from the sample has at least one of the features “brown hair” and “blue eyes”, is 1011.

In our next example, we will calculate compound probabilities for the results of two spinners.

Example 3: Compound Probability for Two Spinners

If these two spinners are spun, what is the probability that the sum of the numbers the arrows land on is a multiple of 5?

Answer

To find the probability that the sum of the numbers the two arrows land on is a multiple of 5, let us first put all possible outcomes into a table.

All possibilities for spinner 2 are in the blue block across the top and those for spinner 1 are in the red block down the left-hand side. The sums of the values on the two spinners are in the body of the table. So, for example, if spinner 2 lands with the pointer on 8 and spinner 1 lands with the pointer on 7, the sum is 15, which is below the column headed “8” and across the row labeled “7.”

To find the probability that the sum of the numbers on the two spinners is a multiple of 5, we need to know two things:

  1. How many of the results are multiples of 5,
  2. How many possibilities there are in total.

First, considering multiples of 5, we note that the range of values in the table is from 3 to 18. Between these two, the multiples of 5 are 5, 10, and 15. Let us mark the instances of 5, 10, and 15 in our table and count how many we have.

As we can see, there is one instance of the number 5 and two instances each of the numbers 10 and 15. We therefore have 1+2+2=5 “favorable outcomes,” that is, 5 instances where the sum on the two spinners is a multiple of 5.

We have not quite finished yet, however. Remember, we are working out the probability that the sum of the two numbers is a multiple of 5. The second number we need in order to calculate this is the number of possibilities in total. We can find this by multiplying the number of rows by the number of columns in the body of our spinner results table (the number of rows corresponds to the number of segments on spinner 1 and the number of columns is the number of segments on spinner 2).

The total number of possible outcomes is therefore 3×7=21; hence, 𝑃()==521.sumisamultipleof5numberofwayssumisamultipleof5totalpossibleoutcomes

The probability that the sum on the two spinners is a multiple of 5 is therefore 521.

When we throw a coin, what happens on our second throw is not affected by what happened on our first throw. Whether we threw a heads or a tails on the first throw makes no difference to the probability of the outcome of the second throw, whereas when we take a ball from a bag, there is one less ball in the bag for our second selection. This affects the probability of the outcome of our second selection. The events “first ball red“ and “second ball red” are therefore not independent events.

Let us define “independence” of events before we look at some examples.

Definition: Independent and Dependent Events

Two events 𝐴 and 𝐵 are

  • independent if the fact that 𝐴 occurs does not affect the probability of 𝐵 occurring,
  • dependent if the fact that 𝐴 occurs does affect the probability of 𝐵 occurring.

For independent events, 𝑃(𝐴𝐵)=𝑃(𝐴)×𝑃(𝐵).

Our next example is a straightforward example of calculating probability for independent events.

Example 4: Calculating Probability for Independent Events

𝐴 and 𝐵 are independent events, where 𝑃(𝐴)=13 and 𝑃(𝐵)=25. What is the probability that events 𝐴 and 𝐵 both occur?

Answer

Given that events 𝐴 and 𝐵 are independent, the probability that they both occur is 𝑃(𝐴𝐵)=𝑃(𝐴)×𝑃(𝐵)=13×25=215.

Our next example shows how to calculate probabilities for selection without replacement, that is, for dependent events.

Example 5: Compound Probability: Taking Two Balls from a Bag without Replacement

A bag contains 22 red balls and 9 green balls. One red ball is removed from the bag and then a ball is drawn at random. Find the probability that the drawn ball is red.

Answer

To find the probability of drawing a second red ball from the bag, we note first that since there are 22 red balls and 9 green ones, there are 22+9=31 balls in total. The probability of drawing a red ball from the bag on our first pick is, therefore, 𝑃()==2231.Rednumberofredballstotalnumberofballs

Since we are keeping the first ball out of the bag, there is one less ball in the bag in total, so there are now 30 balls in the bag. And of those 30, there is one less red since the ball we took out was red. Hence, the number of red balls is now 21.