The probability of event `A` is `3//4`. The probability of event `B`, given that event `A` occurs is `1//4`. The probability of event `A`, given that event `B` occurs is `2//3`. The probability that neither event occurs is

Options

A. `(1)/(6)`
B. `(27)/(112)`
C. `(5)/(32)`
D. `(1)/(8)`

Explanation

To solve the problem step by step, we will use the given probabilities and the relationships between them.

### Step 1: Identify the given probabilities
- Probability of event A, \( P(A) = \frac{3}{4} \)
- Probability of event B given A, \( P(B|A) = \frac{1}{4} \)
- Probability of event A given B, \( P(A|B) = \frac{2}{3} \)

### Step 2: Use the formula for conditional probability to find \( P(A \cap B) \)
The formula for conditional probability states:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
From this, we can rearrange to find \( P(A \cap B) \):
\[ P(A \cap B) = P(B|A) \times P(A) \]
Substituting the known values:
\[ P(A \cap B) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} \]

### Step 3: Use the formula for conditional probability to find \( P(B) \)
Now, we can use the other conditional probability:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Rearranging gives us:
\[ P(B) = \frac{P(A \cap B)}{P(A|B)} \]
Substituting the known values:
\[ P(B) = \frac{\frac{3}{16}}{\frac{2}{3}} = \frac{3}{16} \times \frac{3}{2} = \frac{9}{32} \]

### Step 4: Calculate the probability of \( P(A \cup B) \)
To find the probability that neither event occurs, we first need to find \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values we have:
\[ P(A \cup B) = \frac{3}{4} + \frac{9}{32} - \frac{3}{16} \]
To perform this calculation, we need a common denominator. The least common multiple of 4, 32, and 16 is 32:
\[ P(A) = \frac{3}{4} = \frac{24}{32}, \quad P(A \cap B) = \frac{3}{16} = \frac{6}{32} \]
Now substituting:
\[ P(A \cup B) = \frac{24}{32} + \frac{9}{32} - \frac{6}{32} = \frac{24 + 9 - 6}{32} = \frac{27}{32} \]

### Step 5: Calculate the probability that neither event occurs
The probability that neither event A nor event B occurs is given by:
\[ P(\text{neither A nor B}) = 1 - P(A \cup B) \]
Substituting the value we found:
\[ P(\text{neither A nor B}) = 1 - \frac{27}{32} = \frac{32 - 27}{32} = \frac{5}{32} \]

### Final Answer
The probability that neither event occurs is \( \frac{5}{32} \).

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