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May 15, 2024**Mutually Exclusive Events:** In the theory of probability, two events are said to be mutually exclusive events if they cannot occur simultaneously or at the same time. Mutually exclusive events are called disjoint events. If the probability of happening the two events at the same time is zero, then they are known as mutually exclusive events.

For example, turning towards the left and towards the right cannot happen at the same time; they are known as mutually exclusive events. In this article, we will discuss events and specifically mutually exclusive events.

In the theory of probability, two events are said to be mutually exclusive events if they cannot occur simultaneously or at the same time. Mutually Exclusive events are termed as disjoint events.

In simple words, mutually exclusive events are the events or actions which are opposite to each other.

Two events are said to be mutually exclusive events if they cannot occur simultaneously or at the same time.

- We cannot run forwards and backwards at the same time. The events “running backwards” and “running forwards” are mutually exclusive events.

2. Tossing a coin is a mutually exclusive events type. We cannot get both heads and tails while tossing a coin. So “getting a head” and “getting a tail” are mutually exclusive events.

3. When a dice is rolled, we cannot get the numbers \(2\) and \(5\) at the same time. Thus, the events of getting numbers \(2\) and \(5\) on a die are mutually exclusive events.

4. We can not worry, and we can not feel happy at the same time. The occurrence of one event prevents the occurrence of another event. So, the events of worry and happiness are mutually exclusive events.

The mutually exclusive events probability is zero. Before, going through this topic will discuss some important term or relations related to it.

The probability of an event is the number of chances of occurring that event. The probability defines the ratio of the number of favourable outcomes to the total outcomes. The probability of occurrence of an event \(E\) is given by

\({{P}}\left({{E}} \right) = \frac{{{\mathbf{number}}\,{\mathbf{of}}\,{\mathbf{favourable}}\,{\mathbf{outcomes}}\,{\mathbf{of}}\,{\mathbf{an}}\,{\mathbf{event}}}}{{{\mathbf{Total}}\,{\mathbf{number}}\,{\mathbf{of}}\,{\mathbf{outcomes}}}}\)

Example: Probability of getting head when a coin is tossed is \(P(E)= \frac {1}{2}.\)

Some important formulas related to probability are

1. The probability of the sure or certain event is one.

2. The probability of that event cannot happen is zero.

3. The sum of the probability of all the elementary events is one.

4. \(0 ≤ P (E) ≤ 1\)

Venn diagram representation of the union of sets is given below:

Intersection of sets \(A\),\(B\) is denoted by \(A ∩ B\), mathematically. The Venn diagram representation of the intersection of two sets is shown below:

If \(A\) and \(B\) are two events, then the probability of happening both of them is zero.

\(P (A ∩ B) = P(A\) and \(B)= 0\)

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

For two mutually exclusive events (say \(A\) and \(B\)), the probability of happening the event \(A\) or event \(B\) is given by

\(P (A ∪ B) = P (A) + P (B)\)

The probability of non-mutual exclusive events (\(A\) and \(B\)) is given by using the formula

\(P(A ∪ B) = P (A) + P (B) – P (A ∩ B)\)

The mutually exclusive events are shown as there is no common shaded portion of the events in the Venn diagram representation.

The probability rule of mutually exclusive events is

\(P(A ∪ B) = P(A) + P(B)\)

Two events are said to be mutually exclusive events if they cannot occur simultaneously or at the same time. The probability of mutually exclusive events is zero. There are specific rules or mutually exclusive events formulas are associated with the probability of mutually exclusive events, which are derived by using the definitions of mutually exclusive events.

- Addition Rule:

\(P(A + B) = 1\) - Subtraction Rule:

\(P(A ∪ B)’= 0\) - Multiplication Rule:

\(P(A ∩ B) = 0\)

By using the multiplication rule of the probability of mutually exclusive events, \(P (A ∩ B) = 0\).

1. \(P\left( {B|A} \right) = 0\)

2. \(P\left( {A|B} \right) = 0\)

*Q.1. Asit is playing with the dice. What is the probability of a dice showing the numbers \(2\) or \(5\)?*

** Ans:** The total number of faces on the dice is six. Hence, the total number of outcomes obtained throwing dice is \(6\).

We know that the probability of an event is the ratio of the number of favourable outcomes to the total number of outcomes.

The probability of getting a number \(2\) on throwing dice is

\(P\left(2 \right) = \frac{1}{6}\)

The probability of getting a number \(5\) on throwing dice is

\(P\left(5 \right) = \frac{1}{6}\)

Getting the numbers \(2\) or \(5\) on throwing the dice cannot occur at a time. So, they are mutually exclusive events.

The probability of getting the mutually exclusive events A or B is given by the formula,

\(P(A ∪ B) = P(A) + P(B)\)

\( \Rightarrow P(2\) or \(5) = \frac{1}{6} + \frac{1}{6}\)

\( \Rightarrow P(2\) or \(5) = \frac{2}{6}\)

\( \Rightarrow P(2\) or \(5) = \frac{1}{3}\)

Hence, the probability of getting \(2\) or \(5\) on throwing dice is \(\frac{1}{3}.\)

*Q.2. Avinash is selecting the hearts or spades cards from the deck of \(52\) cards. Find the probability of drawing hearts or spades from the well-shuffled cards.*

** Ans:** A well-shuffled deck of cards contains \(52\) cards.

So, the total number of outcomes is \(52\).

In well-shuffled cards, there are \(13\) hearts and \(13\) spades.

We know that the probability of an event is the ratio of the number of favourable outcomes to the total number of outcomes.

The probability of drawing hearts is \(\frac{{13}}{{52}} = \frac{1}{4}.\)

The probability of drawing spades is \(\frac{{13}}{{52}} = \frac{1}{4}.\)

The events of drawing the hearts and spades are mutually exclusive events.

The probability of getting the mutually exclusive events \(A\) or \(B\) is given by the formula,

\(P(A ∪ B) = P(A) + P(B)\)

\(⇒ P(\rm{hearts}\,\rm{or}\,\rm{spades}) = P(\rm{Hearts}) + P(\rm{Spades})\)

\(⇒ P(\rm{hearts}\,\rm{or}\,\rm{spades}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\)

Hence, the probability of drawing hearts and spades from the well-shuffled cards is \(\frac{1}{2}.\)

*Q.3. A coin is tossed once. Find the probability of getting head or tail?*

** Ans:** A coin is tossed, will get only either head or tail as the outcome.

The total number of outcomes is \(2\).

Getting the head or tail are mutually exclusive events.

The probability of getting head while tossing a coin is \(\frac{1}{2}.\)

\(P\left(H \right) = \frac{1}{2}\)

The probability of getting a tail while tossing a coin is \(\frac{1}{2}.\)

\(P\left(T \right) = \frac{1}{2}\)

The probability of getting the mutually exclusive events \(A\) or \(B\) is given by the formula,

\(P(A ∪ B) = P(A) + P(B)\)

\(P(H\) or \(T) = P(H) + P(T) = \frac{1}{2} + \frac{1}{2} = 1\)

Hence, the probability of getting head or tail while tossing a coin is one.

*Q.4. A box contains \(4\) red balls and \(6\) white balls. A ball is taken out from the box. What is the probability of getting a ball is red or white.*

** Ans:** Given: A box contains \(4\) red balls and \(6\) white balls.

The total number of balls in a box is \(10\).

The probability of getting a red ball is given by \(P(R) = \frac{4}{10}\)

The probability of getting a White ball is given by \(P(W) = \frac{6}{10}\)

Drawing a red or white ball from the box is a mutually exclusive event.

The probability of getting the mutually exclusive events \(A\) or \(B\) is given by the formula,

\(P(A ∪ B) = P(A) + P(B)\)

\( ⇒ P(R\) or \(W) = P(R) + P(W) = \frac{4}{10} + \frac{6}{10} = \frac{10}{10} = 1\)

Hence, the probability of drawing a red or white ball is one.

*Q.5. Rachana takes one fish away from the tank with \(5\) male fish and \(3\) female fish. What is the probability of selecting the male and female fish?*

** Ans:** Given, a tank has \(5\) male fish and \(3\) female fish.

Rachna was taking only one fish from the tank. So, the possible chance is either it is a male fish or female fish.

But, both female and male fish is not possible.

So, they are mutually exclusive events.

We know that the probability of mutually exclusive events is zero.

Hence, the probability of selecting male and female fish from the tank is zero.

In this article, we have studied the definition of mutually exclusive events, which tells that two mutually exclusive events cannot occur at the same time. We studied the different examples of mutually exclusive events.

This article also tells the probability of mutually exclusive events and probability formulas of mutually exclusive events such as addition, subtraction and multiplication rules. We also studied the conditional probability of mutually exclusive events. This article also gives the solved examples of mutually exclusive events for better understanding the concept.

*Q.1. What is an example of mutually exclusive?*** Ans:** Two events are said to be mutually exclusive events if they cannot occur simultaneously.

*Q.2. How do you know if an event is mutually exclusive?*** Ans:** Two events are said to be mutually exclusive events if they cannot occur simultaneously. An event is said to be a mutually exclusive event if it stops the occurrence of another event at the same time.

*Q.3. What is the probability formula of mutually exclusive events?*** Ans:** The probability of two mutually exclusive events (Say \(A\) and \(B\)) is zero.

\(P(A ∩ B) = 0\); \(P(A ∪ B) = P(A) + P(B)\)

*Q.4. Are mutually exclusive events are dependent events?*** Ans:** Two mutually exclusive events are neither dependent events nor independent events.

*Q.5. What is the difference between mutually exclusive and exhaustive events?*** Ans:** Events are said to be mutually exclusive events if they cannot occur simultaneously. Events are said to be exhaustive if at least one of the events must occur.

*We hope you find this detailed article on mutually exclusive events helpful. If you have any doubts or queries regarding this topic, feel to ask us in the comment section and we will be more than happy to assist you.*