Probability of Mutually Exclusive Events

Two Events Cannot Occur in the Same Trial

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.

How Mutually Exclusive Events Behave

Mutually exclusive events share no outcome, so their overlap contributes nothing to the total. The three checks below describe that idea in terms of outcomes and probabilities.

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:

The symbol $\emptyset$ indicates an empty set, meaning there are no common elements between the two events.

Outcomes That Cannot Occur Together

The following pairs cannot occur in the same trial:

• 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

The general addition rule subtracts the overlap $P(A \cap B)$. Mutually exclusive events have $P(A \cap B)=0$, so that term disappears:

This formula shows that the probability of event $A$ or event $B$ occurring equals the sum of the individual probabilities of each event.

Mutually exclusive does not mean independent. If $P(A) > 0$ and $P(B) > 0$, learning that $A$ occurred makes $B$ impossible. Thus $P(A \cap B) = 0$, while independent events would require $P(A \cap B) = P(A)P(B) > 0$.

Zero Intersection in the Addition Rule

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:

Becomes:

Calculations with the Addition Rule

Each calculation below adds the probabilities of two events that cannot occur together. The examples start with a single die and then extend the addition rule to two dice.

Dice Rolling

A die is rolled once. Determine the probability of getting a prime number or a number greater than $4$.

Solution:

Sample space: $S = \{1, 2, 3, 4, 5, 6\}$

• Event $A$ (prime number): $A = \{2, 3, 5\}$
• Event $B$ ($\text{number} > 4$): $B = \{5, 6\}$

Check intersection: $A \cap B = \{5\}$

Since the number $5$ lies in both events, they are not mutually exclusive. The original question still has a valid answer, but it requires subtracting the overlap:

The next example shows when the shorter mutually exclusive rule applies.

Contrasting Mutually Exclusive Example:

• Event $A$ (odd number): $A = \{1, 3, 5\}$
• Event $C$ (even number): $C = \{2, 4, 6\}$

Check intersection: $A \cap C = \emptyset$

Since there is no intersection, the two events are mutually exclusive.

The odd and even outcomes together cover all six faces of the die, so their union has probability $1$.

Rolling Two Dice

Two dice are rolled simultaneously. Determine the probability of getting a sum of $5$ or a sum of $7$.

Solution:

The table lists all $36$ ordered outcomes of the two dice:

Total possibilities is $6 \times 6 = 36$

Event identification:

Event $A$ ($\text{sum} = 5$):

From the table above, pairs that produce sum $5$ are:

• $(1,4): 1 + 4 = 5$
• $(2,3): 2 + 3 = 5$
• $(3,2): 3 + 2 = 5$
• $(4,1): 4 + 1 = 5$

So there are $4 \text{ ways}$ to get sum $5$.

Event $B$ ($\text{sum} = 7$):

From the table above, pairs that produce sum $7$ are:

• $(1,6): 1 + 6 = 7$
• $(2,5): 2 + 5 = 7$
• $(3,4): 3 + 4 = 7$
• $(4,3): 4 + 3 = 7$
• $(5,2): 5 + 2 = 7$
• $(6,1): 6 + 1 = 7$

So there are $6 \text{ ways}$ to get sum $7$.

Check intersection: It's impossible for the sum to be both $5$ and $7$, so $A \cap B = \emptyset$

Both events are mutually exclusive.

Solution for fraction addition:

To add $\frac{1}{9} + \frac{1}{6}$, we need the least common multiple of $9$ and $6$.

$\operatorname{lcm}(9, 6) = 18$, so:

How to Work Through a Mutually Exclusive Problem

Before adding two probabilities, check whether the events share an outcome. The steps below put that test first and confirm the result afterwards. Adding without that check counts any shared outcome twice.

Check the Intersection Before Adding

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 $P(A \cup B) = P(A) + P(B)$

Checks for the Final Answer

Check three points before accepting the answer:

• Draw the event sets in a diagram or table when their intersection is not yet clear.
• Verify that the final probability lies between $0$ and $1$.
• Read "or" as union unless the problem explicitly states otherwise.

Events with a Nonempty Intersection

Two events are not mutually exclusive when their intersection contains at least one outcome. The following examples have a nonempty intersection:

Example 1: Dice Rolling

• Event $A$: Getting a prime number is $\{2, 3, 5\}$
• Event $B$: Getting an odd number is $\{1, 3, 5\}$

Common mistake: "Prime and odd are different, so they are mutually exclusive" Reality: $A \cap B = \{3, 5\} \ne \emptyset$, so not mutually exclusive

Example 2: Card Drawing

• Event $A$: Drawing a red card
• Event $B$: 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: Two Student Criteria

• Event $A$: Students who are tall ($>160 \text{ cm}$)
• Event $B$: Students whose score is above $80$

Common mistake: “Height and test score are different traits, so the events are mutually exclusive” Reality: A student can satisfy both criteria, so the events are 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

Try each exercise before reading its worked solution. Write both sets explicitly and check their intersection before choosing the addition rule.

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.

Worked Solutions

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:

   $$P(A) = \frac{2}{6} = \frac{1}{3}$$

   $$P(B) = \frac{1}{6}$$

   $$\begin{aligned} P(A \cup B) &= P(A) + P(B) \\ &= \frac{1}{3} + \frac{1}{6} \\ &= \frac{2}{6} + \frac{1}{6} \\ &= \frac{3}{6} \\ &= \frac{1}{2} \end{aligned}$$

2. Solution:

   Total cards is $52$

   • Event $A$ (Ace card): $4 \text{ Aces}$
   • Event $B$ (King card): $4 \text{ Kings}$

   Check intersection: No card is simultaneously an Ace and a King, so $A \cap B = \emptyset$

   Both events are mutually exclusive.

   $$P(A) = \frac{4}{52} = \frac{1}{13}$$

   $$P(B) = \frac{4}{52} = \frac{1}{13}$$

   $$\begin{aligned} P(A \cup B) &= P(A) + P(B) \\ &= \frac{1}{13} + \frac{1}{13} \\ &= \frac{2}{13} \end{aligned}$$

3. Solution:

   Use $0$ for heads and $1$ for tails. The sample space is $\{(0,0), (0,1), (1,0), (1,1)\}$, with $4$ outcomes.

   • Event $A$ (exactly one tail): $\{(0,1), (1,0)\}$, so $n(A) = 2$
   • Event $B$ (exactly two heads): $\{(0,0)\}$, so $n(B) = 1$

   Check intersection: $A \cap B = \emptyset$ (impossible to have exactly one tail and exactly two heads simultaneously)

   $$P(A) = \frac{2}{4} = \frac{1}{2}$$

   $$P(B) = \frac{1}{4}$$

   $$\begin{aligned} P(A \cup B) &= P(A) + P(B) \\ &= \frac{1}{2} + \frac{1}{4} \\ &= \frac{2}{4} + \frac{1}{4} \\ &= \frac{3}{4} \end{aligned}$$

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)

   $$P(A) = \frac{5}{10} = \frac{1}{2}$$

   $$P(B) = \frac{3}{10}$$

   $$\begin{aligned} P(A \cup B) &= P(A) + P(B) \\ &= \frac{1}{2} + \frac{3}{10} \end{aligned}$$

   Solution for fraction addition:

   To add $\frac{1}{2} + \frac{3}{10}$, we need to equalize the denominators.

   $\operatorname{lcm}(2, 10) = 10$, so:

   $$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$$

   $$P(A \cup B) = \frac{5}{10} + \frac{3}{10} = \frac{8}{10} = \frac{4}{5}$$