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# Disjoint Events

Probability theory is a very important subject in mathematics in which the students study about the possible chances of a random phenomenon. A phenomenon may be referred as an experiment. In probability, there is an extremely useful concept called event. The event is the outcomes of a particular experiment. More specifically, an event may be defined as a possible outcome or a set of outcomes of some experiment or phenomenon.

The sample space is the collection of all the possible outcomes of some experiment. An event is said to be a subset of the sample space for an experiment. An event may be a single outcomes or a set of more than one outcomes. One outcome is said to be an element of various different events. For example -

The sample space (set of all possible outcomes) for the throw of a dice would be {1, 2, 3, 4, 5, 6}, but the events may the subset of this set such as - {5}, {1, 4, 5}, {2, 3} etc.

Getting a head as a result of toss of coin in also an example of an event.

There are different types of events in probability theory. These are as follows :
1) Elementary Event
2) Compound Event
3) Sure Event
4) Impossible Event
5) Independent Event
6) Dependent Event
7) Complementary Event
8) Disjoint Event

In this chapter, we shall go ahead and learn about Disjoint event which is known as a mutually exclusive event.

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## Definition

In probability theory, the concept of disjoint events is used quite often in various experiments. Two events are said to be disjoint if there is no common element in them. The term mutually exclusive events is also used for disjoint events. Two events are statistically disjoint or mutually exclusive when the happening of one is not at all compatible with the happening of the other. In other words, the disjoint events are the events that cannot occur simultaneously.

In the language of statistics, we can say that two events are defined to be disjoint if their intersection comes out to an empty set. Two events, say P and Q, will be disjoint if and only if:
P $\cap$ Q = $\phi$
In other words,
pr (P $\cap$ Q) = 0
i.e. P and Q cannot happen together.
For Example:
During the toss of a coin, getting head and tail are two disjoint events.
While drawing a card from the deck of 52, obtaining an Ace and a king are also disjoint events.
When a die is rolled, the probability of getting 3 or a multiple of 2 are two mutually exclusive events.

In probability theory, the addition rule plays a vital role. Addition rules gives a formula for the union of two events. Let us suppose that there are two events P and Q. Then, the general addition rule is given by the following equation:

Pr (P $\cup$ Q) = Pr (P) + Pr (Q) - Pr (P $\cap$ Q)

If P and Q are two disjoint or mutually exclusive events, then by the definition:
Pr (P $\cap$ Q) = 0

Hence, Addition rule for disjoint event is given below:
Pr (P $\cup$ Q) = Pr (P) + Pr (Q)(Here, P and Q are necessarily be disjoint events.)

## Disjoint Vs Independent Event

Sometimes, being disjoint is confused with being independent. There is definitely a big difference between these two. Two events are said to be disjoint when they cannot occur together, i.e, they cannot overlap. On the other hand, two events are termed as independent events in case if one event does not affect another. In other words, two events will be defined as independent events when the occurrence of one event does not affect or change the probability of occurrence of the other.

For Example: The probability of getting a 3 on rolling a die will never affect on getting a tail on flipping a coin. Both of them do not affect each other. So, these are two independent events.
While, on a die, getting 5 and getting a multiple of 3 are two disjoint events because they can not happen simultaneously.

Also, if two events are independent, then the probability of their happening is equal to the product of the probabilities of both the events separately. i.e. the joint probability of two independent events is same as the product of probabilities of the events. Let P and Q be two independent events, then
Pr (P $\cap$ Q) = Pr (P) . Pr (Q)
While, for two disjoint events :
Pr (P $\cap$ Q) = 0
Thus, if two events P and Q are independent, then occurrence of P does not given any information about Q and vice versa.

## Can Disjoint Events be Independent

"Disjoint" and 'independent" are two types of events in the field of probability. The disjoint events are the events that cannot occur together at the same time. While, the independent events are the events that are independent to each other. This means that the happening of one does not affect on the happening of another. But clearly, both event can occur together.

In this way, the answer to our question "if two disjoint event can be independent', is "no". Two disjoint events P and Q are never said to be independent events since the fact that P and Q are disjoint implies that if P occurs, then Q cannot occur i.e. between two disjoint events P and Q, either P or Q exists. Hence disjoint events can not be referred to independent events.

## Examples

Few examples based on disjoint events are as follows:

Example 1: Which of the following two are disjoint and why?
On a roll of a die,

a) getting a 4 and getting less than 5.
b) getting two multiple of 3 and getting a 5.

Solution:
a) Let A = getting a 4 and B = getting less than 5
Favorable outcomes for getting a 4 = {4}
Favorable outcomes for getting less than 5 = {1, 2, 3, 4}
A $\cap$ B = {4} $\neq$ $\phi$
Therefore, these events are not disjoint.

(a) Let A = getting two multiple of 3 and B = getting a 5
Favorable outcomes for getting a multiple of 3 = {3, 6}
Favorable outcomes for getting a 5 = {5}
A $\cap$ B = $\phi$
Therefore, these events are disjoint.

Example 2: When a pair of dice is thrown, What would be the probability getting the sum of the numbers be either 6 or 11?

Solution: Total number of possible outcomes, i.e. sample space = 36
Outcomes that have sum of 6 are :
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)

P(Sum of 6) = $\frac{5}{36}$

Outcomes that have sum of 11 are:
(5, 6), (6, 5)
P(Sum of 11) = $\frac{2}{36}$

These two events are disjoint since the sum of numbers can never be 6 and 11 simultaneously. So, by the definition of disjoint events
P(Sum of 6 or Sum of 11) = P(Sum of 6) + P(Sum of 11)

= $\frac{5}{36}$ + $\frac{2}{36}$

= $\frac{7}{36}$

= 0.194 = 0.2 approx.

Example 3: Jennifer and Bill draw a card from the deck of 52. What is the probability that they get either Ace or king.

Solution: Sample space = 52
Probability of getting an ace is:
P(Ace) = $\frac{1}{52}$

Probability of getting an ace is:

P(King) = $\frac{1}{52}$

The two given events are disjoint because ace and king can not be drawn at the same time. Thus, by the definition of disjoint events, we have
P(Ace or King) = P(Ace) + P(King)
= $\frac{1}{52}$+$\frac{1}{52}$

= $\frac{2}{52}$

= $\frac{1}{26}$

Example 4: 100 lottery tickets were sold. Out of which 3 were purchased by Ben and 2 by Peter. What is the probability that one of them should win.

Solution: Sample space = 100
Probability of Ben to win:

P(Ben) = $\frac{3}{100}$

Probability of Peter to win:
P(Peter) = $\frac{2}{100}$

Above two events are disjoint event as both of them cannot win simultaneously. Therefore, according to the definition of disjoint events:
P(B or P) = P(B) + P(P)

= $\frac{3}{100}$ + $\frac{2}{100}$

= $\frac{5}{100}$

= $\frac{1}{20}$
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