In a series of three trials , the probability of two successes is 9 times the probability of three successes . Then the probability of success in each trial is :

Options

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

Answer

To solve the problem, we need to find the probability of success in each trial, denoted as \( p \), given that the probability of two successes in three trials is nine times the probability of three successes.

### Step-by-Step Solution:

1. **Understand the Problem**: We are given that the probability of two successes is nine times the probability of three successes in three trials.

2. **Use the Binomial Probability Formula**: The probability of getting exactly \( k \) successes in \( n \) trials is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

3. **Set Up the Equations**:
- For two successes (\( k = 2 \)):
\[ P(X = 2) = \binom{3}{2} p^2 (1-p)^{1} \]
- For three successes (\( k = 3 \)):
\[ P(X = 3) = \binom{3}{3} p^3 (1-p)^{0} \]

4. **Calculate the Binomial Coefficients**:
- \( \binom{3}{2} = 3 \)
- \( \binom{3}{3} = 1 \)

5. **Substitute the Values into the Equations**:
- The probability of two successes becomes:
\[ P(X = 2) = 3 p^2 (1-p) \]
- The probability of three successes becomes:
\[ P(X = 3) = 1 \cdot p^3 = p^3 \]

6. **Set Up the Given Relationship**:
According to the problem, we have:
\[ P(X = 2) = 9 \cdot P(X = 3) \]
Substituting the expressions we derived:
\[ 3 p^2 (1-p) = 9 p^3 \]

7. **Simplify the Equation**:
Rearranging gives:
\[ 3 p^2 (1-p) = 9 p^3 \]
Dividing both sides by \( p^2 \) (assuming \( p \neq 0 \)):
\[ 3(1-p) = 9p \]

8. **Solve for \( p \)**:
Expanding and rearranging:
\[ 3 - 3p = 9p \]
\[ 3 = 12p \]
\[ p = \frac{3}{12} = \frac{1}{4} \]

9. **Conclusion**:
The probability of success in each trial is:
\[ p = \frac{1}{4} \]