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In an experiment, the probability that event AA occurs is 58\frac{5}{8}, the probability that event BB occurs is 14\frac{1}{4}, and the probability that events AA and BB both occur is 19\frac{1}{9}. What is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.

Full solution

Q. In an experiment, the probability that event AA occurs is 58\frac{5}{8}, the probability that event BB occurs is 14\frac{1}{4}, and the probability that events AA and BB both occur is 19\frac{1}{9}. What is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.
  1. Identify 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(A and B)P(A \text{ and } B) and P(B)P(B): We know P(A and B)=19P(A \text{ and } B) = \frac{1}{9} and P(B)=14P(B) = \frac{1}{4}. So, P(AB)=1914P(A|B) = \frac{\frac{1}{9}}{\frac{1}{4}}.
  3. Calculate P(AB)P(A|B): To divide fractions, we multiply by the reciprocal of the divisor. So, P(AB)=19×41=49P(A|B) = \frac{1}{9} \times \frac{4}{1} = \frac{4}{9}.

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