Question

Let \(A\) and \(B\) be two events of a sample space such that \[ P(A)=\frac{1}{2},\qquad P(A\mid B)=\frac{1}{4},\qquad P(B\mid A)=\frac{1}{2}. \] Then, \[ P(\overline{A}\mid\overline{B})=\_\_\_\_\_, \] where \(\overline{A}\) denotes the complement of \(A\).

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{3}{4} \)
• D. 0

Hint:

Whenever a conditional probability is asked, always verify that the conditioning event has non-zero probability. If \(P(B)=1\), then \(P(\bar B)=0\), making \(P(\bar A|\bar B)\) undefined.

Answer 1

The Correct Option is D

Concept: Conditional probability is defined as \[ P(A|B)=\frac{P(A\cap B)}{P(B)}, \qquad P(B|A)=\frac{P(A\cap B)}{P(A)}. \] To evaluate conditional probabilities involving complements, the denominator must have a non-zero probability.
• First determine the intersection probability.
• Then compute the probability of event \(B\).
• Finally check whether \(P(\bar{B})\neq0\).

Step 1: Find \(P(A\cap B)\).
Using \[ P(B|A)=\frac{P(A\cap B)}{P(A)}, \] we get \[ P(A\cap B) =P(B|A)\times P(A) =\frac12\times\frac12 =\frac14. \] Thus, \[ \boxed{P(A\cap B)=\frac14.} \]

Step 2: Find \(P(B)\).
Using \[ P(A|B)=\frac{P(A\cap B)}{P(B)}, \] we obtain \[ \frac14=\frac{\frac14}{P(B)}. \] Therefore, \[ \boxed{P(B)=1.} \] Hence, \[ P(\bar B)=1-P(B)=0. \]

Step 3: Evaluate \(P(\bar A|\bar B)\).
By definition, \[ P(\bar A|\bar B) = \frac{P(\bar A\cap\bar B)}{P(\bar B)}. \] Since \[ P(\bar B)=0, \] the denominator becomes zero. Therefore, \[ \boxed{P(\bar A|\bar B)\text{ is undefined}.} \]