Compound Probability
When two events both happen, you multiply their probabilities — adjusting for whether the second event depends on the first.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
For independent "AND" events, multiply. For dependent events (without replacement), the second probability uses an updated count.
How to solve one, step by step
Example: draw $2$ red from $3$ red of $5$ marbles, without replacement.
- First red: $\frac{3}{5}$.
- Second red (one fewer of each): $\frac{2}{4}$.
- Multiply: $\frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$.
The mistakes that cost points
- Adding for AND events. "Both happen" multiplies probabilities; adding is for "either/or."
- Not adjusting the denominator. Without replacement, the total drops for the second draw.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
A fair coin is flipped and a fair six-sided die is rolled. What is the probability of getting heads AND rolling a 4?
- A$\frac{1}{12}$
- B$\frac{1}{12}$ (but computed as $\frac{1}{2} \times \frac{1}{6}$ treating as dependent)
- C$\frac{1}{11}$
- D$\frac{1}{2} + \frac{1}{6} = \frac{2}{3}$
Show solution
Answer: A, $\frac{1}{12}$. The events are independent. $P(\text{heads}) = \frac{1}{2}$ and $P(\text{rolling 4}) = \frac{1}{6}$. For independent events: $P(A \text{ and } B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.
Medium
Two cards are drawn from a standard deck of 52 cards without replacement. What is the probability that both cards are aces?
- A$\frac{4}{52} \times \frac{4}{52} = \frac{1}{169}$
- B$\frac{4}{52} + \frac{3}{51}$
- C$\frac{4}{52} \times \frac{3}{52} = \frac{12}{2704}$
- D$\frac{1}{221}$
Show solution
Answer: D, $\frac{1}{221}$. First card: $P(\text{ace}) = \frac{4}{52} = \frac{1}{13}$. After removing one ace, 3 aces remain in 51 cards. Second card: $P(\text{ace}) = \frac{3}{51} = \frac{1}{17}$. Combined: $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.
Hard
A committee of 2 is chosen at random from a group of 4 men and 6 women (10 people total). What is the probability that both chosen people are women?
- A$\frac{1}{3}$
- B$\frac{6}{10} \times \frac{5}{10} = \frac{3}{10}$
- C$\frac{6}{10} + \frac{5}{9} = \frac{27}{30}$
- D$\frac{6}{10} \times \frac{6}{10} = \frac{9}{25}$
Show solution
Answer: A, $\frac{1}{3}$. $P(\text{1st is woman}) = \frac{6}{10} = \frac{3}{5}$. After selecting a woman, 5 women remain out of 9 people. $P(\text{2nd is woman}) = \frac{5}{9}$. Combined: $\frac{3}{5} \times \frac{5}{9} = \frac{15}{45} = \frac{1}{3}$.
Find your exact gaps
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