A student appears for tests I, II and III. The student is successful if he passes either in tests I

Q: Wait, why do we use the distributive law here? Can't I just add the probabilities of the two events?

That is a classic trap! If you simply add $P(A \cap B)$ and $P(A \cap C)$, you are double-counting the region where the student passes all three tests ($A \cap B \cap C$). The distributive law $A \cap (B \cup C)$ elegantly avoids this by grouping the logic correctly before you even touch the probability values.

Q: How do I know for sure that the tests are independent?

In probability problems of this type, unless stated otherwise, we assume that the outcome of one test does not influence the outcome of another. This independence is what allows us to write $P(A \cap B) = P(A)P(B)$ and $P(B \cap C) = P(B)P(C)$. Without this assumption, the problem would be unsolvable with the given information.

Q: Why do we check the options instead of solving for $p$ and $q$ directly?

Great observation! We have one equation: $p(q + 1) = 1$, but two variables, $p$ and $q$. This is an underdetermined system. However, we have a hidden constraint: $0 \le p \le 1$ and $0 \le q \le 1$. By testing the options, we find the only pair that satisfies both the equation and the fundamental axioms of probability.

Question

A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are $p, q$ and $\frac{1}{2}$ respectively. If the probability that the student is successful is $\frac{1}{2}$, then

Options:

(A) $p = q = 1$

(B) $p = q = \frac{1}{2}$

(C) $p = 1, q = 0$

(D) $p = 1, q = \frac{1}{2}$

(E) none of these

Sigma Percentile · Accepted Answer

### The Geometry of Success: A Probability Odyssey

Welcome, future engineer! Today, we are not just solving a probability problem; we are dissecting the anatomy of success. In the world of JEE Advanced, probability is often less about complex formulas and more about the clarity of your visualization.

Let us embark on this journey together.

### Phase 1: Defining the Landscape

Imagine you are a student preparing for three distinct tests. We define our events simply: let $A$, $B$, and $C$ be the events of passing tests I, II, and III, respectively.

We are given the probabilities: $P(A) = p$, $P(B) = q$, and $P(C) = \frac{1}{2}$. This is our starting point. We have three independent variables, and our goal is to find the relationship between $p$ and $q$ that satisfies the condition of success.

### Phase 2: The Success Condition

The problem states that the student is successful if they pass (I and II) OR (I and III). In the language of set theory, this is the union of two intersections: $P(	ext{Success}) = P((A \cap B) \cup (A \cap C))$.

If you try to calculate this directly, you might fall into the trap of double-counting the scenario where the student passes all three tests. This is where the beauty of set theory comes to our rescue. We use the **Distributive Law of Sets** to simplify our expression:

$$(A \cap B) \cup (A \cap C) = A \cap (B \cup C)$$

Look at that! By factoring out $A$, we have transformed a complex union of intersections into a simple intersection of $A$ with the union of $B$ and $C$. Physically, this means the student must pass Test I, AND they must pass at least one of the other two tests (II or III).

### Phase 3: The Power of Independence

Since the tests are independent, the probability of the intersection is the product of the probabilities. Thus, our success probability becomes:

$$P(	ext{Success}) = P(A) 	imes P(B \cup C)$$

Now, we need to calculate $P(B \cup C)$. Using the addition theorem, we know that $P(B \cup C) = P(B) + P(C) - P(B \cap C)$.

Because $B$ and $C$ are independent, $P(B \cap C) = P(B)P(C)$. Substituting our known values, $P(B) = q$ and $P(C) = \frac{1}{2}$, we get:

$$P(B \cup C) = q + \frac{1}{2} - \left( q 	imes \frac{1}{2} ight) = q + \frac{1}{2} - \frac{q}{2} = \frac{q + 1}{2}$$

### Phase 4: The Final Equation

We are almost there. Substituting this back into our success equation, we have:

$$P(	ext{Success}) = p 	imes \frac{q + 1}{2}$$

We are given that the probability of success is $\frac{1}{2}$. Therefore:

$$\frac{p(q + 1)}{2} = \frac{1}{2} \implies p(q + 1) = 1$$

### Phase 5: The Boundary of Reality

We have one equation, $p(q + 1) = 1$, and two variables. We must remember the fundamental constraint of probability: $0 \le p \le 1$ and $0 \le q \le 1$.

If we look at the equation $p(q + 1) = 1$, we can rewrite it as $p = \frac{1}{q + 1}$. Since $q$ is a probability, $0 \le q \le 1$, which means $1 \le q + 1 \le 2$.

Consequently, $\frac{1}{2} \le p \le 1$. By testing the given options, we find that only **$p=1$ and $q=0$** satisfy the equation $1(0 + 1) = 1$. This is the only valid pair.

### Conclusion

Mathematics is often about finding the most elegant path through a forest of variables. By using the distributive law, we turned a potentially messy calculation into a simple, solvable equation. Keep this mindset—visualize, simplify, and then calculate. You are ready for the next challenge!

Select Answer:

Visualized Solution

Defining the Events

The Success Condition

Simplifying with Set Laws

Probability of Independent Events

The Addition Theorem

Substituting the Probabilities

Simplifying the Union

The Final Equation

Solving the Equation

Analyzing Probability Constraints

Testing the Options

The Sigma Insight: Addition and Multiplication Theorems

The Geometry of Success

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Conclusion

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