Drawing Cards Without Replacement

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

Find P(first heart): There are 13 hearts out of 52 cards = $P(H_1) = \frac{13}{52} = \frac{1}{4}$

Find P(second heart | first was heart): After drawing 1 heart: 12 hearts remain, 51 cards total = $P(H_2 | H_1) = \frac{12}{51}$

Multiply the probabilities: $P(H_1 \text{ and } H_2) = \frac{13}{52} \times \frac{12}{51}$ = $= \frac{156}{2652}$

Simplify: $\frac{156}{2652} = \frac{1}{17}$ = $\approx 0.059$ or about $5.9\%$

Answer: The probability of drawing 2 hearts in a row is $\frac{1}{17}$ or approximately $5.9\%$

Selecting from a Group

A bag contains 6 red marbles and 4 blue marbles. You draw 2 marbles without replacement. What is the probability of drawing a red marble first, then a blue marble?

Find P(red first): 6 red marbles out of 10 total = $P(R) = \frac{6}{10} = \frac{3}{5}$

Find P(blue second | red first): After removing 1 red: 4 blue out of 9 remaining = $P(B | R) = \frac{4}{9}$

Multiply the probabilities: $P(R \text{ then } B) = \frac{3}{5} \times \frac{4}{9}$ = $= \frac{12}{45}$

Simplify: $\frac{12}{45} = \frac{4}{15}$ = $\approx 0.267$ or about $26.7\%$

Answer: The probability is $\frac{4}{15}$ or approximately $26.7\%$

Three Dependent Events

A jar has 5 red, 3 green, and 2 yellow candies. You pick 3 candies without replacement. What is the probability of picking red, then green, then yellow?

Find P(red first): 5 red out of 10 total = $P(R) = \frac{5}{10} = \frac{1}{2}$

Find P(green second | red first): 3 green out of 9 remaining = $P(G | R) = \frac{3}{9} = \frac{1}{3}$

Find P(yellow third | red then green): 2 yellow out of 8 remaining = $P(Y | R,G) = \frac{2}{8} = \frac{1}{4}$

Multiply all probabilities: $P(R,G,Y) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}$ = $= \frac{1}{24} \approx 4.2\%$

Answer: The probability is $\frac{1}{24}$ or approximately $4.2\%$

Mistake: Using the same denominator for both events

Why: 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.

Mistake: Forgetting to adjust the numerator

Why: 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.

Mistake: Treating dependent events as independent

Why: Using $P(A) \times P(B)$ instead of $P(A) \times P(B|A)$ 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.

Quality Control Testing

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

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

Card Game Strategies

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

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

Selecting Committee Members

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

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

Q: How do I know if events are dependent or independent?

A: 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).

Q: What does 'without replacement' mean?

A: 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.

Q: Can events be dependent even with replacement?

A: 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.