Exhaustive Events

Last Updated : 04 Mar, 2024

Exhaustive Events are a set of events where at least one of the events must occur while performing an experiment. Exhaustive events are a set of events whose union makes up the complete sample space of the experiment.

In this article, we will understand the meaning of exhaustive events, its definition, Venn diagram of exhaustive events, collective exhaustive events, and examples of exhaustive events.

Table of Content

What are Exhaustive Events?
Exhaustive Event Venn Diagram
Collectively Exhaustive Events
Examples of Exhaustive Events
Calculation of Probability for Exhaustive Events

What are Exhaustive Events?

Exhaustive events, in the context of probability, refer to a set of events that collectively cover all possible outcomes of an experiment or situation. In other words, when we say that a set of events is exhaustive, it means that one of those events must occur. There are no other possible outcomes left.

For example, consider flipping a fair coin. The possible outcomes are heads (H) or tails (T). In this case, “getting heads” and “getting tails” are exhaustive events because they cover all possible outcomes when you flip the coin. The sample space, in this case, is {H, T}, and the events “getting heads” and “getting tails” are exhaustive since there are no other possible outcomes.

Mathematically, if E₁, E₂, ……., E_n are exhaustive events, their union (E₁ ∪ E₂ ∪ …… ∪E_n) equals the entire sample space (S).

Definition of Exhaustive Events

Exhaustive events in probability refer to a collection of events that together cover all possible outcomes of an experiment or situation.

In simpler terms, exhaustive events ensure that one of the events in the collection must occur, leaving no room for other outcomes. This concept is fundamental in probability theory as it ensures a comprehensive understanding and analysis of possible outcomes.

What are Events in Probability?

In probability theory, events refer to outcomes or occurrences that can happen as a result of an experiment or observation. A simple example of rolling a six-sided die. The possible outcomes when rolling the die are the numbers 1, 2, 3, 4, 5, and 6.

The two events:

Event A: Rolling an even number.
Event B: Rolling a number greater than 4.

For Event A, the possible outcomes are 2, 4, and 6. So, if you roll the die and get any of these numbers, you have experienced Event A.

For Event B, the possible outcomes are 5 and 6. If you roll the die and get either 5 or 6, you have experienced Event B.

Events in probability are essentially sets of outcomes, and their occurrence depends on the specific conditions or criteria set for each event. Analyzing events helps in calculating probabilities and understanding the likelihood of different outcomes in a given experiment or situation.

Collectively Exhaustive Events

Collectively exhaustive events, in probability theory, refer to a set of events that cover all possible outcomes and no outcome is counted more than once across the events. Here, the events do not overlap with each other.

Consider flipping a fair coin. The possible outcomes are either heads (H) or tails (T). In this case, the events “getting heads” and “getting tails” are collectively exhaustive because one of these outcomes must happen when the coin is flipped. There are no other possible outcomes besides heads or tails.

Mutually Exclusive and Exhaustive Events

Mutually exclusive events and exhaustive events are two important concepts in probability theory.

Aspect	Mutually Exclusive Events	Exhaustive Events
Definition	Events that cannot occur simultaneously.	Events that cover all possible outcomes of an experiment.
Occurrence	Cannot occur together in the same trial.	At least one of the events must occur.
Probability Relationship	The intersection of mutually exclusive events has zero probability i.e., P(A ⋂ B) = 0.	The union of exhaustive events covers the entire sample space i.e., A₁ ⋃ A₂ ⋃ … ⋃ A_n = S.
Example	Rolling an even number and rolling an odd number on a fair six-sided die.	Rolling a number less than 3 and rolling a number greater than or equal to 3 on a fair six-sided die.

Example: Consider rolling a fair six-sided die. The possible outcomes are numbers 1 through 6. Now, let’s define two events:

Event A: Rolling an odd number {1, 3, 5}
Event B: Rolling an even number {2, 4, 6}

Solution:

Events A and B are mutually exclusive because if you roll an odd number (Event A), you cannot roll an even number (Event B), and vice versa.
Events A and B are also collectively exhaustive because together, they cover all possible outcomes of rolling the die. You will always roll either an odd number (Event A) or an even number (Event B).

Exhaustive Event Venn Diagram

Exhaustive events in probability cover all possible outcomes of an experiment, leaving no room for other outcomes. In a Venn diagram, representing exhaustive events involves using circles to encompass all possible outcomes.

Consider the exhaustive events when tossing a fair coin. The possible outcomes are either getting heads (H) or tails (T). In the Venn diagram:

Event A represents the event of getting heads (H).
Event B represents the event of getting tails (T).

Exhaustive-Events

Both events together cover all possible outcomes (H and T), making them exhaustive events. The entire rectangle enclosing them represents the sample space, indicating all potential outcomes of the coin toss.

Examples of Exhaustive Events

Examples of Exhaustive Events can be coin tossing, rolling a dice and drawing cards from a deck of cards. Explanation of each of them is given below:

Coin Tossing

When tossing a fair coin, the possible outcomes are either heads (H) or tails (T).
The events “getting heads” and “getting tails” are collectively exhaustive because one of these outcomes must occur.
For example, if you toss a coin, you will either get heads or tails. There are no other possible outcomes.

Rolling a Dice

When rolling a fair six-sided die, the possible outcomes are numbers 1 through 6.
The events “rolling a 1,” “rolling a 2,” and so on up to “rolling a 6” are collectively exhaustive because one of these outcomes must occur.
For instance, if you roll a die, you will get a number between 1 and 6. There are no other possible outcomes besides these six numbers.

Drawing Cards from a Deck

When drawing cards from a standard deck of 52 playing cards, the possible outcomes include the various ranks (2 through 10, Jack, Queen, King, Ace) and suits (hearts, diamonds, clubs, spades).
The events “drawing a heart,” “drawing a diamond,” “drawing a club,” and “drawing a spade” are collectively exhaustive because one of these outcomes must occur.
Additionally, the events “drawing a 2,” “drawing a 3,” and so on up to “drawing an Ace” for each suit are also collectively exhaustive.
For example, if you draw a card from a deck, it will be either a heart, diamond, club, or spade, and it will also be one of the ranks from 2 to Ace. There are no other possible outcomes besides these.

Calculation of Probability for Exhaustive Events

Calculating the probability for exhaustive events involves determining the likelihood of each individual event occurring and then using the concept of exhaustiveness to find the probability of at least one of the events happening. The steps to calculate probability for exhaustive events are given below:

Step 1: Determine Individual Probabilities

Calculate the probability of each individual event occurring. This can be done by dividing the number of favorable outcomes (for each event) by the total number of possible outcomes in the sample space.

Step 2: Verify Exhaustiveness

Ensure that the events are collectively exhaustive, meaning that together they cover all possible outcomes.

Step 3: Use the Principle of Exhaustiveness

Since the events are collectively exhaustive, the probability of at least one of them occurring is equal to 1 (or 100%). This is because if one event doesn’t happen, then another event in the set must happen. Therefore, the sum of the probabilities of all exhaustive events equals 1.

Example: Consider rolling a fair six-sided die. The possible outcomes are numbers 1 through 6.

Events A: Rolling an odd number {1, 3, 5}
Events B: Rolling an even number {2, 4, 6}

Solution:

Step 1: Calculate Individual Probabilities

Probability of Event A:
P(A) = 3/6 = 1/2

Probability of Event B:
P(B) = 3/6 =1/2

Step 2: Verify Exhaustiveness

Together, Events A and B cover all possible outcomes (1 through 6), making them collectively exhaustive.

Step 3: Use the Principle of Exhaustiveness

Since the events are collectively exhaustive, the probability of at least one of them occurring is 1 (or 100%).

So, in this example, the probability of rolling either an odd number or an even number (or both) on a fair six-sided die is 1. This means that you are guaranteed to roll an odd or an even number when you roll the die.

Read More,

Probability Distribution
Types of Events in Probability
Dependent and Independent Events

Solved Examples on Exhaustive Events

Example 1: You roll two fair six-sided dice. Define events for the sum of the numbers rolled (2 through 12). Are these events collectively exhaustive? What is the probability of getting any sum from 6 to 12?

Solution:

Sum of the numbers rolled can range from 2 (when both dice show a 1) to 12 (when both dice show a 6).

Here are the events for each possible sum:

Event for the sum of 2: {(1, 1)}
Event for the sum of 3: {(1, 2), (2, 1)}
Event for the sum of 4: {(1, 3), (2, 2), (3, 1)}
Event for the sum of 5: {(1, 4), (2, 3), (3, 2), (4, 1)}
Event for the sum of 6: {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Event for the sum of 7: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Event for the sum of 8: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
Event for the sum of 9: {(3, 6), (4, 5), (5, 4), (6, 3)}
Event for the sum of 10: {(4, 6), (5, 5), (6, 4)}
Event for the sum of 11: {(5, 6), (6, 5)}
Event for the sum of 12: {(6, 6)}
These events cover all possible sums from 2 to 12. Therefore, they are collectively exhaustive.

To find the probability of getting any sum from 6 to 12, we need to calculate the probability of each event and then add them up.

Since each die is fair and has 6 sides, there are a total of (6 × 6 = 36) equally likely outcomes when rolling two dice.

calculate the probabilities for each event:

Probability of sum = 6: P(6) = 5/36
Probability of sum = 7: P(7) = 6/36
Probability of sum = 8: P(8) = 5/36
Probability of sum = 9: P(9) = 4/36
Probability of sum = 10: P(10) = 3/36
Probability of sum = 11: P(11) = 2/36
Probability of sum = 12: P(12) = 1/36
Now, to find the probability of getting any sum from 6 to 12, we add up the probabilities of all these events:

P(sum from 6 to 12)=P(6)+P(7)+P(8)+…+P(12)

⇒ P(sum from 6 to 12)= [Tex]\frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} + \frac{2}{36}+ \frac{1}{36} [/Tex]

⇒ P(sum from 6 to 12)= [Tex]\frac{5 + 6 + 5 + 4 +3 + 2 + 1}{36} [/Tex]

⇒ P(sum from 6 to 12)=26/36

⇒ P(sum from 6 to 12)=13/18

So, the probability of getting any sum from 6 to 12 when rolling two fair six-sided dice is 13/18 .

Example 2: You flip three fair coins. Define events for the number of heads obtained (0, 1, 2, or 3). Are these events collectively exhaustive? What is the probability of getting any number of heads from 0 to 3?

Solution:

To define events for the number of heads obtained when flipping three fair coins, we need to consider all possible combinations of outcomes. Each coin can either show heads (H) or tails (T).

Here are the events for each possible number of heads:

Event for 0 heads (all tails): {TTT}
Event for 1 head: {HTT, THT, TTH}
Event for 2 heads: {HHT, HTH, THH}
Event for 3 heads (all heads): {HHH}
These events cover all possible outcomes when flipping three fair coins and represent the number of heads obtained (0, 1, 2, or 3). Therefore, they are collectively exhaustive.

Calculate the probability of getting any number of heads from 0 to 3.

Since each coin is fair and has 2 equally likely outcomes (heads or tails), there are a total of 23=8 equally likely outcomes when flipping three coins.

Probability of getting 0 heads: There is only 1 outcome (TTT) with 0 heads. P(0 heads)= 1/8
Probability of getting 1 head: There are 3 outcomes (HTT, THT, TTH) with 1 head. P(1 head)= 3/8
Probability of getting 2 heads: There are 3 outcomes (HHT, HTH, THH) with 2 heads. P(2 heads)=3/8
Probability of getting 3 heads: There is only 1 outcome (HHH) with 3 heads. P(3 heads)= 1/8
To find the probability of getting any number of heads from 0 to 3, we add up the probabilities of these events:

[Tex](\text{0 to 3 heads}) = P(\text{0 heads}) + P(\text{1 head}) + P(\text{2 heads}) + P(\text{3 heads}) [/Tex]

⇒ [Tex]P(\text{0 to 3 heads}) = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} [/Tex]

⇒ [Tex]P(\text{0 to 3 heads}) = \frac{8}{8} = 1 [/Tex]

So, the probability of getting any number of heads from 0 to 3 when flipping three fair coins is 1 or 100%. This is because one of these outcomes is guaranteed to occur when you flip the coins.

Practice Questions on Exhaustive Events

Question 1: You flip a fair coin. Define two events:

Event A: Getting heads (H)
Event B: Getting tails (T)

Are these events collectively exhaustive? What is the probability of getting either heads or tails?

Question 2: You roll a fair six-sided die. Define six events, one for each possible outcome (numbers 1 through 6). Are these events collectively exhaustive? What is the probability of rolling any number from 1 to 6?

Question 3: You draw a card from a standard deck of 52 playing cards. Define four events:

Event A: Drawing a heart
Event B: Drawing a diamond
Event C: Drawing a club
Event D: Drawing a spade

Are these events collectively exhaustive? What is the probability of drawing a card of any suit?

Question 4: You flip two fair coins. Define three events:

Event A: Getting two heads (HH)
Event B: Getting two tails (TT)
Event C: Getting one head and one tail (HT or TH)

Are these events collectively exhaustive? What is the probability of getting either two heads, two tails, or one head and one tail?