In a hurdle race, a runner has probability `p` of jumping over a specific hurdle. Given that in `5` trials, the runner succeeded `3` times, the conditional probability that the runner had succeeded in the first trial is

Options

A. `3//5`
B. `2//5`
C. `1//5`
D. None of these

Explanation

To solve the problem, we need to find the conditional probability that the runner succeeded in the first trial given that he succeeded exactly 3 times in 5 trials. We will denote the events as follows:

- Let \( A \) be the event that the runner succeeded exactly 3 times in 5 trials.
- Let \( B \) be the event that the runner succeeded in the first trial.

We need to find \( P(B|A) \), which is the conditional probability of \( B \) given \( A \). According to the definition of conditional probability:

\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]

### Step 1: Calculate \( P(A \cap B) \)

The event \( A \cap B \) means that the runner succeeded in the first trial and succeeded exactly 3 times in total out of 5 trials. Since the first trial is a success, we need to find the probability that the runner succeeds in 2 out of the remaining 4 trials.

1. The probability of success in the first trial is \( p \).
2. We need to choose 2 successes from the remaining 4 trials. The number of ways to choose 2 successes from 4 trials is given by \( \binom{4}{2} \).
3. The probability of success in the 2 chosen trials is \( p^2 \) and the probability of failure in the remaining 2 trials is \( (1-p)^2 \).

Thus, we can express \( P(A \cap B) \) as:

\[ P(A \cap B) = p \cdot \binom{4}{2} \cdot p^2 \cdot (1-p)^2 \]

Calculating \( \binom{4}{2} = 6 \):

\[ P(A \cap B) = p \cdot 6 \cdot p^2 \cdot (1-p)^2 = 6p^3(1-p)^2 \]

### Step 2: Calculate \( P(A) \)

The event \( A \) means that the runner succeeded exactly 3 times in 5 trials. The total number of ways to choose 3 successes from 5 trials is given by \( \binom{5}{3} \).

Thus, we can express \( P(A) \) as:

\[ P(A) = \binom{5}{3} \cdot p^3 \cdot (1-p)^2 \]

Calculating \( \binom{5}{3} = 10 \):

\[ P(A) = 10 \cdot p^3 \cdot (1-p)^2 \]

### Step 3: Calculate \( P(B|A) \)

Now we can substitute \( P(A \cap B) \) and \( P(A) \) into the formula for conditional probability:

\[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{6p^3(1-p)^2}{10p^3(1-p)^2} \]

The \( p^3(1-p)^2 \) terms cancel out:

\[ P(B|A) = \frac{6}{10} = \frac{3}{5} \]

### Final Answer

Thus, the conditional probability that the runner succeeded in the first trial given that he succeeded exactly 3 times in 5 trials is:

\[ \boxed{\frac{3}{5}} \]