A candidate takes three tests in succession and the probability of passing the first test is p. The probability of passing each succeeding test is p or `(p)/(2)` according as he passes or fails in the preceding one. The candidate is selected, it he passes atleast two tests. The probability that the candidate is selected, is

Options

A. `p^(2)(2-p)`
B. `p(2-p)`
C. `p+p^(2)+p^(3)`
D. `p^(2)(1-p)`

Explanation

To solve the problem, we need to calculate the probability that the candidate passes at least 2 out of the 3 tests. We will break down the problem step by step.

### Step 1: Define the probabilities
- Let \( P \) be the probability of passing the first test.
- If the candidate passes the first test, the probability of passing the second test is \( P \).
- If the candidate fails the first test, the probability of passing the second test is \( \frac{P}{2} \).
- If the candidate passes the second test, the probability of passing the third test is \( P \).
- If the candidate fails the second test, the probability of passing the third test is \( \frac{P}{2} \).

### Step 2: Calculate the probabilities for each scenario
We will consider the different scenarios based on the results of the tests.

1. **Scenario 1: Pass, Pass, Pass (PPP)**
- Probability: \( P \times P \times P = P^3 \)

2. **Scenario 2: Pass, Pass, Fail (PPF)**
- Probability: \( P \times P \times (1 - P) = P^2(1 - P) \)

3. **Scenario 3: Pass, Fail, Pass (PFP)**
- Probability: \( P \times (1 - P) \times P = P^2(1 - P) \)

4. **Scenario 4: Pass, Fail, Fail (PFF)**
- Probability: \( P \times (1 - P) \times (1 - \frac{P}{2}) = P(1 - P)(1 - \frac{P}{2}) \)

5. **Scenario 5: Fail, Pass, Pass (FPP)**
- Probability: \( (1 - P) \times \frac{P}{2} \times P = \frac{P^2(1 - P)}{2} \)

6. **Scenario 6: Fail, Pass, Fail (FPF)**
- Probability: \( (1 - P) \times \frac{P}{2} \times (1 - P) = \frac{P(1 - P)^2}{2} \)

7. **Scenario 7: Fail, Fail, Pass (FFP)**
- Probability: \( (1 - P) \times (1 - \frac{P}{2}) \times \frac{P}{2} = (1 - P)(1 - \frac{P}{2})\frac{P}{2} \)

8. **Scenario 8: Fail, Fail, Fail (FFF)**
- Probability: \( (1 - P) \times (1 - \frac{P}{2}) \times (1 - \frac{P}{2}) = (1 - P)(1 - \frac{P}{2})^2 \)

### Step 3: Calculate the total probability of passing at least 2 tests
To find the total probability of passing at least 2 tests, we sum the probabilities of scenarios 1, 2, 3, 5, and 7:

\[ P(\text{selected}) = P^3 + P^2(1 - P) + P^2(1 - P) + \frac{P^2(1 - P)}{2} + (1 - P)(1 - \frac{P}{2})\frac{P}{2} \]

### Step 4: Simplify the expression
Combine the terms:

\[ P(\text{selected}) = P^3 + 2P^2(1 - P) + \frac{P^2(1 - P)}{2} + (1 - P)(1 - \frac{P}{2})\frac{P}{2} \]

This expression can be simplified further, but the exact simplification will depend on the value of \( P \).

### Final Answer
The final probability that the candidate is selected is given by the above expression, which can be computed for specific values of \( P \).