For a student to qualify he must pass at least two out of three exams. The probability that he will

Q: Wait, why do we add these probabilities? Shouldn't we multiply them?

That is a fantastic question! We multiply probabilities when events happen in sequence *within* a single path (like Pass AND Pass AND Pass). We add them when we are looking at *different, mutually exclusive* ways to achieve the same goal (like Path 1 OR Path 2 OR Path 3). Since you cannot pass all three exams AND pass exactly two at the same time, these paths are mutually exclusive, so we add them together.

Q: Why does the probability change after a failure? Is this realistic?

In the world of JEE probability, we model systems with 'memory.' This is a classic conditional probability setup. The problem explicitly states that a failure impacts the next attempt's probability. Always read the problem statement for these 'memory' rules—they are the keys to the kingdom! It models the psychological impact of failure, which is a very real phenomenon in competitive exams.

Q: Can I use the complement rule (1 minus the probability of failing) to solve this?

You absolutely can! The complement of 'passing at least two' is 'passing zero' or 'passing exactly one.' However, be careful: you must calculate the probabilities of $FFF, FFP, FPF, PFF$ correctly, keeping the conditional rules in mind. Sometimes the direct method is safer to avoid missing a case, but using the complement is a great way to verify your answer!

Question

For a student to qualify he must pass at least two out of three exams. The probability that he will pass the $1^{st}$ exam is $p$. If he fails in one of the exams then the probability of his passing in the next exam is $p/2$ otherwise it remains the same. Find the probability that he will qualify.

Sigma Percentile · Accepted Answer

### The Anatomy of Success: Navigating Conditional Probability

Welcome, future engineer. Today, we are not just solving a probability problem; we are dissecting the anatomy of success. In the high-stakes environment of the JEE, probability is not just about rolling dice or flipping coins—it is about understanding the flow of events and how the past dictates the future.

This problem is a beautiful exercise in conditional probability, where the outcome of one exam ripples forward to affect the next. Let us embark on this journey together.

### Phase 1: Defining the Goal

Imagine you are the student sitting in that exam hall. The criteria for qualification is clear: you must pass at least two out of three exams. In the language of sets and logic, 'at least two' means you can pass exactly two, or you can pass all three.

Let us map out the universe of success. We denote a Pass as $P$ and a Failure as $F$. To qualify, our successful paths are:

1. $PPP$ (The perfect run)
2. $PPF$ (The stumble at the end)
3. $PFP$ (The recovery)
4. $FPP$ (The comeback)

Any other combination, such as $FFP$ or $FFF$, leads to disqualification. Our mission is to calculate the probability of each of these four mutually exclusive paths and sum them up. This is the bedrock of our strategy.

### Phase 2: The Conditional Chain

Here is where the problem gets interesting. We are told that the probability of passing the first exam is $p$. However, the system has 'memory.'

If you fail, your confidence—and thus your probability of passing the next exam—drops to $\frac{p}{2}$. If you pass, you maintain your momentum, and the probability remains $p$. This is a chain of dependent events.

### Phase 3: Calculating the Four Paths

**Case 1: The Perfect Run ($PPP$)**
This is the simplest path. You pass the first exam with probability $p$. Since you passed, the probability of passing the second remains $p$, and the probability of passing the third remains $p$.

$$P(PPP) = p \cdot p \cdot p = p^3$$

**Case 2: The Stumble at the End ($PPF$)**
Here, you pass the first ($p$), pass the second ($p$), but fail the third. Since you passed the second, the probability of passing the third was $p$, which means the probability of failing is $(1 - p)$.

$$P(PPF) = p \cdot p \cdot (1 - p) = p^2 - p^3$$

**Case 3: The Recovery ($PFP$)**
This is where the conditional rule kicks in. You pass the first ($p$) and fail the second ($1 - p$). Because you failed the second, the probability of passing the third drops to $\frac{p}{2}$.

$$P(PFP) = p \cdot (1 - p) \cdot \frac{p}{2} = \frac{p^2}{2} - \frac{p^3}{2}$$

**Case 4: The Comeback ($FPP$)**
Finally, the comeback. You fail the first exam with probability $(1 - p)$. Because of this failure, the probability of passing the second exam drops to $\frac{p}{2}$. You pass the second, and since you passed the second, the probability of passing the third returns to the base probability of $p$.

$$P(FPP) = (1 - p) \cdot \frac{p}{2} \cdot p = \frac{p^2}{2} - \frac{p^3}{2}$$

### Phase 4: The Grand Summation

Now, we bring it all together. Since these paths are mutually exclusive, we sum their probabilities to find the total probability of qualifying:

$$P(	ext{Qualify}) = P(PPP) + P(PPF) + P(PFP) + P(FPP)$$

Substituting our values:

$$P(	ext{Qualify}) = p^3 + (p^2 - p^3) + \left(\frac{p^2}{2} - \frac{p^3}{2}ight) + \left(\frac{p^2}{2} - \frac{p^3}{2}ight)$$

Let us group the terms with elegance. First, the $p^2$ terms:

$$p^2 + \frac{p^2}{2} + \frac{p^2}{2} = 2p^2$$

Next, the $p^3$ terms:

$$p^3 - p^3 - \frac{p^3}{2} - \frac{p^3}{2} = -p^3$$

Combining these, we arrive at our final, beautiful result:

**$$P(	ext{Qualify}) = 2p^2 - p^3$$**

### Final Thoughts

Look at that result. It is clean, elegant, and derived from a logical progression. When you face such problems in the exam, do not rush. Visualize the paths, respect the conditional rules, and trust your algebra. You have just mastered the art of conditional probability.

Visualized Solution

Understanding the Qualification Criteria

The Rules of Probability Shift

Case 1: Pass, Pass, Pass ( $P P P$ )

Case 2: Pass, Pass, Fail ( $P P F$ )

Case 3: Pass, Fail, Pass ( $P F P$ )

Case 4: Fail, Pass, Pass ( $F P P$ )

Combining All Mutually Exclusive Cases

Simplifying the Expression

The Sigma Insight: Addition and Multiplication Theorems

The Anatomy of Success

Phase 1

Phase 2

Phase 3

Phase 4

Final Thoughts

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