Question

If A and B are two events such that \(P(A)=\frac{1}{2}\), \(P(B)=\frac{1}{3}\) and \(P(A \cap B)=\frac{1}{4}\), then \(P\left(\frac{A'}{B}\right)=\)

• A. \(\frac{3}{4}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{5}{8}\)

Hint:

Use \(P(A'|B)=\frac{P(B)-P(A\cap B)}{P(B)}\).

Answer 1

The Correct Option is B

Step 1: Find P(A|B) first:
Here we use the complement rule for conditional probability: $P(A'|B)=1-P(A|B)$.
Start with the ordinary conditional probability: \[ P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{1/4}{1/3} \]
Dividing by a fraction means multiplying by its reciprocal, so \[ P(A|B)=\frac{1}{4}\times 3=\frac{3}{4} \]

Step 2: Use the complement rule:
Given that B has happened, either A happens or A does not happen. These two cases add up to probability 1. So \[ P(A'|B)=1-\frac{3}{4}=\frac{1}{4} \]

Step 3: Cross-check with counting outcomes:
Imagine 12 equally likely outcomes. Then B has $12\times\frac{1}{3}=4$ outcomes and $A\cap B$ has $12\times\frac{1}{4}=3$ outcomes.
So only 1 of the 4 outcomes of B lies outside A. That gives $\frac{1}{4}$ again.

Step 4: Match with the options:
The value $\frac{3}{4}$ in option 1 is the trap: it is $P(A|B)$, not $P(A'|B)$. The value $\frac{1}{4}$ is in option 2.

Final Answer:
Both methods give the same value, so option 2 is correct. \[ \boxed{P(A'|B)=\frac{1}{4}} \]