Question

Let A and B be two events such that $P(A)=\frac{1}{8}$ and $P(A/B)=\frac{1}{4}$, $P(B/A)=\frac{2}{5}$. Then $P(B)$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{1}{15}$
• D. $\frac{2}{15}$
• E. $\frac{1}{3}$

Hint:

Math Tip: You can combine the two conditional probability formulas into Bayes' Theorem to solve it in one line: $P(B) = \frac{P(B|A) \cdot P(A)}{P(A|B)}$. Plugging the numbers in gives $\frac{(2/5) \cdot (1/8)}{1/4} = 1/5$ instantly!

Answer 1

The Correct Option is A

Concept:
Probability - Conditional Probability.
The formula for conditional probability is: $$ P(E|F) = \frac{P(E \cap F)}{P(F)} $$ This can be rearranged to find the intersection: $P(E \cap F) = P(E|F) \cdot P(F)$.
Step 1: List all given probabilities.

• $P(A) = \frac{1}{8}$
• $P(A|B) = \frac{1}{4}$
• $P(B|A) = \frac{2}{5}$

Step 2: Find the intersection probability $P(A \cap B)$.
Use the conditional probability formula oriented around $P(B|A)$: $$ P(B|A) = \frac{P(A \cap B)}{P(A)} $$ Rearrange to solve for $P(A \cap B)$: $$ P(A \cap B) = P(B|A) \cdot P(A) $$ Substitute the known values: $$ P(A \cap B) = \left(\frac{2}{5}\right) \cdot \left(\frac{1}{8}\right) $$ $$ P(A \cap B) = \frac{2}{40} = \frac{1}{20} $$
Step 3: Calculate $P(B)$ using the other conditional probability.
Now use the conditional probability formula oriented around $P(A|B)$: $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$ Rearrange to solve for $P(B)$: $$ P(B) = \frac{P(A \cap B)}{P(A|B)} $$
Step 4: Substitute and solve.
Substitute $P(A \cap B) = \frac{1}{20}$ and $P(A|B) = \frac{1}{4}$ into the equation: $$ P(B) = \frac{\frac{1}{20}}{\frac{1}{4}} $$ Multiply by the reciprocal of the denominator: $$ P(B) = \frac{1}{20} \cdot 4 $$ $$ P(B) = \frac{4}{20} = \frac{1}{5} $$