A-level Mathematics

12.1 Mutually exclusive and independent events — Learn

Prerequisites: finding the probability of a single event, using the complement rule P(not A)=1−P(A)P(\text{not } A) = 1 - P(A), and listing equally likely outcomes from GCSE.

By the end of this section you can: use the addition rule for mutually exclusive events; use the general addition rule for any two events; use the multiplication rule for independent events; test whether two events are independent from a two-way table; and explain the difference between mutually exclusive and independent events.

1. Events and the sample space

The sample space is the set of all possible outcomes of an experiment. For a single roll of a fair six-sided die, the sample space is

{1,2,3,4,5,6}.\{1, 2, 3, 4, 5, 6\}.

An event is a set of outcomes. The event "the number is even" is the set {2,4,6}\{2, 4, 6\}. In this chapter, an event usually has a capital letter as its name.

For equally likely outcomes, the probability of an event is

P(A)=number of outcomes in Atotal number of outcomes.P(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes}}.

So the probability of an even number is 36=12\dfrac{3}{6} = \dfrac{1}{2}.

2. Mutually exclusive events

Two events are mutually exclusive when they cannot both happen at the same time. Rolling a 5 and rolling an even number are mutually exclusive, because 5 is not even.

When two events are mutually exclusive, the overlap is empty and P(A and B)=0P(A \text{ and } B) = 0. The probability that one or the other happens is the sum of the two probabilities:

P(A or B)=P(A)+P(B).P(A \text{ or } B) = P(A) + P(B).

This is the addition rule for mutually exclusive events. In probability, "or" includes the case where both happen, unless the question says otherwise.

Example 1. A fair six-sided die is rolled once. Find the probability of an even number or a 5.

The event "even" is {2,4,6}\{2, 4, 6\}, so P(even)=36P(\text{even}) = \dfrac{3}{6}. The event "a 5" is {5}\{5\}, so P(5)=16P(5) = \dfrac{1}{6}. The two events cannot happen together, so add:

P(even or 5)=36+16=46=23.P(\text{even or } 5) = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}.

3. The general addition rule

Most pairs of events can happen together. If you add their probabilities, you count the outcomes in both events twice. Subtract the overlap once:

P(A or B)=P(A)+P(B)−P(A and B).P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).

This is the general addition rule. It works for every pair of events, and it reduces to the mutually exclusive rule when the overlap is empty.

Example 2. A fair six-sided die is rolled once. Find the probability of an even number or a prime number.

The evens are {2,4,6}\{2, 4, 6\}, so P(even)=36P(\text{even}) = \dfrac{3}{6}. The primes are {2,3,5}\{2, 3, 5\}, so P(prime)=36P(\text{prime}) = \dfrac{3}{6}. The number 2 is in both events, so P(even and prime)=16P(\text{even and prime}) = \dfrac{1}{6}.

Substitute into the general addition rule:

P(even or prime)=36+36−16=56.P(\text{even or prime}) = \frac{3}{6} + \frac{3}{6} - \frac{1}{6} = \frac{5}{6}.

The union is {2,3,4,5,6}\{2, 3, 4, 5, 6\}, which has five outcomes, so the answer agrees with the definition.

4. Independent events

Two events are independent when the chance of one does not change when you know the other has happened. A coin toss and a die roll are independent, because the coin does not affect the die.

For independent events, multiply the probabilities to find the chance that both happen:

P(A and B)=P(A)×P(B).P(A \text{ and } B) = P(A) \times P(B).

This is the multiplication rule.

Example 3. A fair coin is tossed and a fair six-sided die is rolled. Find the probability of a head and a 6.

The coin and the die are independent, so multiply:

P(head and 6)=P(head)×P(6)=12×16=112.P(\text{head and } 6) = P(\text{head}) \times P(6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}.

When items are drawn with replacement, the draws are independent, because the first item goes back before the second is taken. When items are drawn without replacement, the draws are not independent. Section 12.2 handles that case with conditional probability.

5. Testing independence

You can test independence without knowing why the events happen. Find P(A and B)P(A \text{ and } B) and compare it with P(A)×P(B)P(A) \times P(B):

A two-way table is often the quickest way to read the values.

Example 4. A survey of 300 adults records whether each owns a car and whether each owns a bicycle. The results are in the table.

Owns a bicycle Does not own a bicycle Total
Owns a car 72 108 180
Does not own a car 48 72 120
Total 120 180 300

Show that owning a car and owning a bicycle are independent.

Read the three probabilities from the table:

P(car)=180300=35,P(bicycle)=120300=25,P(\text{car}) = \frac{180}{300} = \frac{3}{5}, \qquad P(\text{bicycle}) = \frac{120}{300} = \frac{2}{5}, P(car and bicycle)=72300=625.P(\text{car and bicycle}) = \frac{72}{300} = \frac{6}{25}.

Now multiply the first two probabilities:

P(car)×P(bicycle)=35×25=625.P(\text{car}) \times P(\text{bicycle}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}.

The product equals P(car and bicycle)P(\text{car and bicycle}), so the two events are independent. In this survey, owning a bicycle does not change the chance of owning a car.

Example 5. A second survey of 250 people records whether each owns a cat and whether each owns a dog. Here 150 own a cat, 100 own a dog, and 30 own both. Show that the two events are not independent.

P(cat)=150250=35,P(dog)=100250=25.P(\text{cat}) = \frac{150}{250} = \frac{3}{5}, \qquad P(\text{dog}) = \frac{100}{250} = \frac{2}{5}.

The product is 35×25=625=0.24\dfrac{3}{5} \times \dfrac{2}{5} = \dfrac{6}{25} = 0.24. The joint probability from the data is

P(cat and dog)=30250=325=0.12.P(\text{cat and dog}) = \frac{30}{250} = \frac{3}{25} = 0.12.

The two values differ, so owning a cat and owning a dog are not independent. Fewer people own both than independence would predict.

6. Mutually exclusive, or independent?

These two words describe different things, and mixing them up is the most common error in this section.

If two events are mutually exclusive and both have a probability greater than zero, they are not independent. Knowing that one happened makes the other impossible, which changes its probability to zero.

Example 6. On a single roll of a fair die, let AA be "even" and let BB be "a 5". The events are mutually exclusive. Check independence.

P(A)=12,P(B)=16,P(A)×P(B)=112.P(A) = \frac{1}{2}, \qquad P(B) = \frac{1}{6}, \qquad P(A) \times P(B) = \frac{1}{12}.

But P(A and B)=0P(A \text{ and } B) = 0, because the events cannot happen together. The values 00 and 112\dfrac{1}{12} differ, so the events are not independent.

7. Try it: when are two events independent?

Predict first. The figure shows a two-way table with 100 people in each row and 100 people in each column, so P(A)=0.5P(A) = 0.5 and P(B)=0.5P(B) = 0.5. The control sets the number of people in both AA and BB. Before you move the control, predict the value that makes AA and BB independent.

Test. Move the control to your predicted value. Watch P(A and B)P(A \text{ and } B) and P(A)×P(B)P(A) \times P(B), and watch P(A∣B)P(A \mid B) against P(A)P(A). Then set the overlap to 20 and to 80 and compare.

Explain. Write two sentences. The first explains why P(A and B)P(A \text{ and } B) falls as the overlap falls. The second explains why AA and BB are independent exactly at the value you found.

Watch out. Three mistakes are common.

Adding probabilities without checking for an overlap. The addition rule P(A)+P(B)P(A) + P(B) is only for mutually exclusive events. When the events can happen together, subtract the overlap. For two events from a single die roll, the overlap is often not zero.

Calling mutually exclusive events independent. They are opposites in effect. If AA and BB cannot both happen, then knowing AA happened tells you that BB did not, so the events are as dependent as they can be.

Multiplying for events that are not independent. The multiplication rule needs independence. Drawing two counters from a bag without replacement is the usual trap: the first draw changes the second.

Re-check: after any addition, ask whether the events can overlap. After any multiplication, ask whether the events are independent.

Progress check

Return to the chapter problem. Records for the warehouse cover 100000 mornings. The table shows how many mornings had a fire and how many had an alarm.

Alarm sounds Alarm does not sound Total
Fire 95 5 100
No fire 1998 97902 99900
Total 2093 97907 100000
  1. Find P(fire)P(\text{fire}).
  2. Find P(alarm)P(\text{alarm}).
  3. Work out P(fire)×P(alarm)P(\text{fire}) \times P(\text{alarm}).
  4. Read P(fire and alarm)P(\text{fire and alarm}) from the table.
  5. Are the events "fire" and "alarm" independent? Explain your answer.

Compare your answers with a partner. The events are not independent, and the direction of the difference is what makes an alarm useful.

You should now be able to

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