Question

If $P(A/B)=\frac{1}{2}, P(B/A)=\frac{1}{3}$ and $P(A\cap B)=\frac{1}{6}$ then $P(A\cup B)$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{9}$
• D. $\frac{1}{6}$
• E. $\frac{2}{3}$

Hint:

Logic Tip: Ensure you maintain consistent notation. Many exams use $P(A/B)$ to represent conditional probability $P(A|B)$. Remember that the event given *after* the slash or pipe is the one that goes in the denominator.

Answer 1

Answer: (E)

Concept:
The formula for conditional probability is $P(X|Y) = \frac{P(X\cap Y)}{P(Y)}$. To find the probability of the union of two events, use the Addition Rule: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$.
Step 1: Calculate P(B) using the first conditional probability.
We are given $P(A/B) = \frac{1}{2}$ and $P(A\cap B) = \frac{1}{6}$. Substitute into the conditional probability formula: $$P(A/B) = \frac{P(A\cap B)}{P(B)}$$ $$\frac{1}{2} = \frac{\frac{1}{6}}{P(B)}$$ Solving for $P(B)$: $$P(B) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{6} = \frac{1}{3}$$
Step 2: Calculate P(A) using the second conditional probability.
We are given $P(B/A) = \frac{1}{3}$. Substitute into the conditional probability formula: $$P(B/A) = \frac{P(A\cap B)}{P(A)}$$ $$\frac{1}{3} = \frac{\frac{1}{6}}{P(A)}$$ Solving for $P(A)$: $$P(A) = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \times \frac{3}{1} = \frac{3}{6} = \frac{1}{2}$$
Step 3: Apply the Addition Rule to find the union.
Now we have all the components needed to find $P(A\cup B)$: $P(A) = \frac{1}{2}$ $P(B) = \frac{1}{3}$ $P(A\cap B) = \frac{1}{6}$ Substitute into the Addition Rule formula: $$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$ $$P(A\cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6}$$ Find a common denominator (6): $$P(A\cup B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6}$$ $$P(A\cup B) = \frac{4}{6} = \frac{2}{3}$$