If A and B are events such that `P(A uu B) = (3)//(4), P(A nn B) = (1)//(4)` and `P(A^(c)) = (2)//(3)`, then find
(a) P(A) (b) P(B)
(c ) `P(A nn B^(c )) (d) P(A^(c ) nn B)`

Explanation

To solve the problem step by step, we will use the given probabilities and the formulas of probability.

### Given:
1. \( P(A \cup B) = \frac{3}{4} \)
2. \( P(A \cap B) = \frac{1}{4} \)
3. \( P(A^c) = \frac{2}{3} \)

### Step 1: Find \( P(A) \)
We know that:
\[ P(A^c) = 1 - P(A) \]
Substituting the value of \( P(A^c) \):
\[ \frac{2}{3} = 1 - P(A) \]
Rearranging gives:
\[ P(A) = 1 - \frac{2}{3} = \frac{1}{3} \]

### Step 2: Find \( P(B) \)
Using the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ \frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4} \]
To solve for \( P(B) \), first, convert all fractions to a common denominator (12):
\[ \frac{3}{4} = \frac{9}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12} \]
Now substituting:
\[ \frac{9}{12} = \frac{4}{12} + P(B) - \frac{3}{12} \]
This simplifies to:
\[ \frac{9}{12} = \frac{1}{12} + P(B) \]
Thus:
\[ P(B) = \frac{9}{12} - \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \]

### Step 3: Find \( P(A \cap B^c) \)
Using the formula:
\[ P(A \cap B^c) = P(A) - P(A \cap B) \]
Substituting the known values:
\[ P(A \cap B^c) = \frac{1}{3} - \frac{1}{4} \]
Converting to a common denominator (12):
\[ P(A \cap B^c) = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \]

### Step 4: Find \( P(A^c \cap B) \)
Using the formula:
\[ P(A^c \cap B) = P(B) - P(A \cap B) \]
Substituting the known values:
\[ P(A^c \cap B) = \frac{2}{3} - \frac{1}{4} \]
Converting to a common denominator (12):
\[ P(A^c \cap B) = \frac{8}{12} - \frac{3}{12} = \frac{5}{12} \]

### Summary of Results:
(a) \( P(A) = \frac{1}{3} \)
(b) \( P(B) = \frac{2}{3} \)
(c) \( P(A \cap B^c) = \frac{1}{12} \)
(d) \( P(A^c \cap B) = \frac{5}{12} \)