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In an experiment, the probability that event AA occurs is 59\frac{5}{9}, the probability that event BB occurs is 78\frac{7}{8}, and the probability that events AA and BB both occur is 12\frac{1}{2}.\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 59\frac{5}{9}, the probability that event BB occurs is 78\frac{7}{8}, and the probability that events AA and BB both occur is 12\frac{1}{2}.\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)=12P(A \text{ and } B) = \frac{1}{2} and P(B)=78P(B) = \frac{7}{8}. So, P(AB)=1278P(A|B) = \frac{\frac{1}{2}}{\frac{7}{8}}.
  3. Multiply by Reciprocal: To divide the fractions, we multiply by the reciprocal of the second fraction: (12)×(87)(\frac{1}{2}) \times (\frac{8}{7}).
  4. Simplify the Fraction: Now, multiply the numerators and the denominators: 1×82×7\frac{1 \times 8}{2 \times 7}.
  5. Reduce the Fraction: This simplifies to 814\frac{8}{14}, which can be further reduced to 47\frac{4}{7}.

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