Question

If $\mathrm{P(A)} = \frac{3}{10}$, $\mathrm{P(B)} = \frac{2}{5}$, and $\mathrm{P(A \cup B)} = \frac{3}{5}$, then $\mathrm{P(A / B) \times P(B / A)} = $

• A. $\frac{1}{3}$
• B. $\frac{1}{12}$
• C. $\frac{1}{10}$
• D. $\frac{1}{4}$

Hint:

When computing conditional products like $\mathrm{P(A/B) \times P(B/A)}$, you can write out the expansion directly as $\frac{\mathrm{P(A \cap B)}^2}{\mathrm{P(A) \cdot P(B)}}$. Keeping fractions in terms of a common denominator (like base 10) makes cross-multiplications and final fraction reductions much simpler to track!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given individual probabilities for two events $A$ and $B$, as well as the probability of their union $\mathrm{P(A \cup B)}$. We need to compute the product of their conditional probabilities, $\mathrm{P(A / B) \times P(B / A)}$.

Step 2: Key Formula or Approach:
According to the definition of conditional probability: $$\mathrm{P(A / B)} = \frac{\mathrm{P(A \cap B)}}{\mathrm{P(B)}} \quad \text{and} \quad \mathrm{P(B / A)} = \frac{\mathrm{P(A \cap B)}}{\mathrm{P(A)}}$$ We can determine the intersection probability $\mathrm{P(A \cap B)}$ using the general addition rule of probability: $$\mathrm{P(A \cup B)} = \mathrm{P(A)} + \mathrm{P(B)} - \mathrm{P(A \cap B)}$$

Step 3: Detailed Explanation:
First, find the common intersection probability by substituting our given values into the addition rule: $$\frac{3}{5} = \frac{3}{10} + \frac{2}{5} - \mathrm{P(A \cap B)}$$ Isolate $\mathrm{P(A \cap B)}$: $$\mathrm{P(A \cap B)} = \frac{3}{10} + \frac{4}{10} - \frac{6}{10} = \frac{3 + 4 - 6}{10} = \frac{1}{10}$$ Now, set up the targeted product of conditional probabilities: $$\text{Product} = \mathrm{P(A / B) \times P(B / A)} = \frac{\mathrm{P(A \cap B)}}{\mathrm{P(B)}} \times \frac{\mathrm{P(A \cap B)}}{\mathrm{P(A)}} = \frac{[\mathrm{P(A \cap B)}]^2}{\mathrm{P(A) \times P(B)}}$$ Substitute the known fraction values into this product layout: $$\text{Product} = \frac{\left(\frac{1}{10}\right)^2}{\frac{3}{10} \times \frac{2}{5}} = \frac{\frac{1}{100}}{\frac{6}{50}} = \frac{1}{100} \times \frac{50}{6} = \frac{50}{600} = \frac{1}{12}$$ This matches option (B).

Step 4: Final Answer:
The product of the conditional probabilities is $\frac{1}{12}$, which corresponds to option (B).