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URL: https://nakafa.com/en/subjects/mathematics/combinatorics/probability-of-mutually-exclusive-events
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Understand mutually exclusive events in probability, calculate their probabilities, and apply the concept to examples.
---
## Understanding Mutually Exclusive Events
In everyday life, we often face situations where **two events cannot occur simultaneously**. For example, when flipping a coin, we cannot get both heads and tails at the same time in a single flip. Events like these are called mutually exclusive events.
**Mutually exclusive events** are two or more events that cannot occur simultaneously in a single experiment. If one event occurs, then the other event will definitely not occur.
As a simple illustration, imagine you draw one card from a deck. The card you draw cannot be both red and black at the same time. These two events are mutually exclusive because there is no card that has both colors.
## Characteristics of Mutually Exclusive Events
### No Intersection
The main characteristic of mutually exclusive events is **having no intersection or common elements**. In mathematical notation, if $$A$$ and $$B$$ are mutually exclusive events, then:
Visible text: The main characteristic of mutually exclusive events is **having no intersection or common elements**. In mathematical notation, if and are mutually exclusive events, then:
```math
A \cap B = \emptyset
```
The symbol $$\emptyset$$ indicates an empty set, meaning there are no common elements between the two events.
Visible text: The symbol indicates an empty set, meaning there are no common elements between the two events.
### Examples in Real Life
Several examples of mutually exclusive events that are easy to understand:
- In dice rolling: getting an even number and getting an odd number
- In card drawing: drawing an Ace and drawing a King in a single draw
- In a race: placing first and placing second simultaneously
### Identifying Mutually Exclusive Events
To identify whether two events are mutually exclusive, ask yourself: "Can these two events occur simultaneously in one experiment?" If the answer is no, then the two events are mutually exclusive.
## Formula for Probability of Mutually Exclusive Events
Since mutually exclusive events have no intersection, calculating their probability becomes simpler. The **basic formula** for the probability of mutually exclusive events is:
```math
P(A \cup B) = P(A) + P(B)
```
This formula shows that the probability of event $$A$$ or event $$B$$ occurring equals the sum of the individual probabilities of each event.
Visible text: This formula shows that the probability of event or event occurring equals the sum of the individual probabilities of each event.
### Why This Formula Works
Unlike events that are not mutually exclusive, in mutually exclusive events we don't need to subtract the intersection because $$P(A \cap B) = 0$$. Therefore, the general formula:
Visible text: Unlike events that are not mutually exclusive, in mutually exclusive events we don't need to subtract the intersection because . Therefore, the general formula:
```math
P(A \cup B) = P(A) + P(B) - P(A \cap B)
```
Becomes:
```math
P(A \cup B) = P(A) + P(B) - 0 = P(A) + P(B)
```
## Application in Calculations
### Dice Rolling
A die is rolled once. Determine the probability of getting a prime number or a number greater than $$4$$.
Visible text: A die is rolled once. Determine the probability of getting a prime number or a number greater than .
**Solution:**
Sample space: $$S = \{1, 2, 3, 4, 5, 6\}$$
Visible text: Sample space:
- Event $$A$$ (prime number): $$A = \{2, 3, 5\}$$
- Event $$B$$ ($$\text{number} > 4$$): $$B = \{5, 6\}$$
Visible text: - Event (prime number):
- Event ():
Check intersection: $$A \cap B = \{5\}$$
Visible text: Check intersection:
Since there is an intersection (number $$5$$), the two events are **not mutually exclusive**. Let's use an example that is truly mutually exclusive.
Visible text: Since there is an intersection (number ), the two events are **not mutually exclusive**. Let's use an example that is truly mutually exclusive.
**Correct Example:**
- Event $$A$$ (odd number): $$A = \{1, 3, 5\}$$
- Event $$C$$ (even number): $$C = \{2, 4, 6\}$$
Visible text: - Event (odd number):
- Event (even number):
Check intersection: $$A \cap C = \emptyset$$
Visible text: Check intersection:
Since there is no intersection, the two events are **mutually exclusive**.
Component: MathContainer
Children:
```math
P(A) = \frac{3}{6} = \frac{1}{2}
```
```math
P(C) = \frac{3}{6} = \frac{1}{2}
```
```math
P(A \cup C) = P(A) + P(C) = \frac{1}{2} + \frac{1}{2} = 1
```
This result makes sense because in dice rolling, either an odd or even number will definitely appear (all possibilities are covered).
### Rolling Two Dice
Two dice are rolled simultaneously. Determine the probability of getting a sum of $$5$$ or a sum of $$7$$.
Visible text: Two dice are rolled simultaneously. Determine the probability of getting a sum of or a sum of .
**Solution:**
To understand more clearly, let's look at all possible outcomes of rolling two dice in the following table:
| Die $$1$$ \\ Die $$2$$ | $$1$$ | $$2$$ | $$3$$ | $$4$$ | $$5$$ | $$6$$ |
|------------------|---|---|---|---|---|---|
| $$1$$ | $$(1,1)$$ | $$(1,2)$$ | $$(1,3)$$ | $$(1,4)$$ | $$(1,5)$$ | $$(1,6)$$ |
| $$2$$ | $$(2,1)$$ | $$(2,2)$$ | $$(2,3)$$ | $$(2,4)$$ | $$(2,5)$$ | $$(2,6)$$ |
| $$3$$ | $$(3,1)$$ | $$(3,2)$$ | $$(3,3)$$ | $$(3,4)$$ | $$(3,5)$$ | $$(3,6)$$ |
| $$4$$ | $$(4,1)$$ | $$(4,2)$$ | $$(4,3)$$ | $$(4,4)$$ | $$(4,5)$$ | $$(4,6)$$ |
| $$5$$ | $$(5,1)$$ | $$(5,2)$$ | $$(5,3)$$ | $$(5,4)$$ | $$(5,5)$$ | $$(5,6)$$ |
| $$6$$ | $$(6,1)$$ | $$(6,2)$$ | $$(6,3)$$ | $$(6,4)$$ | $$(6,5)$$ | $$(6,6)$$ |
Visible text: | Die \\ Die | | | | | | |
|------------------|---|---|---|---|---|---|
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
Total possibilities is $$6 \times 6 = 36$$
Visible text: Total possibilities is
**Event identification:**
**Event $$A$$** ($$\text{sum} = 5$$):
Visible text: **Event ** ():
From the table above, pairs that produce sum $$5$$ are:
Visible text: From the table above, pairs that produce sum are:
- $$(1,4): 1 + 4 = 5$$
- $$(2,3): 2 + 3 = 5$$
- $$(3,2): 3 + 2 = 5$$
- $$(4,1): 4 + 1 = 5$$
Visible text: -
-
-
-
So there are **$$4 \text{ ways}$$** to get sum $$5$$.
Visible text: So there are **** to get sum .
**Event $$B$$** ($$\text{sum} = 7$$):
Visible text: **Event ** ():
From the table above, pairs that produce sum $$7$$ are:
Visible text: From the table above, pairs that produce sum are:
- $$(1,6): 1 + 6 = 7$$
- $$(2,5): 2 + 5 = 7$$
- $$(3,4): 3 + 4 = 7$$
- $$(4,3): 4 + 3 = 7$$
- $$(5,2): 5 + 2 = 7$$
- $$(6,1): 6 + 1 = 7$$
Visible text: -
-
-
-
-
-
So there are **$$6 \text{ ways}$$** to get sum $$7$$.
Visible text: So there are **** to get sum .
**Check intersection:** It's impossible for the sum to be both $$5$$ and $$7$$, so $$A \cap B = \emptyset$$
Visible text: **Check intersection:** It's impossible for the sum to be both and , so
Both events are **mutually exclusive**.
Component: MathContainer
Children:
```math
P(A) = \frac{4}{36} = \frac{1}{9}
```
```math
P(B) = \frac{6}{36} = \frac{1}{6}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{9} + \frac{1}{6}
```
**Solution for fraction addition:**
To add $$\frac{1}{9} + \frac{1}{6}$$, we need to find the LCM of $$9$$ and $$6$$.
Visible text: To add , we need to find the LCM of and .
$$LCM(9, 6) = 18$$, so:
Visible text: , so:
Component: MathContainer
Children:
```math
\frac{1}{9} = \frac{1 \times 2}{9 \times 2} = \frac{2}{18}
```
```math
\frac{1}{6} = \frac{1 \times 3}{6 \times 3} = \frac{3}{18}
```
```math
P(A \cup B) = \frac{2}{18} + \frac{3}{18} = \frac{5}{18}
```
## Problem Solving Strategy
### Systematic Steps
To solve probability problems for mutually exclusive events, follow these steps:
1. **Identify the sample space** and determine the total possible outcomes
2. **Define the events** mentioned in the problem clearly
3. **Check intersection** between events to ensure they are mutually exclusive
4. **Calculate the probability of each** event separately
5. **Apply the formula** $$P(A \cup B) = P(A) + P(B)$$
Visible text: 1. **Identify the sample space** and determine the total possible outcomes
2. **Define the events** mentioned in the problem clearly
3. **Check intersection** between events to ensure they are mutually exclusive
4. **Calculate the probability of each** event separately
5. **Apply the formula**
### Practical Tips
Several useful tips:
- **Visualize** events using diagrams or tables when possible
- **Double-check** whether the final result makes sense (probability must be between $$0$$ and $$1$$)
- **Ensure interpretation** of the word "or" in the problem corresponds to the union operation
Visible text: - **Visualize** events using diagrams or tables when possible
- **Double-check** whether the final result makes sense (probability must be between and )
- **Ensure interpretation** of the word "or" in the problem corresponds to the union operation
### Beware of Events That Appear Mutually Exclusive
Many students incorrectly identify mutually exclusive events. Here are examples of events that **appear** mutually exclusive but **actually are not**:
**Example $$1$$: Dice Rolling**
Visible text: **Example : Dice Rolling**
- Event $$A$$: Getting a prime number is $$\{2, 3, 5\}$$
- Event $$B$$: Getting an odd number is $$\{1, 3, 5\}$$
Visible text: - Event : Getting a prime number is
- Event : Getting an odd number is
**Common mistake**: "Prime and odd are different, so they are mutually exclusive"
**Reality**: $$A \cap B = \{3, 5\} \ne \emptyset$$, so **not mutually exclusive**
Visible text: **Common mistake**: "Prime and odd are different, so they are mutually exclusive"
**Reality**: , so **not mutually exclusive**
**Example $$2$$: Card Drawing**
Visible text: **Example : Card Drawing**
- Event $$A$$: Drawing a red card
- Event $$B$$: Drawing an Ace
Visible text: - Event : Drawing a red card
- Event : Drawing an Ace
**Common mistake**: "Color and card type are different, so they are mutually exclusive"
**Reality**: There are red Aces (Ace of hearts and Ace of diamonds), so **not mutually exclusive**
**Example $$3$$: Student Characteristics**
Visible text: **Example : Student Characteristics**
- Event $$A$$: Students who are tall ($$>160 \text{ cm}$$)
- Event $$B$$: Students who are smart ($$\text{score} > 80$$)
Visible text: - Event : Students who are tall ()
- Event : Students who are smart ()
**Common mistake**: "Height and intelligence are unrelated"
**Reality**: There can be students who are both tall and smart, so **not mutually exclusive**
**Identification Strategy:**
1. **Ask**: "Can one element satisfy both criteria simultaneously?"
2. **Find intersection**: Identify elements that belong to both events
3. **If there is intersection**: Events are **not mutually exclusive**
4. **If there is no intersection**: Events are **mutually exclusive**
## Exercises
1. A die is rolled once. Determine the probability of getting a number less than $$3$$ or a number greater than $$5$$.
2. From a standard deck of bridge cards, one card is drawn randomly. Calculate the probability of drawing an Ace or a King.
3. Two coins are flipped simultaneously. Determine the probability of getting exactly $$1$$ tail or exactly $$2$$ heads.
4. In a box there are $$10 \text{ balls}$$ numbered $$1$$ to $$10$$. A ball is drawn randomly. Calculate the probability of drawing an even-numbered ball or an odd prime-numbered ball.
Visible text: 1. A die is rolled once. Determine the probability of getting a number less than or a number greater than .
2. From a standard deck of bridge cards, one card is drawn randomly. Calculate the probability of drawing an Ace or a King.
3. Two coins are flipped simultaneously. Determine the probability of getting exactly tail or exactly heads.
4. In a box there are numbered to . A ball is drawn randomly. Calculate the probability of drawing an even-numbered ball or an odd prime-numbered ball.
### Answer Key
1. **Solution:**
Sample space: $$S = \{1, 2, 3, 4, 5, 6\}$$, so $$n(S) = 6$$
- Event $$A$$ ($$\text{number} < 3$$): $$A = \{1, 2\}$$, so $$n(A) = 2$$
- Event $$B$$ ($$\text{number} > 5$$): $$B = \{6\}$$, so $$n(B) = 1$$
Check intersection: $$A \cap B = \emptyset$$ (no number simultaneously satisfies $$x < 3$$ and $$x > 5$$)
Since they are mutually exclusive, use the formula:
```math
P(A) = \frac{2}{6} = \frac{1}{3}
```
```math
P(B) = \frac{1}{6}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
```
2. **Solution:**
Total cards is $$52$$
- Event $$A$$ (Ace card): $$4$$ Ace cards
- Event $$B$$ (King card): $$4$$ King cards
Check intersection: No card is simultaneously an Ace and a King, so $$A \cap B = \emptyset$$
Both events are mutually exclusive.
```math
P(A) = \frac{4}{52} = \frac{1}{13}
```
```math
P(B) = \frac{4}{52} = \frac{1}{13}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}
```
3. **Solution:**
Sample space for flipping two coins: $$\{(H,H), (H,T), (T,H), (T,T)\}$$, total is $$4$$
- Event $$A$$ (exactly one tail): $$\{(H,T), (T,H)\}$$, so $$n(A) = 2$$
- Event $$B$$ (exactly two heads): $$\{(H,H)\}$$, so $$n(B) = 1$$
Check intersection: $$A \cap B = \emptyset$$ (impossible to have exactly one tail and exactly two heads simultaneously)
```math
P(A) = \frac{2}{4} = \frac{1}{2}
```
```math
P(B) = \frac{1}{4}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}
```
4. **Solution:**
Odd prime is a prime number that is also odd. Prime numbers from $$1\text{-}10$$ are $$\{2, 3, 5, 7\}$$, so odd primes are $$\{3, 5, 7\}$$.
Sample space: $$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$, so $$n(S) = 10$$
- Event $$A$$ (even number): $$A = \{2, 4, 6, 8, 10\}$$, so $$n(A) = 5$$
- Event $$B$$ (odd prime number): $$B = \{3, 5, 7\}$$, so $$n(B) = 3$$
Check intersection: $$A \cap B = \emptyset$$ (no number is simultaneously even and odd prime)
```math
P(A) = \frac{5}{10} = \frac{1}{2}
```
```math
P(B) = \frac{3}{10}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{2} + \frac{3}{10}
```
**Solution for fraction addition:**
To add $$\frac{1}{2} + \frac{3}{10}$$, we need to equalize the denominators.
$$LCM(2, 10) = 10$$, so:
```math
\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}
```
```math
P(A \cup B) = \frac{5}{10} + \frac{3}{10} = \frac{8}{10} = \frac{4}{5}
```
Visible text: 1. **Solution:**
Sample space: , so
- Event (): , so
- Event (): , so
Check intersection: (no number simultaneously satisfies and )
Since they are mutually exclusive, use the formula:
2. **Solution:**
Total cards is
- Event (Ace card): Ace cards
- Event (King card): King cards
Check intersection: No card is simultaneously an Ace and a King, so
Both events are mutually exclusive.
3. **Solution:**
Sample space for flipping two coins: , total is
- Event (exactly one tail): , so
- Event (exactly two heads): , so
Check intersection: (impossible to have exactly one tail and exactly two heads simultaneously)
4. **Solution:**
Odd prime is a prime number that is also odd. Prime numbers from are , so odd primes are .
Sample space: , so
- Event (even number): , so
- Event (odd prime number): , so
Check intersection: (no number is simultaneously even and odd prime)
**Solution for fraction addition:**
To add , we need to equalize the denominators.
, so: