Question

If $M$ and $N$ are events such that $P(M\cup N)=\frac{3}{4}$, $P(M\cap N)=\frac{1}{4}$, and $P(\overline{M})=\frac{2}{3}$, then $P(\overline{M}\cap N)$ is

• A. $\frac{15}{12}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{1}{4}$
• E. $\frac{5}{12}$

Hint:

Break probabilities using union + intersection identities step-by-step.

Answer 1

Answer: (E)

Step 1: Find $P(M)$ \[ P(\overline{M}) = \frac{2}{3} \Rightarrow P(M) = 1 - \frac{2}{3} = \frac{1}{3} \]

Step 2: Use union formula \[ P(M\cup N) = P(M) + P(N) - P(M\cap N) \] \[ \frac{3}{4} = \frac{1}{3} + P(N) - \frac{1}{4} \]

Step 3: Solve for $P(N)$ \[ \frac{3}{4} = \frac{1}{12} + P(N) \] \[ P(N) = \frac{3}{4} - \frac{1}{12} = \frac{9}{12} - \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \]

Step 4: Use decomposition \[ P(N) = P(M\cap N) + P(\overline{M}\cap N) \] \[ \frac{2}{3} = \frac{1}{4} + P(\overline{M}\cap N) \] \[ P(\overline{M}\cap N) = \frac{2}{3} - \frac{1}{4} \] \[ = \frac{8}{12} - \frac{3}{12} = \frac{5}{12} \] \[ \boxed{\frac{5}{12}} \]