ISEE Upper Level Quantitative Reasoning Flashcards: Multi Event Probability

What this deck covers

This deck focuses on Multi Event Probability, giving you a quick way to review the definitions, rules, and examples that matter most for ISEE Upper Level Quantitative Reasoning.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: State the conditional probability formula for $P(A\mid B)$.

Answer: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Conditional probability is the joint probability divided by the probability of the conditioning event.

Flashcard 2: A fair die is rolled twice. What is the probability both rolls are even?

Answer: $\frac{1}{4}$. Each roll has $P(\text{even})=\frac{3}{6}=\frac{1}{2}$, and independent, so $\left(\frac{1}{2}\right)^2$.

Flashcard 3: A fair die is rolled twice. What is the probability of at least one $6$?

Answer: $\frac{11}{36}$. Complement of no sixes: $1 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36}$.

Flashcard 4: State De Morgan's law for the complement of a union: $(A\cup B)^c$.

Answer: $(A\cup B)^c=A^c\cap B^c$. De Morgan's law equates the complement of a union to the intersection of the complements.

Flashcard 5: A card is drawn from a $52$-card deck. What is $P(\text{heart or king})$?

Answer: $\frac{4}{13}$. Addition rule: $P(\text{heart}) + P(\text{king}) - P(\text{heart and king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}$.

Flashcard 6: A bag has $3$ red and $2$ blue marbles. Two are drawn without replacement. What is $P(\text{one red and one blue})$?

Answer: $\frac{3}{5}$. Sum of red-then-blue and blue-then-red: $\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}$.

Flashcard 7: Two cards are drawn without replacement. What is $P(\text{first ace, second king})$?

Answer: $\frac{4}{663}$. Sequential probabilities: $P(\text{first ace}) = \frac{4}{52}$, $P(\text{second king} \mid \text{first ace}) = \frac{4}{51}$, product simplifies to $\frac{4}{663}$.

Flashcard 8: If $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.1$, what is $P(A\cup B)$?

Answer: $0.8$. Use the addition rule: $0.4 + 0.5 - 0.1$.

Flashcard 9: If $P(A)=0.3$ and $P(B)=0.5$ and $A,B$ are independent, what is $P(A\cap B)$?

Answer: $0.15$. Since events are independent, multiply their probabilities: $0.3 \times 0.5$.

Flashcard 10: A fair coin is flipped twice. What is the probability of getting $2$ heads?

Answer: $\frac{1}{4}$. Each flip is independent with $P(H)=\frac{1}{2}$, so multiply: $\left(\frac{1}{2}\right)^2$.

Flashcard 11: State the addition rule for mutually exclusive events $A$ and $B$.

Answer: $P(A\cup B)=P(A)+P(B)$. Mutually exclusive events have no overlap, so their union probability is simply the sum of their individual probabilities.

Flashcard 12: What condition must hold for events $A$ and $B$ to be independent?

Answer: $P(A\cap B)=P(A)P(B)$. Independence requires that the joint probability equals the product of the marginal probabilities.

Flashcard 13: If $P(A)=0.35$, what is $P(A^c)$?

Answer: $0.65$. Apply the complement rule: $1 - 0.35$.

Flashcard 14: A bag has $3$ red and $2$ blue marbles. Two are drawn without replacement. What is $P(\text{both red})$?

Answer: $\frac{3}{10}$. Multiplication rule: $P(\text{first red}) = \frac{3}{5}$, $P(\text{second red} \mid \text{first red}) = \frac{2}{4}$, product is $\frac{3}{10}$.

Flashcard 15: A fair coin is flipped $3$ times. What is the probability of at least one head?

Answer: $\frac{7}{8}$. Complement of all tails: $1 - \left(\frac{1}{2}\right)^3 = 1 - \frac{1}{8}$.

Flashcard 16: Two cards are drawn without replacement. What is $P(\text{both aces})$?

Answer: $\frac{1}{221}$. Multiplication rule without replacement: $\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$.

Flashcard 17: State the multiplication rule for independent events $A$ and $B$.

Answer: $P(A \cap B)=P(A)P(B)$. For independent events, the probability of their intersection equals the product of their individual probabilities.

Flashcard 18: If $P(A\cap B)=0.12$ and $P(B)=0.3$, what is $P(A\mid B)$?

Answer: $0.4$. Use conditional probability: $\frac{0.12}{0.3}$.

Flashcard 19: If $P(A)=0.6$ and $P(B\mid A)=0.2$, what is $P(A\cap B)$?

Answer: $0.12$. Apply the multiplication rule: $P(A) \times P(B \mid A) = 0.6 \times 0.2$.

Flashcard 20: State the addition rule for any events $A$ and $B$.

Answer: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The addition rule accounts for overlap by subtracting the intersection probability from the sum of individual probabilities.

Flashcard 21: What condition must hold for events $A$ and $B$ to be mutually exclusive?

Answer: $P(A\cap B)=0$. Mutually exclusive events cannot occur together, so their intersection probability is zero.

Flashcard 22: State the complement rule for an event $A$.

Answer: $P(A^c)=1-P(A)$. The complement rule states that the probability of an event not occurring is one minus the probability of it occurring.

Flashcard 23: State De Morgan's law for the complement of an intersection: $(A\cap B)^c$.

Answer: $(A\cap B)^c=A^c\cup B^c$. De Morgan's law equates the complement of an intersection to the union of the complements.

Flashcard 24: If $P(A)=0.25$ and $P(B\mid A)=0.6$, what is $P(B\cap A)$?

Answer: $0.15$. Apply the multiplication rule: $0.25 \times 0.6$.

Flashcard 25: State the general multiplication rule for events $A$ and $B$ using conditional probability.

Answer: $P(A \cap B)=P(A)P(B\mid A)$. The general rule expresses the joint probability as the product of one event's probability and the conditional probability of the other given the first.