If A and B are two events such that `P(AcupB)=(5)/(6), P(A)=(1)/(3) and P(B)=(3)/(4)`, then A and B are

Options

A. mutually exclusive
B. dependent
C. independent
D. None of these

Answer

To solve the problem, we need to determine the relationship between the two events A and B based on the given probabilities.

### Given:
- \( P(A \cup B) = \frac{5}{6} \)
- \( P(A) = \frac{1}{3} \)
- \( P(B) = \frac{3}{4} \)

### Step 1: Use the formula for the probability of the union of two events.
The formula for the probability of the union of two events is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

### Step 2: Substitute the known values into the formula.
Substituting the values we have:
\[ \frac{5}{6} = \frac{1}{3} + \frac{3}{4} - P(A \cap B) \]

### Step 3: Convert fractions to a common denominator.
The common denominator for \( \frac{1}{3} \) and \( \frac{3}{4} \) is 12. We convert:
\[ \frac{1}{3} = \frac{4}{12}, \quad \frac{3}{4} = \frac{9}{12} \]
Thus, we rewrite the equation:
\[ \frac{5}{6} = \frac{4}{12} + \frac{9}{12} - P(A \cap B) \]

### Step 4: Simplify the equation.
Now, simplify the left side:
\[ \frac{5}{6} = \frac{13}{12} - P(A \cap B) \]
Next, convert \( \frac{5}{6} \) to a fraction with a denominator of 12:
\[ \frac{5}{6} = \frac{10}{12} \]
So we have:
\[ \frac{10}{12} = \frac{13}{12} - P(A \cap B) \]

### Step 5: Solve for \( P(A \cap B) \).
Rearranging gives us:
\[ P(A \cap B) = \frac{13}{12} - \frac{10}{12} = \frac{3}{12} = \frac{1}{4} \]

### Step 6: Determine if A and B are mutually exclusive, independent, or dependent.
- **Mutually Exclusive:** If \( P(A \cap B) = 0 \), then A and B are mutually exclusive. Here, \( P(A \cap B) = \frac{1}{4} \neq 0 \), so they are not mutually exclusive.
- **Independent:** If \( P(A \cap B) = P(A) \cdot P(B) \), then A and B are independent.
- Calculate \( P(A) \cdot P(B) \):
\[ P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4} \]
Since \( P(A \cap B) = \frac{1}{4} \), A and B are independent.

### Conclusion:
Thus, A and B are independent events.