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In an experiment, the probability that event AA occurs is 13\frac{1}{3}, the probability that event BB occurs is 79\frac{7}{9}, and the probability that events AA and BB both occur is 27\frac{2}{7}.\newlineWhat is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.\newline____\newline

Full solution

Q. In an experiment, the probability that event AA occurs is 13\frac{1}{3}, the probability that event BB occurs is 79\frac{7}{9}, and the probability that events AA and BB both occur is 27\frac{2}{7}.\newlineWhat is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.\newline____\newline
  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. Identify Given Probabilities: We know P(A and B)=27P(A \text{ and } B) = \frac{2}{7} and P(B)=79P(B) = \frac{7}{9}. So, P(AB)=2779P(A|B) = \frac{\frac{2}{7}}{\frac{7}{9}}.
  3. Calculate Conditional Probability: To divide the fractions, we multiply by the reciprocal of the second fraction: (27)×(97)(\frac{2}{7}) \times (\frac{9}{7}).
  4. Simplify Fraction: Multiplying the numerators and denominators, we get (2×9)/(7×7)=18/49(2 \times 9) / (7 \times 7) = 18/49.

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