Question

If A and B are two events such that \(P(A) = \frac{2}{3}\), \(P(B) = \frac{1}{2}\) and \(P(A|B) = \frac{2}{3}\), then \(P(A^'\cup B)+P(A\cup B^') =\)

• A. \(\frac{2}{3}\)
• B. \(1\)
• C. \(\frac{3}{2}\)
• D. \(\frac{5}{6}\)

Hint:

Find P(A and B) = 1/3 first.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need \(P(A\cap B)\) first. From the conditional probability, \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}\).

Step 2: Find the intersection:
\[ P(A\cap B)=\frac23\times\frac12=\frac13 \]

Step 3: First term:
\(P(A'\cup B)=1-P(A\cap B')\). Now \(P(A\cap B')=P(A)-P(A\cap B)=\frac23-\frac13=\frac13\). So \(P(A'\cup B)=\frac23\).

Step 4: Second term:
\(P(A\cup B')=1-P(A'\cap B)\). Now \(P(A'\cap B)=P(B)-P(A\cap B)=\frac12-\frac13=\frac16\). So \(P(A\cup B')=\frac56\).

Step 5: Add:
\[ \frac23+\frac56=\frac{4+5}{6}=\frac32 \]
Option (C).

Final Answer:
The sum is 2/3 + 5/6 = 3/2. \[ \boxed{\frac32} \]