If S is the sample space of random experiment `S = A cup B and P(A) = 1/3 P(B)` where A and B are mutually exclusive events then P(A) is equal to

Options

A. `1/2`
B. `3/4`
C. `1/4`
D. `3/8`

Answer

To solve the problem step by step, we will use the information given in the question and apply the properties of probabilities.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We have two events A and B that are mutually exclusive, meaning they cannot occur at the same time. We are given that:
\[ P(A) = \frac{1}{3} P(B) \]
and the sample space \( S = A \cup B \).

2. **Using the Property of Sample Space:**
Since \( S \) is the sample space, the total probability of all outcomes in the sample space is 1. Thus, we can write:
\[ P(A \cup B) = P(A) + P(B) \]
Since A and B are mutually exclusive, we do not subtract \( P(A \cap B) \) (which is 0).

3. **Substituting the Known Relationship:**
From the relationship \( P(A) = \frac{1}{3} P(B) \), we can express \( P(B) \) in terms of \( P(A) \):
\[ P(B) = 3P(A) \]

4. **Substituting into the Total Probability:**
Now we substitute \( P(B) \) into the equation for the total probability:
\[ P(A) + P(B) = 1 \]
becomes:
\[ P(A) + 3P(A) = 1 \]
Simplifying this gives:
\[ 4P(A) = 1 \]

5. **Solving for \( P(A) \):**
To find \( P(A) \), we divide both sides by 4:
\[ P(A) = \frac{1}{4} \]

### Final Answer:
Thus, the probability of event A is:
\[ P(A) = \frac{1}{4} \]