Multi-Event Probability - ISEE Upper Level Quantitative Reasoning

  1. State the conditional probability formula for P(A∣B)P(A\mid B).
    P(A∣B)=P(A∩B)P(B)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.
  2. A fair die is rolled twice. What is the probability both rolls are even?
    14\frac{1}{4}. Each roll has P(even)=36=12P(\text{even})=\frac{3}{6}=\frac{1}{2}, and independent, so (12)2\left(\frac{1}{2}\right)^2.
  3. A fair die is rolled twice. What is the probability of at least one 66?
    1136\frac{11}{36}. Complement of no sixes: 1−(56)2=1−25361 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36}.
  4. State De Morgan's law for the complement of a union: (A∪B)c(A\cup B)^c.
    (A∪B)c=Ac∩Bc(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.
  5. A card is drawn from a 5252-card deck. What is P(heart or king)P(\text{heart or king})?
    413\frac{4}{13}. Addition rule: P(heart)+P(king)−P(heart and king)=1352+452−152=1652P(\text{heart}) + P(\text{king}) - P(\text{heart and king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}.
  6. A bag has 33 red and 22 blue marbles. Two are drawn without replacement. What is P(one red and one blue)P(\text{one red and one blue})?
    35\frac{3}{5}. Sum of red-then-blue and blue-then-red: 35×24+25×34=310+310=35\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}.
  7. Two cards are drawn without replacement. What is P(first ace, second king)P(\text{first ace, second king})?
    4663\frac{4}{663}. Sequential probabilities: P(first ace)=452P(\text{first ace}) = \frac{4}{52}, P(second king∣first ace)=451P(\text{second king} \mid \text{first ace}) = \frac{4}{51}, product simplifies to 4663\frac{4}{663}.
  8. If P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, and P(A∩B)=0.1P(A\cap B)=0.1, what is P(A∪B)P(A\cup B)?
    0.80.8. Use the addition rule: 0.4+0.5−0.10.4 + 0.5 - 0.1.
  9. If P(A)=0.3P(A)=0.3 and P(B)=0.5P(B)=0.5 and A,BA,B are independent, what is P(A∩B)P(A\cap B)?
    0.150.15. Since events are independent, multiply their probabilities: 0.3×0.50.3 \times 0.5.
  10. A fair coin is flipped twice. What is the probability of getting 22 heads?
    14\frac{1}{4}. Each flip is independent with P(H)=12P(H)=\frac{1}{2}, so multiply: (12)2\left(\frac{1}{2}\right)^2.
  11. State the addition rule for mutually exclusive events AA and BB.
    P(A∪B)=P(A)+P(B)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.
  12. What condition must hold for events AA and BB to be independent?
    P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B). Independence requires that the joint probability equals the product of the marginal probabilities.
  13. If P(A)=0.35P(A)=0.35, what is P(Ac)P(A^c)?
    0.650.65. Apply the complement rule: 1−0.351 - 0.35.
  14. A bag has 33 red and 22 blue marbles. Two are drawn without replacement. What is P(both red)P(\text{both red})?
    310\frac{3}{10}. Multiplication rule: P(first red)=35P(\text{first red}) = \frac{3}{5}, P(second red∣first red)=24P(\text{second red} \mid \text{first red}) = \frac{2}{4}, product is 310\frac{3}{10}.
  15. A fair coin is flipped 33 times. What is the probability of at least one head?
    78\frac{7}{8}. Complement of all tails: 1−(12)3=1−181 - \left(\frac{1}{2}\right)^3 = 1 - \frac{1}{8}.
  16. Two cards are drawn without replacement. What is P(both aces)P(\text{both aces})?
    1221\frac{1}{221}. Multiplication rule without replacement: 452×351=122652=1221\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}.
  17. State the multiplication rule for independent events AA and BB.
    P(A∩B)=P(A)P(B)P(A \cap B)=P(A)P(B). For independent events, the probability of their intersection equals the product of their individual probabilities.
  18. If P(A∩B)=0.12P(A\cap B)=0.12 and P(B)=0.3P(B)=0.3, what is P(A∣B)P(A\mid B)?
    0.40.4. Use conditional probability: 0.120.3\frac{0.12}{0.3}.
  19. If P(A)=0.6P(A)=0.6 and P(B∣A)=0.2P(B\mid A)=0.2, what is P(A∩B)P(A\cap B)?
    0.120.12. Apply the multiplication rule: P(A)×P(B∣A)=0.6×0.2P(A) \times P(B \mid A) = 0.6 \times 0.2.
  20. State the addition rule for any events AA and BB.
    P(A∪B)=P(A)+P(B)−P(A∩B)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.
  21. What condition must hold for events AA and BB to be mutually exclusive?
    P(A∩B)=0P(A\cap B)=0. Mutually exclusive events cannot occur together, so their intersection probability is zero.
  22. State the complement rule for an event AA.
    P(Ac)=1−P(A)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.
  23. State De Morgan's law for the complement of an intersection: (A∩B)c(A\cap B)^c.
    (A∩B)c=Ac∪Bc(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.
  24. If P(A)=0.25P(A)=0.25 and P(B∣A)=0.6P(B\mid A)=0.6, what is P(B∩A)P(B\cap A)?
    0.150.15. Apply the multiplication rule: 0.25×0.60.25 \times 0.6.
  25. State the general multiplication rule for events AA and BB using conditional probability.
    P(A∩B)=P(A)P(B∣A)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.

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