Question

If events A and B are such that \(P(A) = \frac{1}{4}\), \(P(A\cap B) = \frac{1}{10}\) and \(P(A/B) = 2P(B/A)\) then \(P(B) =\)...........

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{5}\)
• C. \(\frac{1}{8}\)
• D. \(\frac{2}{5}\)

Hint:

Write both conditional probabilities with P(A and B) and compare.

Answer 1

The Correct Option is C

Step 1: Write the conditional probabilities:
\(P(A/B)=\dfrac{P(A\cap B)}{P(B)}\) and \(P(B/A)=\dfrac{P(A\cap B)}{P(A)}\).

Step 2: Use the given relation:
\(\dfrac{P(A\cap B)}{P(B)}=2\cdot\dfrac{P(A\cap B)}{P(A)}\). Since \(P(A\cap B)=\dfrac1{10}\ne0\), cancel it: \(\dfrac1{P(B)}=\dfrac2{P(A)}\), so \(P(A)=2P(B)\).

Step 3: Substitute P(A):
\(P(B)=\dfrac{P(A)}{2}=\dfrac{1}{8}\). Option C.

Step 4: Why the other options are wrong.
The relation forces \(P(B)=\dfrac{P(A)}{2}=\dfrac18\), so \(\dfrac12\), \(\dfrac15\) and \(\dfrac25\) cannot hold. As a check, \(P(A\cap B)=\dfrac1{10}\leq\dfrac18\), which is consistent.

Final Answer:
P(B) = 1/8. \[ \boxed{\text{(C) }\dfrac18} \]