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In an experiment, the probability that event AA occurs is 23\frac{2}{3}, the probability that event BB occurs is 37\frac{3}{7}, and the probability that events AA and BB both occur is 14\frac{1}{4}.\newlineWhat is the probability that AA occurs given that BB occurs?\newlineSimplify any fractions.\newline____

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Q. In an experiment, the probability that event AA occurs is 23\frac{2}{3}, the probability that event BB occurs is 37\frac{3}{7}, and the probability that events AA and BB both occur is 14\frac{1}{4}.\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)=14P(A \text{ and } B) = \frac{1}{4} and P(B)=37P(B) = \frac{3}{7}. So, P(AB)=1437P(A|B) = \frac{\frac{1}{4}}{\frac{3}{7}}.
  3. Multiply Fractions: To divide the fractions, we multiply by the reciprocal of the second fraction: (14)×(73)(\frac{1}{4}) \times (\frac{7}{3}).
  4. Simplify Result: Now, multiply the numerators and denominators: 1×74×3\frac{1 \times 7}{4 \times 3}.
  5. Simplify Result: Now, multiply the numerators and denominators: (1×7)/(4×3)(1 \times 7) / (4 \times 3).This simplifies to 7/127/12. So, P(AB)=7/12P(A|B) = 7/12.

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