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In an experiment, the probability that event AA occurs is 14\frac{1}{4}, the probability that event BB occurs is 38\frac{3}{8}, and the probability that events AA and BB both occur is 17\frac{1}{7}. What is the probability that AA occurs given that BB occurs? Simplify any fractions.

Full solution

Q. In an experiment, the probability that event AA occurs is 14\frac{1}{4}, the probability that event BB occurs is 38\frac{3}{8}, and the probability that events AA and BB both occur is 17\frac{1}{7}. What is the probability that AA occurs given that BB occurs? Simplify any fractions.
  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)=17P(A \text{ and } B) = \frac{1}{7} and P(B)=38P(B) = \frac{3}{8}. So, P(AB)=1738P(A|B) = \frac{\frac{1}{7}}{\frac{3}{8}}.
  3. Divide Fractions: To divide fractions, we multiply by the reciprocal of the divisor. So, P(AB)=17×83P(A|B) = \frac{1}{7} \times \frac{8}{3}.
  4. Multiply Numerators and Denominators: Now, multiply the numerators and denominators: (1×8)/(7×3)=8/21(1 \times 8) / (7 \times 3) = 8 / 21.

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