Probability of Compound Events

Combining a Die Result with a Coin Result

Many probability questions combine two or more events. A die and a coin may be tossed together, or two cards may be drawn from one deck. These are compound-event situations.

A compound event combines two or more simple events within one probability model. For example, we can study one condition that includes both a die result and a coin result.

Suppose you toss a die and a coin. A simple event might focus only on the die or only on the coin. A compound event considers both, such as “the die shows $3$ AND the coin shows tails.”

Types of Compound Events

Two events either share outcomes or exclude each other, and that difference decides which formula applies. Checking one outcome that belongs to both events is usually enough to tell the two cases apart before you calculate anything.

Mutually Exclusive Events

Two events are mutually exclusive when they cannot occur in the same trial. If event $A$ occurs, event $B$ cannot occur, and vice versa.

A standard example is one die roll. “The result is even” and “the result is odd” are mutually exclusive because one result cannot be both.

For mutually exclusive events, the probability formula is:

Calculation Example:

A die is rolled once. Find the probability of rolling an even number or rolling $1$ or $3$.

Solution:

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

• Event $A$ (even numbers): $A = \{2, 4, 6\}$
• Event $B$ (numbers $1$ or $3$): $B = \{1, 3\}$

Check intersection: $A \cap B = \emptyset$ (empty set)

Since there is no intersection, both events are mutually exclusive.

Not Mutually Exclusive Events

Events are not mutually exclusive when they can occur in the same trial. Their intersection is not empty.

For example, “draw a red card” and “draw an Ace” are not mutually exclusive. The Ace of Hearts and Ace of Diamonds satisfy both conditions.

For non-mutually exclusive events, the formula is:

We subtract $P(A \cap B)$ because the outcomes in the intersection were counted once in $P(A)$ and again in $P(B)$.

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 $= 52$

• Event $A$ (red cards): $26 \text{ cards}$ ($13 \text{ Hearts} + 13 \text{ Diamonds}$)
• Event $B$ (face cards): $12 \text{ cards}$ ($4 \text{ Jacks} + 4 \text{ Queens} + 4 \text{ Kings}$)

Intersection (red and face cards): $6 \text{ cards}$ (Jack, Queen, King from Hearts and Diamonds)

Independent Events

Two events are independent when learning that one occurred does not change the probability distribution of the other. Check this through probabilities. Events can happen at the same time and still influence each other.

A simple example is tossing two fair coins. Knowing the first coin landed heads does not change the second coin: its probability of tails remains $50\%$.

Formula for independent events:

Calculation Example:

Two fair dice are rolled. Find the probability that the first die shows $3$ and the second die shows an even number.

Solution:

• Event $A$ (number $3$ on the first die): $1 \text{ possibility}$ out of $6$
• Event $B$ (even number on the second die): $\{2, 4, 6\} = 3 \text{ possibilities}$ out of $6$

Since the result of the first die does not affect the second die, both events are independent.

Dependent Events

Events are dependent when the outcome of one changes the probability of the next. The conditions for the second stage are no longer the same as they were at the start.

One common cause is drawing without replacement. The selected object is not returned, so both the total number of objects and possibly the number of favorable objects change.

Formula for dependent events:

In this formula, $P(B|A)$ is the probability of event $B$ occurring given that event $A$ 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.

Solution:

Initial total balls is $8 + 4 = 12 \text{ balls}$

• 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 $11 \text{ total balls}$

Since drawing is without replacement, conditions change after the first draw, so this is a dependent event.

Union and Intersection Calculations

Union and intersection answer different questions about the same two events. For a union, an outcome in either event counts, while an intersection keeps only outcomes that lie in both.

Union Operation

To find the probability of “event $A$ OR event $B$,” use the union. Here “or” is inclusive: at least one event occurs, and an overlap belongs to the union too.

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

To find the probability of “event $A$ AND event $B$,” use the intersection. Every counted outcome must satisfy both conditions.

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

Start a compound-event problem by identifying the relationship between the events. Keywords help, but the probability mechanism is the final test:

1. "or" indicates union operation
2. "and" indicates intersection operation
3. "without replacement" indicates dependent events
4. "with replacement" often preserves independence, but verify that one outcome truly does not change the other event's probability

Then check whether both events can occur together. Choose the formula only after those two questions are settled, and verify that no overlap is counted twice.

Exercises

Try each problem before opening its worked solution. Name the events, decide whether the problem asks for a union or intersection, and justify whether the events are mutually exclusive, independent, or dependent.

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.

Worked Solutions

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:

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

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

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

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

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

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.

   $$P(A) = \frac{6}{10} = \frac{3}{5}$$

   $$P(B|A) = \frac{5}{9}$$

   $$P(A \cap B) = P(A) \times P(B|A) = \frac{3}{5} \times \frac{5}{9} = \frac{15}{45} = \frac{1}{3}$$

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 $11$.

   $$P(A) = \frac{6}{36} = \frac{1}{6}$$

   $$P(B) = \frac{2}{36} = \frac{1}{18}$$

   $$P(A \cup B) = P(A) + P(B) = \frac{1}{6} + \frac{1}{18} = \frac{3}{18} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}$$

4. Solution:

   Total cards $= 52$

   • Event $A$ (Ace cards): $4 \text{ Aces}$
   • Event $B$ (black cards): $26 \text{ cards}$ (Spades and Clubs)

   Intersection (black Aces): Ace of Spades and Ace of Clubs is $2 \text{ cards}$

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

   $$P(B) = \frac{26}{52} = \frac{1}{2}$$

   $$P(A \cap B) = \frac{2}{52} = \frac{1}{26}$$

   $$P(A \cup B) = \frac{1}{13} + \frac{1}{2} - \frac{1}{26} = \frac{2}{26} + \frac{13}{26} - \frac{1}{26} = \frac{14}{26} = \frac{7}{13}$$