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In an experiment, the probability that event AA occurs is 56\frac{5}{6}, the probability that event BB occurs is 78\frac{7}{8}, and the probability that events AA and BB both occur is 45\frac{4}{5}.\newlineWhat is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.\newline____

Full solution

Q. In an experiment, the probability that event AA occurs is 56\frac{5}{6}, the probability that event BB occurs is 78\frac{7}{8}, and the probability that events AA and BB both occur is 45\frac{4}{5}.\newlineWhat is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.\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. Calculate P(AB)P(A|B): We know P(A and B)=45P(A \text{ and } B) = \frac{4}{5} and P(B)=78P(B) = \frac{7}{8}. So, P(AB)=4578P(A|B) = \frac{\frac{4}{5}}{\frac{7}{8}}.
  3. Multiply Fractions: To divide the fractions, we multiply by the reciprocal of the second fraction: (45)×(87)(\frac{4}{5}) \times (\frac{8}{7}).
  4. Simplify Result: Now, multiply the numerators and denominators: (4×8)/(5×7)=32/35(4 \times 8) / (5 \times 7) = 32/35.

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