Probability of Mutually Exclusive Events

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:

The symbol $\emptyset$ 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:

This formula shows that the probability of event A or event B 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:

Becomes:

Application in Calculations

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 (number > 4): B = $\{5, 6\}$

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

Since there is an intersection (number 5), 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 = $\{2, 4, 6\}$

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

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

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.

Solution:

To understand more clearly, let's look at all possible outcomes of rolling two dice in the following table:

Total possibilities = $6 \times 6 = 36$

Event identification:

Event A (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 ways to get sum 5.

Event B (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 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 to find the LCM of 9 and 6.

$LCM(9, 6) = 18$, so:

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

Practical Tips

Several tips to facilitate understanding:

• 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

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

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

Common mistake: "Prime and odd are different, so they are mutually exclusive" Reality: $A \cap B = \{3, 5\}$ ≠ ∅, 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: Student Characteristics

• Event A: Students who are tall (> 160 cm)
• Event B: Students who are smart (score > 80)

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 one tail or exactly two heads.

4. In a box there are 10 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.

Answer Key

1. Solution:

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

   • Event A (number < 3): A = $\{1, 2\}$, so $n(A) = 2$
   • Event B (number > 5): B = $\{6\}$, so $n(B) = 1$

   Check intersection: $A \cap B = \emptyset$ (no number is simultaneously < 3 and > 5)

   Since they are mutually exclusive, use the formula:

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

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

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

2. Solution:

   Total cards = $52$

   • 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.

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

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

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

3. Solution:

   Sample space for flipping two coins: $\{(H,H), (H,T), (T,H), (T,T)\}$, total = $4$

   • 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)

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

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

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

4. Solution:

   Odd prime is a prime number that is also odd. Prime numbers from 1-10 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)

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

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

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

   Solution for fraction addition:

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

   $LCM(2, 10) = 10$, so:

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

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