Compound Events and Independence

A compound event involves two or more individual events happening together or in sequence. When you need to find the probability that event A and event B both occur, you use the multiplication rule. The key question is whether the events are independent or dependent — this determines which version of the rule to apply.

Independent vs Dependent Events

Two events are independent if the outcome of one event has no effect on the probability of the other. Mathematically, events A and B are independent when:

$P(B \mid A) = P(B)$

This means knowing that A happened does not change the likelihood of B.

Two events are dependent if the outcome of the first event changes the probability of the second event. When events are dependent:

$P(B \mid A) \neq P(B)$

Here are some examples to build your intuition:

Multiplication Rule for Independent Events

When events A and B are independent, the probability that both occur is:

$P(A \text{ and } B) = P(A) \cdot P(B)$

This extends to any number of independent events:

$P(A \text{ and } B \text{ and } C) = P(A) \cdot P(B) \cdot P(C)$

Example 1: Two Coin Flips

What is the probability of getting heads on both flips of a fair coin?

Each flip has $P(\text{heads}) = \frac{1}{2}$, and the flips are independent.

$P(\text{heads and heads}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$

Answer: The probability of two heads in a row is $\frac{1}{4}$, or 25%.

Example 2: Rolling Two Dice

What is the probability of rolling a 6 on the first die AND an even number on the second die?

The two dice are independent. There is 1 way to roll a 6 out of 6 outcomes, and there are 3 even numbers (2, 4, 6) out of 6 outcomes.

$P(6 \text{ on first}) = \frac{1}{6}$

$P(\text{even on second}) = \frac{3}{6} = \frac{1}{2}$

$P(6 \text{ and even}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \approx 0.0833$

Answer: The probability is $\frac{1}{12}$, or about 8.3%.

Multiplication Rule for Dependent Events

When events are dependent, the probability that both occur uses conditional probability:

$P(A \text{ and } B) = P(A) \cdot P(B \mid A)$

Here $P(B \mid A)$ is the probability of B occurring given that A has already occurred. The first event changes the conditions for the second event.

Example 3: Drawing Cards Without Replacement

What is the probability of drawing two aces in a row from a standard 52-card deck, without replacing the first card?

Step 1: Find the probability of drawing an ace on the first draw.

$P(\text{first ace}) = \frac{4}{52} = \frac{1}{13}$

Step 2: Given that the first card was an ace, find the probability of drawing another ace. Now only 3 aces remain out of 51 total cards.

$P(\text{second ace} \mid \text{first ace}) = \frac{3}{51} = \frac{1}{17}$

Step 3: Multiply.

$P(\text{two aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.00452$

Answer: The probability of drawing two aces in a row without replacement is $\frac{1}{221}$, or about 0.45%.

With Replacement vs Without Replacement

Many probability problems involve drawing items from a collection. The crucial distinction is whether you put the item back before the next draw.

Example 4: Colored Marbles

A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles (10 total). You draw one marble, note its color, then draw a second marble. Find the probability of drawing a red marble first and a blue marble second.

With replacement (put the first marble back):

The draws are independent. The total stays at 10 for both draws.

$P(\text{red then blue}) = \frac{5}{10} \times \frac{3}{10} = \frac{15}{100} = \frac{3}{20} = 0.15$

Without replacement (keep the first marble out):

The draws are dependent. After removing a red marble, 9 marbles remain.

$P(\text{red then blue}) = \frac{5}{10} \times \frac{3}{9} = \frac{15}{90} = \frac{1}{6} \approx 0.1667$

Answer: With replacement, the probability is $\frac{3}{20} = 0.15$. Without replacement, the probability is $\frac{1}{6} \approx 0.167$.

Notice that the without-replacement probability is slightly higher. This makes sense: after removing a red marble, the remaining pool has a slightly higher proportion of blue marbles (3 out of 9 instead of 3 out of 10).

Testing for Independence

To check whether two events are independent, verify whether this equation holds:

$P(A \text{ and } B) = P(A) \cdot P(B)$

If the equation is true, the events are independent. If not, they are dependent.

Example: A survey of 200 people finds that 80 exercise regularly, 50 are vegetarian, and 30 both exercise regularly and are vegetarian.

$P(\text{exercise}) = \frac{80}{200} = 0.40$

$P(\text{vegetarian}) = \frac{50}{200} = 0.25$

$P(\text{exercise}) \times P(\text{vegetarian}) = 0.40 \times 0.25 = 0.10$

$P(\text{exercise and vegetarian}) = \frac{30}{200} = 0.15$

Since $0.15 \neq 0.10$, the events are not independent. People who exercise regularly are more likely to be vegetarian in this data.

”At Least One” Using the Complement

When a problem asks for the probability of “at least one” occurrence, it is almost always easier to use the complement:

$P(\text{at least one}) = 1 - P(\text{none})$

This avoids having to calculate every possible combination (exactly one, exactly two, exactly three, and so on).

Example 5: Defective Items

A factory’s defect rate is 5%, meaning $P(\text{defective}) = 0.05$ for each item. Three items are randomly selected from a large production run (large enough to treat the draws as independent). What is the probability that at least one item is defective?

Step 1: Find the probability that a single item is NOT defective.

$P(\text{not defective}) = 1 - 0.05 = 0.95$

Step 2: Find the probability that ALL three items are not defective.

$P(\text{none defective}) = 0.95 \times 0.95 \times 0.95 = 0.95^3 = 0.857375$

Step 3: Use the complement.

$P(\text{at least one defective}) = 1 - 0.857375 = 0.142625 \approx 0.143$

Answer: There is about a 14.3% chance that at least one of the three items is defective.

Real-World Application: Nursing — Sequential Diagnostic Tests

In clinical settings, patients often undergo multiple diagnostic tests. Understanding compound events helps nurses and healthcare professionals interpret the combined results.

A patient is given two independent screening tests for a condition. Test A has a sensitivity of 90% (correctly identifies 90% of people with the condition), and Test B has a sensitivity of 85%.

What is the probability that both tests correctly detect the condition in a patient who has it?

Since the tests are independent:

$P(\text{both detect}) = 0.90 \times 0.85 = 0.765 = 76.5\%$

What is the probability that at least one test detects the condition?

$P(\text{neither detects}) = (1 - 0.90) \times (1 - 0.85) = 0.10 \times 0.15 = 0.015$

$P(\text{at least one detects}) = 1 - 0.015 = 0.985 = 98.5\%$

Using two independent tests increases the overall detection rate from 90% (one test alone) to 98.5% (at least one of two tests). This is why clinicians often use multiple screening tools — the compound probability of missing a condition drops dramatically.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Independent events do not affect each other: $P(A \text{ and } B) = P(A) \cdot P(B)$.
• Dependent events require conditional probability: $P(A \text{ and } B) = P(A) \cdot P(B \mid A)$.
• With replacement keeps events independent; without replacement makes them dependent.
• To test for independence, check whether $P(A \text{ and } B) = P(A) \times P(B)$.
• For “at least one” problems, use the complement: $P(\text{at least one}) = 1 - P(\text{none})$.
• These rules extend to any number of events — multiply across the chain.

Return to Statistics for more topics in this section.