Mutually Exclusive Events

Mutually Exclusive Events

Two or more events are said to be mutually exclusive if the occurrence of any one of them means the others will not occur (That is, we cannot have 2 or more such events occurring at the same time).

For example, if we throw a 6-sided die, the events "4" and "5" are mutually exclusive. We cannot get both 4 and 5 at the same time when we throw one die.

If E₁ and E₂ are mutually exclusive events, then E₁ and E₂ will not happen together. So the probabality of the 2 events will be zero:

P(E₁ and E₂) = 0.

Now, suppose "E₁ or E₂" denotes the event that "either E₁ or E₂ both occur", then

(a) If E₁ and E₂ are not mutually exclusive events:

P(E₁ or E₂) = P(E₁) + P(E₂) − P(E₁ and E₂)
We can also write:
P(E₁ ∪ E₂) = P(E₁) + P(E₂) − P(E₁ ∩ E₂)

A diagram for this situation is as follows. We see that there is some overlap between the events E₁ and E₂. The probability of that overlap portion is P(E₁ ∩ E₂).

An example for non-mutually exclusive events could be:

E₁ = students in the swimming team
E₂ = students in the debating team

In this case, the yellow area represents students in the swimming team only, and the darker green area represents students in the debating team only. The light green overlap area represents the students in both the swimming team and the debating team.

(b) If E₁ and E₂ are mutually exclusive events:

P(E₁ or E₂) = P(E₁) + P(E₂)

Our diagram for mutually exclusive events shows that there is no overlap:

An example of mutually exclusive events could be:

E₁ = male students
E₂ = female students

There is no overlap. [Of course, gender is not a simple issue as in fact, some overlap does occur. Don't read too much into it — this is just an example.]

In this case, the intersection E₁ ∩ E₂ is empty, leading to the conclusion:

P(E₁ ∩ E₂) = 0

This explains why, for the mutually exclusive case,

P(E₁ or E₂) = P(E₁) + P(E₂)

Example 1

It is known that the probability of obtaining zero defectives in a sample of $\displaystyle{40}$ items is $\displaystyle{0.34}$ whilst the probability of obtaining $\displaystyle{1}$ defective item in the sample is $\displaystyle{0.46}$. What is the probability of

(a) obtaining not more than $\displaystyle{1}$ defective item in a sample?

(b) obtaining more than $\displaystyle{1}$ defective items in a sample?

ANSWER

"Obtaining not more than one" means we choose either $\displaystyle{0}$ or $\displaystyle{1}$ defective.

Let event E₁ be "obtaining zero defectives" and E₂ be "obtaining $\displaystyle{1}$ defective item".

(a) Events E₁ and E₂ are mutually exclusive, so

P(E₁ or E₂) = P(E₁) + P(E₂) = 0.34 + 0.46 = 0.8

(b) $\displaystyle{P}{\left(\text{more than 1}\right)}={1}-{0.8}={0.2}$

Example 2

The probability that a student passes Mathematics is $\displaystyle\frac{2}{{3}}$ and the probability that he passes English is $\displaystyle\frac{4}{{9}}$. If the probability that he will pass at least one subject is $\displaystyle\frac{4}{{5}}$, what is the probability that he will pass both subjects?

(We assume it is based on probability only.)

ANSWER

It is possible for a student to either:

• Pass math only
• Pass English only
• Pass both math and English

So we conclude that these are not mutually exclusive events. We have:

P(E₁ or E₂) = P(E₁) + P(E₂) − P(E₁ and E₂)
$\displaystyle\frac{4}{{5}}=\frac{2}{{3}}+\frac{4}{{9}}-{P}{\left({E}_{{1}}{\quad\text{and}\quad}{E}_{{2}}\right)}$

So

$\displaystyle{P}{\left({E}_{{1}}{\quad\text{and}\quad}{E}_{{2}}\right)}=\frac{2}{{3}}+\frac{4}{{9}}-\frac{4}{{5}}=\frac{14}{{45}}$