Dependent Events

Learn how one event affects the probability of another when outcomes are connected.

Advanced25 minLesson
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Definition

Dependent events are events where the outcome of one event affects the probability of another event.
When events are dependent:
  • The first event changes the sample space for the second event
  • We use the formula: P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B | A)
The notation P(B∣A)P(B | A) means "the probability of B given that A has occurred."
Key difference from independent events:
  • Independent: P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
  • Dependent: P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B | A)

Try it now

Which scenario describes dependent events?

Worked Examples

A standard deck has 52 cards. You draw 2 cards without replacement. What is the probability that both cards are hearts?
1

Find P(first heart)

There are 13 hearts out of 52 cards → P(H1)=1352=14P(H_1) = \frac{13}{52} = \frac{1}{4}
2

Find P(second heart | first was heart)

After drawing 1 heart: 12 hearts remain, 51 cards total → P(H2∣H1)=1251P(H_2 | H_1) = \frac{12}{51}
3

Multiply the probabilities

P(H1 and H2)=1352×1251P(H_1 \text{ and } H_2) = \frac{13}{52} \times \frac{12}{51} → =1562652= \frac{156}{2652}
4

Simplify

1562652=117\frac{156}{2652} = \frac{1}{17} → ≈0.059\approx 0.059 or about 5.9%5.9\%

Common Mistakes

Using the same denominator for both events

Why it's wrong: When drawing without replacement, the total number of items decreases. Using 52 for both card draws ignores this change.

Correct: After the first draw, reduce the denominator. Second draw from 51 cards, third draw from 50 cards, etc.

Forgetting to adjust the numerator

Why it's wrong: If you draw a heart first, there are only 12 hearts left, not 13. The first event changes what remains.

Correct: If drawing the same category, reduce both numerator AND denominator. If different category, reduce only the denominator.

Treating dependent events as independent

Why it's wrong: Using P(A)×P(B)P(A) \times P(B) instead of P(A)×P(B∣A)P(A) \times P(B|A) gives wrong answers for dependent events.

Correct: Always ask: Does the first event change what's available for the second? If yes, events are dependent.

Interactive Visual

6 red, 4 blue: two draws without replacement

1st draw2nd draw6/105/94/94/106/93/9RedRedBlueBlueRedBlue(Red, Red)30/90(Red, Blue)24/90(Blue, Red)24/90(Blue, Blue)12/90

Multiply the choices at each step: 2 × 2 = 4 (4 outcomes)

All outcome probabilities add up to 1.

Tap an outcome on the right to follow its path.

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Practice Problems

16 problems
Problem 1 of 16
Easy
Which scenario describes dependent events?

Why It Matters

Dependent events appear constantly in real-world scenarios:
  • Card games: Drawing cards without replacement changes the deck composition
  • Quality control: Testing items from a batch affects what remains
  • Weather forecasting: Today's weather affects tomorrow's probability
  • Medical testing: Previous test results influence subsequent diagnoses
  • Sports: A player's fatigue affects performance probability in later games
Understanding dependent events prevents costly mistakes in probability calculations!

Real World Applications

Quality Control Testing

Factories test product samples. Each tested item is removed from the batch, making subsequent selections dependent.

Example:

A box has 100 lightbulbs, 5 defective. Testing 3 without replacement: P(all good)=95100×9499×9398P(\text{all good}) = \frac{95}{100} \times \frac{94}{99} \times \frac{93}{98}

1Try It Yourself

A batch of 20 phones has 2 defective. You test 2 phones without replacement.

What is the probability that both phones tested are NOT defective?

Step 1: Write the mathematical expression

Calculate: 1820×??\frac{18}{20} \times \frac{?}{?}

Card Game Strategies

Professional card players track which cards have been played to calculate remaining probabilities.

Example:

In blackjack, if 3 aces have been dealt from a single deck, the probability of the next card being an ace is 149\frac{1}{49}, not 452\frac{4}{52}.

2Try It Yourself

From a standard deck, 10 cards have been dealt. You saw 3 spades among them.

What is the probability the next card is a spade?

Step 1: Write the mathematical expression

Remaining spades / Remaining cards = ?

Selecting Committee Members

When choosing people for committees, each selection reduces the pool for subsequent choices.

Example:

From 8 men and 6 women, selecting 2 people: P(both women)=614×513P(\text{both women}) = \frac{6}{14} \times \frac{5}{13}

3Try It Yourself

A club has 10 members: 4 seniors and 6 juniors. Two officers are chosen at random.

What is the probability that both officers are seniors?

Step 1: Write the mathematical expression

Calculate the dependent probability:

Key Takeaways

  • 1Dependent events are when one event affects the probability of another
  • 2Use the formula: P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B | A)
  • 3Without replacement means the sample space shrinks after each event
  • 4Always adjust both the numerator (items of interest) and denominator (total items) after each event
  • 5Ask yourself: Does the first outcome change what's possible for the second? If yes, events are dependent

Frequently Asked Questions

Ask: Does the first event change what's possible for the second? Drawing cards without replacement is dependent (deck shrinks). Rolling dice twice is independent (first roll doesn't affect second).
Ask: Does the first event change what's possible for the second? Drawing cards without replacement is dependent (deck shrinks). Rolling dice twice is independent (first roll doesn't affect second).
It means you don't put the item back after selecting it. This changes the probabilities for subsequent selections because both the total count and possibly the count of desired items decrease.
Yes! If the outcome of one event gives you information that changes the probability of another, they're dependent. For example, knowing it rained today affects the probability it rains tomorrow.

Glossary

Dependent events
Events where the outcome of one affects the probability of another
Independent events
Events where the outcome of one has no effect on the probability of another
Conditional probability
The probability of an event given that another event has occurred, written as P(B∣A)P(B|A)
Without replacement
Selecting items without returning them to the original set
Sample space
The set of all possible outcomes
Today's challengeOne problem, three tries

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