Question

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

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{12}\)
• C. \(\frac{3}{16}\)
• D. \(\frac{3}{4}\)

Answer 1

The Correct Option is A

To find the probability of the intersection of the complements of events A and B, we can use the complement rule:
\(P(A' \cap B') = 1 - P(A \cup B)\)
We know that \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
Rearranging the equation, we have:
\(P(A \cap B) = P(A|B) \cdot P(B)\)
Given that \(P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{3}\), and \(P(A|B) = \frac{1}{4}\)
we can substitute these values into the equation:
\(P(A \cap B) = \left(\frac{1}{4}\right) \times \left(\frac{1}{3}\right) = \frac{1}{12}\)
Now, we can find the probability of the union of events A and B: 
\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4}\)
Finally, we can find the probability of the intersection of the complements: 
\(P(A' \cap B') = 1 - P(A \cup B) = 1 - \left(\frac{3}{4}\right) = \frac{1}{4}\)
Therefore, the correct option is (A) \(\frac{1}{4}\)

Answer 2

Given:
\[ P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{3}, \quad P(A|B) = \frac{1}{4} \] From conditional probability formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \] \[ \frac{1}{4} = \frac{P(A \cap B)}{\frac{1}{3}} \Rightarrow P(A \cap B) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} \] Now use the formula: \[ P(A' \cap B') = 1 - P(A \cup B) \] \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} \] \[ = \frac{6 + 4 - 1}{12} = \frac{9}{12} = \frac{3}{4} \] \[ \Rightarrow P(A' \cap B') = 1 - \frac{3}{4} = \frac{1}{4} \] Correct Answer: \(\frac{1}{4}\)

Answer 3

We are given the following probabilities: 

• \(P(A) = \frac{1}{2}\)
• \(P(B) = \frac{1}{3}\)
• \(P(A|B) = \frac{1}{4}\)

We need to find \(P(A' \cap B')\).

First, recall De Morgan's law for events: \(A' \cap B' = (A \cup B)'\).

Using the complement rule, we have:

\[ P(A' \cap B') = P((A \cup B)') = \mathbf{1 - P(A \cup B)} \]

Next, we need to find \(P(A \cup B)\). The formula for the probability of the union of two events is:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

To use this formula, we first need to calculate \(P(A \cap B)\). We can use the definition of conditional probability:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Rearranging this formula gives:

\[ P(A \cap B) = P(A|B) \times P(B) \]

Substitute the given values:

\[ P(A \cap B) = \frac{1}{4} \times \frac{1}{3} = \mathbf{\frac{1}{12}} \]

Now, substitute the values of \(P(A)\), \(P(B)\), and \(P(A \cap B)\) into the formula for \(P(A \cup B)\):

\[ P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} \]

Find a common denominator (12):

\[ P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} \]

\[ P(A \cup B) = \frac{6 + 4 - 1}{12} = \frac{9}{12} = \mathbf{\frac{3}{4}} \]

Finally, calculate \(P(A' \cap B')\):

\[ P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{3}{4} = \mathbf{\frac{1}{4}} \]

Comparing this with the given options, the correct option is:

\(\frac{1}{4}\)