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URL: https://nakafa.com/en/subjects/mathematics/combinatorics/probability-of-compound-events
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Learn compound event probability through independent and dependent events, formulas, and real-world examples.
---
## Understanding Compound Events
In daily life, we often face situations where **two or more events occur simultaneously**. For example, when throwing a dice and a coin at the same time, or drawing two cards from one deck. Situations like this are called compound events.
**Compound events** are combinations of two or more single events that can occur in one experiment or several experiments conducted simultaneously. Unlike single events that only involve one outcome, compound events involve combinations of several outcomes at once.
As a simple illustration, imagine you throw a dice and a coin simultaneously. A single event would only focus on the dice result alone or the coin alone. However, compound events would consider the combination of both results, such as "getting number $$3$$ on the dice AND getting tails on the coin".
Visible text: As a simple illustration, imagine you throw a dice and a coin simultaneously. A single event would only focus on the dice result alone or the coin alone. However, compound events would consider the combination of both results, such as "getting number on the dice AND getting tails on the coin".
## Types of Compound Events
### Mutually Exclusive Events
Two events are said to be **mutually exclusive** when both events cannot occur simultaneously in one experiment. In other words, if event $$A$$ occurs, then event $$B$$ definitely cannot occur, and vice versa.
Visible text: Two events are said to be **mutually exclusive** when both events cannot occur simultaneously in one experiment. In other words, if event occurs, then event definitely cannot occur, and vice versa.
A classic example is throwing one dice. The event of getting an even number and the event of getting an odd number are mutually exclusive, because in one throw it's impossible to get a number that is both even and odd.
For mutually exclusive events, the probability formula is:
```math
P(A \cup B) = P(A) + P(B)
```
**Calculation Example:**
A dice is thrown once. Determine the probability of getting an even number or numbers $$1$$ or $$3$$.
Visible text: A dice is thrown once. Determine the probability of getting an even number or numbers or .
**Solution:**
Sample space: $$S = \{1, 2, 3, 4, 5, 6\}$$
Visible text: Sample space:
- Event $$A$$ (even numbers): $$A = \{2, 4, 6\}$$
- Event $$B$$ (numbers $$1$$ or $$3$$): $$B = \{1, 3\}$$
Visible text: - Event (even numbers):
- Event (numbers or ):
Check intersection: $$A \cap B = \emptyset$$ (empty set)
Visible text: Check intersection: (empty set)
Since there is no intersection, both events are **mutually exclusive**.
Component: MathContainer
Children:
```math
P(A) = \frac{3}{6} = \frac{1}{2}
```
```math
P(B) = \frac{2}{6} = \frac{1}{3}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}
```
### Non-Mutually Exclusive Events
**Non-mutually exclusive events** occur when two events can happen simultaneously in one experiment. This means there is a possibility that both events occur at the same time.
For example, in drawing one card from a standard deck, the event "drawing a red card" and the event "drawing an Ace" are not mutually exclusive. This is because there are cards that are both red and Ace, namely Ace of Hearts and Ace of Diamonds.
For non-mutually exclusive events, the formula is:
```math
P(A \cup B) = P(A) + P(B) - P(A \cap B)
```
The subtraction of $$P(A \cap B)$$ is needed to avoid double counting the intersection of both events.
Visible text: The subtraction of is needed to avoid double counting the intersection of both events.
**Calculation Example:**
From a standard bridge card deck, one card is drawn randomly. Determine the probability of drawing a red card or a face card (Jack, Queen, King).
**Solution:**
Total cards $$= 52$$
Visible text: Total cards
- Event $$A$$ (red cards): $$26 \text{ cards}$$ ($$13 \text{ Hearts} + 13 \text{ Diamonds}$$)
- Event $$B$$ (face cards): $$12 \text{ cards}$$ ($$4 \text{ Jacks} + 4 \text{ Queens} + 4 \text{ Kings}$$)
Visible text: - Event (red cards): ()
- Event (face cards): ()
Intersection (red and face cards): $$6 \text{ cards}$$ (Jack, Queen, King from Hearts and Diamonds)
Visible text: Intersection (red and face cards): (Jack, Queen, King from Hearts and Diamonds)
Component: MathContainer
Children:
```math
P(A) = \frac{26}{52} = \frac{1}{2}
```
```math
P(B) = \frac{12}{52} = \frac{3}{13}
```
```math
P(A \cap B) = \frac{6}{52} = \frac{3}{26}
```
```math
P(A \cup B) = \frac{1}{2} + \frac{3}{13} - \frac{3}{26} = \frac{13}{26} + \frac{6}{26} - \frac{3}{26} = \frac{16}{26} = \frac{8}{13}
```
### Independent Events
Two events are called **independent** when the occurrence of one event does not affect the probability of the other event occurring. The result of the first event does not change the conditions for the second event.
An easy-to-understand example is throwing two coins simultaneously. The result of the first coin toss (for example, heads) will not affect the result of the second coin toss. The probability of getting tails on the second coin remains $$50\%$$ regardless of the first coin's result.
Visible text: An easy-to-understand example is throwing two coins simultaneously. The result of the first coin toss (for example, heads) will not affect the result of the second coin toss. The probability of getting tails on the second coin remains regardless of the first coin's result.
Formula for independent events:
```math
P(A \cap B) = P(A) \times P(B)
```
**Calculation Example:**
Two dice are thrown simultaneously. Determine the probability of getting number $$3$$ on the first dice and an even number on the second dice.
Visible text: Two dice are thrown simultaneously. Determine the probability of getting number on the first dice and an even number on the second dice.
**Solution:**
- Event $$A$$ (number $$3$$ on first dice): $$1 \text{ possibility}$$ out of $$6$$
- Event $$B$$ (even number on second dice): $$\{2, 4, 6\} = 3 \text{ possibilities}$$ out of $$6$$
Visible text: - Event (number on first dice): out of
- Event (even number on second dice): out of
Since the result of the first dice does not affect the second dice, both events are independent.
Component: MathContainer
Children:
```math
P(A) = \frac{1}{6}
```
```math
P(B) = \frac{3}{6} = \frac{1}{2}
```
```math
P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}
```
### Dependent Events
**Dependent events** or conditional events occur when the result of the first event affects the probability of the second event occurring. The conditions after the first event change and affect subsequent calculations.
**Main cause:** Drawing **without replacement** where objects that have been drawn are not returned to their original place, so the total number of objects decreases and changes the probability of subsequent draws.
Formula for dependent events:
```math
P(A \cap B) = P(A) \times P(B|A)
```
where $$P(B|A)$$ is the probability of event $$B$$ occurring given that event $$A$$ has already occurred.
Visible text: where is the probability of event occurring given that event has already occurred.
**Calculation Example:**
In a box there are $$8 \text{ blue balls}$$ and $$4 \text{ yellow balls}$$. Two balls are drawn randomly **without replacement**. Determine the probability that both balls drawn are blue.
Visible text: In a box there are and . Two balls are drawn randomly **without replacement**. Determine the probability that both balls drawn are blue.
**Solution:**
Initial total balls is $$8 + 4 = 12 \text{ balls}$$
Visible text: Initial total balls is
- Event $$A$$ (first ball blue): $$8 \text{ blue balls}$$ out of $$12 \text{ total balls}$$
- Event $$B$$ (second ball blue after $$A$$): Since $$1 \text{ blue ball}$$ has been drawn, $$7 \text{ blue balls}$$ remain out of $$11 \text{ total balls}$$
Visible text: - Event (first ball blue): out of
- Event (second ball blue after ): Since has been drawn, remain out of
Since drawing is without replacement, conditions change after the first draw, so this is a dependent event.
Component: MathContainer
Children:
```math
P(A) = \frac{8}{12} = \frac{2}{3}
```
```math
P(B|A) = \frac{7}{11}
```
```math
P(A \cap B) = P(A) \times P(B|A) = \frac{2}{3} \times \frac{7}{11} = \frac{14}{33}
```
## Application in Calculations
### Union Operation
When we want to know the probability of "event $$A$$ **OR** event $$B$$", we use the union operation. The keyword "or" indicates that we are looking for the probability where at least one event occurs.
Visible text: When we want to know the probability of "event **OR** event ", we use the union operation. The keyword "or" indicates that we are looking for the probability where at least one event occurs.
In dice throwing, if we want to find the probability of getting an odd number or a prime number, we need to consider whether both events are mutually exclusive or not.
### Intersection Operation
Conversely, when looking for the probability of "event $$A$$ **AND** event $$B$$", we use the intersection operation. The keyword "and" indicates that both events must occur simultaneously.
Visible text: Conversely, when looking for the probability of "event **AND** event ", we use the intersection operation. The keyword "and" indicates that both events must occur simultaneously.
In the context of card drawing, if we look for the probability of "drawing a red card AND an even-numbered card", we must count cards that meet both criteria.
## Problem-Solving Strategy
The first step in solving compound event probability problems is **identifying the types of events** involved. Pay attention to keywords in the problem:
1. **"or"** indicates union operation
2. **"and"** indicates intersection operation
3. **"without replacement"** indicates dependent events
4. **"with replacement"** or "simultaneously" indicates independent events
Next, determine whether the events are mutually exclusive or not by checking if there is a possibility that both events occur simultaneously. Finally, choose the appropriate formula and perform calculations carefully.
## Exercises
1. In throwing a dice, determine the probability of getting a prime number or an odd number on the dice.
2. A box contains $$6 \text{ red balls}$$ and $$4 \text{ white balls}$$. Two balls are drawn randomly without replacement. Calculate the probability that both balls drawn are red.
3. Two dice are thrown simultaneously. Determine the probability that the sum of the dice is $$7$$ or $$11$$.
4. From a standard bridge card deck, one card is drawn randomly. Calculate the probability of drawing an Ace or a black card.
Visible text: 1. In throwing a dice, determine the probability of getting a prime number or an odd number on the dice.
2. A box contains and . Two balls are drawn randomly without replacement. Calculate the probability that both balls drawn are red.
3. Two dice are thrown simultaneously. Determine the probability that the sum of the dice is or .
4. From a standard bridge card deck, one card is drawn randomly. Calculate the probability of drawing an Ace or a black card.
### Answer Key
1. **Solution:**
Sample space: $$S = \{1, 2, 3, 4, 5, 6\}$$, so $$n(S) = 6$$
- Event $$A$$ (prime numbers): $$A = \{2, 3, 5\}$$, so $$n(A) = 3$$
- Event $$B$$ (odd numbers): $$B = \{1, 3, 5\}$$, so $$n(B) = 3$$
Intersection $$A$$ and $$B$$: $$A \cap B = \{3, 5\}$$, so $$n(A \cap B) = 2$$
Since there is an intersection, the events are not mutually exclusive. Using the formula:
```math
P(A \cup B) = P(A) + P(B) - P(A \cap B)
```
```math
P(A) = \frac{3}{6} = \frac{1}{2}
```
```math
P(B) = \frac{3}{6} = \frac{1}{2}
```
```math
P(A \cap B) = \frac{2}{6} = \frac{1}{3}
```
```math
P(A \cup B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}
```
2. **Solution:**
Total balls is $$6 + 4 = 10 \text{ balls}$$
- Event $$A$$ (first ball red): $$6 \text{ red balls}$$ out of $$10 \text{ total balls}$$
- Event $$B$$ (second ball red after $$A$$): After $$1 \text{ red ball}$$ is drawn, $$5 \text{ red balls}$$ remain out of $$9 \text{ total balls}$$
Since drawing is **without replacement**, conditions change after the first draw, so this is a dependent event.
```math
P(A) = \frac{6}{10} = \frac{3}{5}
```
```math
P(B|A) = \frac{5}{9}
```
```math
P(A \cap B) = P(A) \times P(B|A) = \frac{3}{5} \times \frac{5}{9} = \frac{15}{45} = \frac{1}{3}
```
3. **Solution:**
Total possibilities is $$6 \times 6 = 36$$
- Event $$A$$ (sum is $$7$$): $$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$$ → $$6 \text{ ways}$$
- Event $$B$$ (sum is $$11$$): $$(5,6), (6,5)$$ → $$2 \text{ ways}$$
Both events are mutually exclusive because it's impossible for the dice sum to be both $$7$$ and $$11$$.
```math
P(A) = \frac{6}{36} = \frac{1}{6}
```
```math
P(B) = \frac{2}{36} = \frac{1}{18}
```
```math
P(A \cup B) = P(A) + P(B) = \frac{1}{6} + \frac{1}{18} = \frac{3}{18} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}
```
4. **Solution:**
Total cards $$= 52$$
- Event $$A$$ (Ace cards): $$4$$ Ace cards
- Event $$B$$ (black cards): $$26 \text{ cards}$$ (Spades and Clubs)
Intersection (black Aces): Ace of Spades and Ace of Clubs is $$2 \text{ cards}$$
```math
P(A) = \frac{4}{52} = \frac{1}{13}
```
```math
P(B) = \frac{26}{52} = \frac{1}{2}
```
```math
P(A \cap B) = \frac{2}{52} = \frac{1}{26}
```
```math
P(A \cup B) = \frac{1}{13} + \frac{1}{2} - \frac{1}{26} = \frac{2}{26} + \frac{13}{26} - \frac{1}{26} = \frac{14}{26} = \frac{7}{13}
```
Visible text: 1. **Solution:**
Sample space: , so
- Event (prime numbers): , so
- Event (odd numbers): , so
Intersection and : , so
Since there is an intersection, the events are not mutually exclusive. Using the formula:
2. **Solution:**
Total balls is
- Event (first ball red): out of
- Event (second ball red after ): After is drawn, remain out of
Since drawing is **without replacement**, conditions change after the first draw, so this is a dependent event.
3. **Solution:**
Total possibilities is
- Event (sum is ): →
- Event (sum is ): →
Both events are mutually exclusive because it's impossible for the dice sum to be both and .
4. **Solution:**
Total cards
- Event (Ace cards): Ace cards
- Event (black cards): (Spades and Clubs)
Intersection (black Aces): Ace of Spades and Ace of Clubs is