Question

Let 𝑀 and 𝑁 be events defined on the sample space 𝑆. If 𝑃(𝑀) = \(\frac{1}{3}\) and 𝑃(𝑁^𝑐 ) = \(\frac{1}{4}\) then which one of the following is necessarily CORRECT?

• A. M and N are disjoint
• B. M and N are not disjoint
• C. M and N are independent
• D. M and N are not independent

Answer 1

The Correct Option is B

To determine the correct answer, we need to analyze the given probabilities and the nature of events \(M\) and \(N\).

We are given:

• \(P(M) = \frac{1}{3}\)
• \(P(N^c) = \frac{1}{4}\)

Since \(P(N^c) = \frac{1}{4}\), we can find \(P(N)\) using the complement rule:

\(P(N) = 1 - P(N^c) = 1 - \frac{1}{4} = \frac{3}{4}\)

We need to determine the relationship between events \(M\) and \(N\). The options are disjoint, not disjoint, independent, and not independent.

1. **Disjoint Events**: Two events \(M\) and \(N\) are disjoint if \(P(M \cap N) = 0\). This implies that \(M\) and \(N\) cannot occur simultaneously.

2. **Independent Events**: Two events \(M\) and \(N\) are independent if \(P(M \cap N) = P(M) \times P(N)\).

Let's evaluate the possible scenarios:

**Are \(M\) and \(N\) disjoint?**

If \(M\) and \(N\) were disjoint, \(P(M \cap N) = 0\). However, this would imply:

\(P(M) + P(N) = P(M \cup N)\)

So:

\(\frac{1}{3} + \frac{3}{4} = \frac{9}{12} + \frac{9}{12} = \frac{18}{12} = 1.5\)

This is not possible as the probability cannot exceed 1. Therefore, \(M\) and \(N\) are not disjoint.

**Conclusion**:

The correct answer is that \(M\) and \(N\) are not disjoint.