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In an experiment, the probability that event AA occurs is 57\frac{5}{7}, the probability that event BB occurs is 47\frac{4}{7}, and the probability that events AA and BB both occur is 59\frac{5}{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 57\frac{5}{7}, the probability that event BB occurs is 47\frac{4}{7}, and the probability that events AA and BB both occur is 59\frac{5}{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)=59P(A \text{ and } B) = \frac{5}{9} and P(B)=47P(B) = \frac{4}{7}. So, P(AB)=5947P(A|B) = \frac{\frac{5}{9}}{\frac{4}{7}}.
  3. Multiply Fractions: To divide the fractions, we multiply by the reciprocal of the second fraction: (59)×(74)(\frac{5}{9}) \times (\frac{7}{4}).
  4. Simplify Result: Now, multiply the numerators and the denominators: (5×7)/(9×4)(5 \times 7) / (9 \times 4).
  5. Correct Multiplication: This simplifies to 3536\frac{35}{36}, which is incorrect because the multiplication was done wrong. It should be (5×7)/(9×4)=3536(5 \times 7) / (9 \times 4) = \frac{35}{36}.

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