A,B,C are three mutually exclusive and exhaustive events associated with random experiment .Find P(A) , it being given that `P(B) = (3)/(2) P(A) and P( C ) =(1)/(2) P(B) `

Options

A. ` (4)/(13)`
B. ` (3)/(13)`
C. ` ( 5)/(13)`
D. None of these

Answer

To solve the problem, we need to find the probability of event A, given the relationships between the probabilities of events A, B, and C. Let's denote:

- \( P(A) = p \)
- \( P(B) = \frac{3}{2} P(A) = \frac{3}{2} p \)
- \( P(C) = \frac{1}{2} P(B) = \frac{1}{2} \left(\frac{3}{2} p\right) = \frac{3}{4} p \)

Since A, B, and C are mutually exclusive and exhaustive events, we know that:

\[ P(A) + P(B) + P(C) = 1 \]

Substituting the expressions for \( P(B) \) and \( P(C) \) in terms of \( P(A) \):

\[ p + \frac{3}{2} p + \frac{3}{4} p = 1 \]

Now, let's combine the terms on the left side. To do this, we need a common denominator. The least common multiple of 1, 2, and 4 is 4. Therefore, we can rewrite the equation as:

\[ p + \frac{3}{2} p + \frac{3}{4} p = \frac{4}{4} p + \frac{6}{4} p + \frac{3}{4} p \]

Now, adding these fractions together:

\[ \frac{4 + 6 + 3}{4} p = \frac{13}{4} p \]

Thus, we have:

\[ \frac{13}{4} p = 1 \]

To find \( p \), we solve for it by multiplying both sides by \( \frac{4}{13} \):

\[ p = \frac{4}{13} \]

Therefore, the probability \( P(A) \) is:

\[ P(A) = \frac{4}{13} \]

### Summary of Steps:
1. Define \( P(A) = p \).
2. Express \( P(B) \) and \( P(C) \) in terms of \( p \).
3. Set up the equation \( P(A) + P(B) + P(C) = 1 \).
4. Substitute the expressions for \( P(B) \) and \( P(C) \).
5. Combine the terms using a common denominator.
6. Solve for \( p \).