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In an experiment, the probability that event AA occurs is 12\frac{1}{2}, the probability that event BB occurs is 34\frac{3}{4}, and the probability that events AA and BB both occur is 49\frac{4}{9}. What is the probability that AA occurs given that BB occurs? Simplify any fractions.

Full solution

Q. In an experiment, the probability that event AA occurs is 12\frac{1}{2}, the probability that event BB occurs is 34\frac{3}{4}, and the probability that events AA and BB both occur is 49\frac{4}{9}. What is the probability that AA occurs given that BB occurs? Simplify any fractions.
  1. Use Conditional Probability Formula: To find the probability that AA occurs given that BB occurs, we use the formula for conditional probability: P(AB)=P(A and B)P(B)P(A|B) = \frac{P(A \text{ and } B)}{P(B)}.
  2. Calculate P(AB)P(A|B): We know P(A and B)=49P(A \text{ and } B) = \frac{4}{9} and P(B)=34P(B) = \frac{3}{4}. So, P(AB)=4934P(A|B) = \frac{\frac{4}{9}}{\frac{3}{4}}.
  3. Multiply by Reciprocal: To divide the fractions, we multiply by the reciprocal of the second fraction: (49)×(43)(\frac{4}{9}) \times (\frac{4}{3}).
  4. Multiply Numerators and Denominators: Now, multiply the numerators and denominators: (4×4)/(9×3)=16/27(4 \times 4) / (9 \times 3) = 16 / 27.

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