PROBABILITY of "A OR B" : \(p\begin{pmatrix} A\cup B \end{pmatrix}\)
Combined Events Part 2

(Disjunction Rule & Addition Rule for Mutually Exclusive Events)

Given an experiment and two of its possible events \(A\) and \(B\), we'll often need to calculate probability of event "A or B" ocurring; that's the probability \(p\begin{pmatrix}A\ \text{or} \ B \end{pmatrix}\). We learn how to calculate such probabilities in this section.

In proper mathematical notation, we'll write the probability of A or B occurring as: \[p\begin{pmatrix}A \cup B \end{pmatrix}\] where you can think of the symbol \(\cup \) as the word "or".

In probability theory the word "or" refers to an inclusive or which means that the event "A or B" occurs when either:

event \(A\) occurs
event \(B\) occurs
both \(A\) and \(B\) occur, that's the event \(A\cap B\).

The following tutorial will provide you with all the essentials to know about the probability of A or B.

The formula for calculating the probability of A or B occurring is known as the disjunction rule and is stated here.

Disjunction Formula - Formula for Probability of "A or B"

Given an experent with, the probability of A or B occurring is given by: \[p\begin{pmatrix}A \cup B\end{pmatrix} = p\begin{pmatrix}A\end{pmatrix} + p\begin{pmatrix}B \end{pmatrix} - p \begin{pmatrix}A \cap B\end{pmatrix}\]

Looking at the Venn diagram we have here, we can see how many elements are in each set, \(A\), \(B\) and the universal set \(U\).

So if we wanted to we could calculate the probability of A or B directly, using the fact that: \[p\begin{pmatrix}A \cup B \end{pmatrix} = \frac{n\begin{pmatrix}A \cup B \end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}\] And in doing so, we would find: \[\begin{aligned} p\begin{pmatrix}A \cup B \end{pmatrix} & = \frac{n\begin{pmatrix}A \cup B \end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} \\ & = \frac{9}{10} \\ p\begin{pmatrix}A \cup B \end{pmatrix} & = 0.9 \end{aligned}\] So, we know that (for this Venn diagram) the probability of A or B occurring is \(0.9\).

Deriving & Using the Formula

To derive the formula the trick is to express \(n\begin{pmatrix}A \cup B \end{pmatrix}\) in terms of \(n\begin{pmatrix}A\end{pmatrix}\), \(n\begin{pmatrix}B\end{pmatrix}\) and \(n\begin{pmatrix}A \cap B\end{pmatrix}\). Using the Venn diagram, we realize that: \[n\begin{pmatrix}A \cup B \end{pmatrix} = n \begin{pmatrix}A\end{pmatrix} + n \begin{pmatrix}B\end{pmatrix} - n \begin{pmatrix}A\cap B \end{pmatrix}\] we subtract \(n \begin{pmatrix}A\cap B \end{pmatrix}\) to not count those elements inside both \(A\) and \(B\) twice.

We can illustrate that result with the following Venn diagrams.

Note: here we can see that if we didn't subtract \(n\begin{pmatrix}A \cap B \end{pmatrix}\) those elements would be counted twice (once when we take the elements (outcomes) in set \(A\) and once when we take the elements in set \(B\)).

Finally, dividing our expression for \(n\begin{pmatrix}A \cup B \end{pmatrix}\) by the total number of elements \(n\begin{pmatrix}U\end{pmatrix}\) we obtain the probability: \[\frac{n\begin{pmatrix}A \cup B \end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} = \frac{n \begin{pmatrix}A\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} + \frac{n \begin{pmatrix}B\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} - \frac{n \begin{pmatrix}A\cap B \end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}\] which becomes the disjunction rule: \[ p\begin{pmatrix}A \cup B \end{pmatrix} = p \begin{pmatrix}A\end{pmatrix} + p \begin{pmatrix}B\end{pmatrix} - p \begin{pmatrix}A\cap B \end{pmatrix} \]

EXAMPLE

For the Venn diagram we have here we calculate \(p\begin{pmatrix}A\cup B\end{pmatrix}\) as follows:

\(n\begin{pmatrix}U\end{pmatrix} = 10\), indeed there are 10 elements (outcomes) in the universal set.
\(n\begin{pmatrix}A\end{pmatrix} = 5\), there are 5 elements (outcomes) in set \(A\)
\(n\begin{pmatrix}B\end{pmatrix} = 6\), there are 6 elements (outcomes) in set \(B\)
\(n\begin{pmatrix}A \cap B\end{pmatrix} = 2\) those are the 2 elements (outcomes) that are in both set \(A\) and \(B\).

We can now calculate each of the probabilities:

\(p\begin{pmatrix}A\end{pmatrix} = \frac{n\begin{pmatrix}A\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}=\frac{5}{10} = 0.5\)
\(p\begin{pmatrix}B\end{pmatrix} = \frac{n\begin{pmatrix}B\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}=\frac{6}{10}=0.6\)
\(p\begin{pmatrix}A \cap B\end{pmatrix} = \frac{n\begin{pmatrix}A\end{pmatrix}}{n\begin{pmatrix}A\cap B \end{pmatrix}}=\frac{2}{10}=0.2\)

Now that we know these probabilities, we can use the disjunction rule and calculate the probability of \(A\) or \(B\): \[\begin{aligned} p\begin{pmatrix}A \cup B \end{pmatrix} & = p \begin{pmatrix}A\end{pmatrix} + p \begin{pmatrix}B\end{pmatrix} - p \begin{pmatrix}A\cap B \end{pmatrix} \\ & = 0.5+0.6 - 0.2 \\ & = 1.1 - 0.2 \\ p\begin{pmatrix}A \cup B \end{pmatrix} & = 0.9 \end{aligned}\] This is the same result as we had found without the disjunction rule, which confirms it works.

Example

At a given school:

the probability that a student studies French is \(0.7\)
the probability that a student studies Spanish is \(0.6\)
the probability that a student studies both French and Spanish is \(0.45\)

A student is taken at random. Find the probability that this student studies French or Spanish.

Solution

It's always good to start by defining the events as well as listing the probabilities we're given:

\(F\): the student studies French, \(p\begin{pmatrix}F\end{pmatrix} = 0.7\)
\(S\): the student studies Spanish, \(p\begin{pmatrix}S\end{pmatrix} = 0.6\)
\(F\cap S\): the student studies both French and Spanish, \(p\begin{pmatrix}F \cap S\end{pmatrix} = 0.45\)

Now that our events are defined, we can find the probability \(p\begin{pmatrix}A\cup B\end{pmatrix}\) using the disjunction rule: \[\begin{aligned} \begin{pmatrix}F \cup S \end{pmatrix} & = p\begin{pmatrix}F\end{pmatrix} + p\begin{pmatrix}S\end{pmatrix} - p\begin{pmatrix}F \cap S\end{pmatrix} \\ & = 0.7 + 0.6 - 0.45 \\ \begin{pmatrix}F \cup S \end{pmatrix} & = 0.85 \end{aligned}\] The probability that the student studies both French and spanish is \(0.85\).

Example

A card is picked, at random, from a regular deck of 52 playing cards. Find the probability that the card is an 8 or a heart.

Solution

We can define the events:

\(A\): picking an 8. Since there are four 8's the probability is \(p\begin{pmatrix}A\end{pmatrix} = \frac{4}{52}\)
\(B\): picking a heart. There are 13 hearts so the probability is \(p\begin{pmatrix}B\end{pmatrix} = \frac{13}{52}\)
\(A\cap B\): picking the 8 of hearts. There is 1 8 of hearts so the probability is \(p\begin{pmatrix}A\cap B\end{pmatrix} = \frac{1}{52}\)

Now, using the disjunction rule: \[\begin{aligned} p\begin{pmatrix}A\cup B \end{pmatrix} &= p\begin{pmatrix}A\end{pmatrix} + p\begin{pmatrix}B\end{pmatrix} - p\begin{pmatrix}A \cap B \end{pmatrix} \\ & = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} \\ & = \frac{4 + 13 - 1}{52} \\ & = \frac{16}{52} \\ p\begin{pmatrix}A\cup B \end{pmatrix} & = \frac{4}{13} \end{aligned}\] So the probability of picking an 8 or a heart is \(\frac{4}{13} \approx 0.308\).

MUTUALLY EXCLUSIVE EVENTS

Given an experiment and two events \(A\) and \(B\), we say that \(A\) and \(B\) are Mutually Exclusive if it is impossible for them to occur simultaneously. In other words the two events cannot both occur simultanesouly, it can only be one or the other, but not both. \[p\begin{pmatrix}A \cap B\end{pmatrix}=0\] When represented on a Venn diagram, as we can see here, the sets representing mutually exclusive events do not overlap (they do not intersect).

Furthermore, since \(p\begin{pmatrix}A \cap B\end{pmatrix}=0\), the disjunction rule (seen further-up) leads to the following result for the probability of A or B for mutually exclsuive events

Addition Rule: Probability of A or B for Mutually Exclusive Events

Given 2 mutually exclusive events \(A\) and \(B\) the probability or \(A\) or \(B\) occurring is: \[p\begin{pmatrix}A \cup B\end{pmatrix} = p\begin{pmatrix}A\end{pmatrix} + p\begin{pmatrix}B \end{pmatrix}\]

Example

A box contains 5 blue, 3 red and 2 green paper slips. A paper slip is picked at random, find the probability that the slip is blue or green.

Solution

Let's start by defining the events and the number of ways each of them can happen:

\(B\): picking a blue slip, \(n\begin{pmatrix}B\end{pmatrix} = 5\)
\(R\): picking a red slip, \(n\begin{pmatrix}R\end{pmatrix} = 3\)
\(G\): picking a green slip, \(n\begin{pmatrix}G\end{pmatrix} = 2\)
\(U\): universal set, \(n\begin{pmatrix}U\end{pmatrix} = 10\)

Each of the events \(B\), \(R\) and \(G\) are mutually exclusive. Indeed when picking a paper slip we can obtain either of the three colors but it is impossible to get two different colors with one paper slip.

We can therefore use the addition rule for mutually exclusive events but first we need to calculate the probability of picking a blue slip and the probability of picking a green slip.

\(p\begin{pmatrix}B\end{pmatrix} = \frac{n\begin{pmatrix}B\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} = \frac{5}{10} = 0.5\)
\(p\begin{pmatrix}G\end{pmatrix} = \frac{n\begin{pmatrix}G\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} = \frac{2}{10} = 0.2\)

Now we can use the addition rule: \[\begin{aligned} p\begin{pmatrix}B \cup G \end{pmatrix} & = p\begin{pmatrix}B \end{pmatrix} + p\begin{pmatrix}G\end{pmatrix} \\ & = 0.5 + 0.2 \\ p\begin{pmatrix}B \cup G \end{pmatrix} & = 0.7 \end{aligned}\] Finally, we can state that the probability of picking a blue or a green paper slip is \(0.7\).

Example

Charlotte is asked to pick a number, at random, between 3 and 12 included. Find the probability that she picks a prime number or an even number.

Solution

We start by defining the events \(A\) and \(B\):

\(A\): she picks a prime number
\(B\): she picks an even number

The events \(A\) and \(B\) are mutually exclusive.

Indeed since all prime numbers (except for 2) are odd: its is impossible for the number Charlotte picks to be both odd and even i.e. \(p\begin{pmatrix}A\cap B \end{pmatrix}=0\).

We can therefore use the addition rule to calculate the probability \(p\begin{pmatrix}A\cup B \end{pmatrix}\). First we need the probabilities \(p\begin{pmatrix}A\end{pmatrix}\) and \(p\begin{pmatrix}B\end{pmatrix}\).

From 3 to 12 (included):

there are 10 numbers so: \(n\begin{pmatrix}U\end{pmatrix}=10\).
there are 4 prime numbers. Indeed 3, 5, 7 and 11 are the only prime numbers between 3 and 12, so: \(n\begin{pmatrix}A\end{pmatrix} = 4\) and \(p\begin{pmatrix}A\end{pmatrix} = \frac{n\begin{pmatrix}A\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} = \frac{4}{10} = 0.4\)
there are 5 even numbers, 4, 6, 8, 10 and 12, so: \(n\begin{pmatrix}B\end{pmatrix}=5\) and \(p\begin{pmatrix}B\end{pmatrix} = \frac{n\begin{pmatrix}B\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}} = \frac{5}{10} = 0.5\)

Finally we use the addition rule: \[\begin{aligned} p\begin{pmatrix}A\cup B\end{pmatrix} & = p\begin{pmatrix}A\end{pmatrix} + p\begin{pmatrix}B\end{pmatrix} \\ & = 0.4 + 0.5 \\ p\begin{pmatrix}A\cup B\end{pmatrix} & = 0.9 \end{aligned}\] The probability that Charlotte picks an even or a prime number is \(0.9\).

1. The numbers of people who participated in the survery equals the total number of crosses we see on the Venn diagram. Since there are \(10\), we can state that the total number of outcomes equals \(10\), which we write: \[n\begin{pmatrix}U \end{pmatrix} = 10\]
2. The number of participants who own a cat is equal to the number of crosses inside the set labelled \(C\), which is \(5\). So we can write: \[n\begin{pmatrix}C \end{pmatrix} = 5\]
3. The people who own both a cat and a dog are represented by the crosses that are in both the set \(C\) and \(D\) (where the two sets overlap, we usually say where they "intersect"). We can see there are \(2\) crosses, so \(2\) people own both a cat and a dog. We write: \[n\begin{pmatrix}C \cap D \end{pmatrix} = 2\]
4. The probability that a person own both a cat and a dog is: \[\begin{aligned} p\begin{pmatrix}C \cap D \end{pmatrix} & = \frac{n \begin{pmatrix} C \cap D \end{pmatrix}}{n \begin{pmatrix}U \end{pmatrix}}\\ & = \frac{2}{10} \\ p\begin{pmatrix}C \cap D \end{pmatrix} & = 0.2 \end{aligned}\]

1. Assuming the two events \(M\) (doing well at her Math test) and \(F\) (doing well at her French test) are independent we can use the formula we saw \[\begin{aligned} p\begin{pmatrix} M \cap F \end{pmatrix} & = 0.8 \times 0.7 \\ p\begin{pmatrix} M \cap F \end{pmatrix} & = 0.56 \end{aligned}\]
2. We start by finding the probability that Helen doesn't do well at her Mathematics test. That's the complement of her doing well at her Mathematics test, so: \[\begin{aligned} p\begin{pmatrix}M'\end{pmatrix} & = 1 - p \begin{pmatrix} M \end{pmatrix} \\ & = 1 - 0.8 \\ p\begin{pmatrix}M'\end{pmatrix} & = 0.2 \end{aligned}\] We now use our formula to calculate \(p\begin{pmatrix} M' \cap F \end{pmatrix}\): \[p\begin{pmatrix} M' \cap F \end{pmatrix} = 0.2 \times 0.7 = 0.14 \]

Assuming that the two events \(C\): "Cathy hits the center of the target" and \(J\): "Jonathan hits the center of the target" are independent, we know that: \[p \begin{pmatrix} C \cap J \end{pmatrix} = p\begin{pmatrix} C \end{pmatrix} \times p\begin{pmatrix} J \end{pmatrix}\] Replacing \(p \begin{pmatrix} C \cap J \end{pmatrix}\) by \(0.2\) and \(p \begin{pmatrix} C \end{pmatrix}\) by \(0.5\) we obtain an equation for the unknown \(p \begin{pmatrix} J \end{pmatrix}\): \[0.2 = 0.5 \times p \begin{pmatrix} J \end{pmatrix}\] which leads to: \[\frac{0.2}{0.5} = p \begin{pmatrix} J \end{pmatrix}\] or \[p \begin{pmatrix} J \end{pmatrix} = \frac{0.2}{0.5} \] Finally, we find: \[p \begin{pmatrix} J \end{pmatrix} = 0.4 \]

1. We define the event \(E\): "rolling an even number" and: \[p \begin{pmatrix}E \end{pmatrix} = \frac{1}{2}\]
2. We define the event \(T\): "flipping tails" and: \[p \begin{pmatrix}T \end{pmatrix} = \frac{1}{2}\]
3. We find the probability of both rolling an even number and flipping tails, \(p\begin{pmatrix}E \cap T \end{pmatrix}\), using our formula for independent events: \[\begin{aligned} p\begin{pmatrix}E \cap T \end{pmatrix} & = p \begin{pmatrix}E \end{pmatrix} \times p\begin{pmatrix}T\end{pmatrix} \\ \frac{1}{2}\times \frac{1}{2} \\ p\begin{pmatrix}E \cap T \end{pmatrix} & = \frac{1}{4} \quad (=0.25) \end{aligned}\]

Scan this QR-Code with your phone/tablet and view this page on your preferred device.

Subscribe Now and view all of our playlists & tutorials.

RadfordMathematics.com

Online Mathematics Book

PROBABILITY of "A OR B" : \(p\begin{pmatrix} A\cup B \end{pmatrix}\)
Combined Events Part 2

(Disjunction Rule & Addition Rule for Mutually Exclusive Events)

Disjunction Formula - Formula for Probability of "A or B"

Example

Solution

Example

Solution

MUTUALLY EXCLUSIVE EVENTS

Addition Rule: Probability of A or B for Mutually Exclusive Events

Example

Solution

Example

Solution

RadfordMathematics.com

Online Mathematics Book

PROBABILITY of "A OR B" : \(p\begin{pmatrix} A\cup B \end{pmatrix}\) Combined Events Part 2

(Disjunction Rule & Addition Rule for Mutually Exclusive Events)

Disjunction Formula - Formula for Probability of "A or B"

Understanding the Disjunction Rule

Deriving & Using the Formula

EXAMPLE

Example

Solution

Example

Solution

MUTUALLY EXCLUSIVE EVENTS

Addition Rule: Probability of A or B for Mutually Exclusive Events

Example

Solution

Example

Solution

Exercise 1 - Answer Key

PROBABILITY of "A OR B" : \(p\begin{pmatrix} A\cup B \end{pmatrix}\)
Combined Events Part 2