A problem in mathematics is given to three students A, B, C and their respective probability of solv

Q: Why can't we just add the probabilities $1/2 + 1/3 + 1/4$?

That is a classic trap! Adding probabilities directly assumes the events are mutually exclusive—meaning if one person solves it, the others cannot. But here, multiple students can solve the problem simultaneously. Adding them would count the overlapping cases multiple times, leading to a probability greater than 1, which is impossible!

Q: How do I know when to use the complement rule?

Whenever you see the phrase 'at least one' in a probability problem, your brain should immediately switch to the complement rule. It is almost always easier to calculate the probability of 'none' and subtract it from 1 than to sum up all the complex 'at least one' scenarios.

Q: Does the order in which they solve the problem matter?

Not at all! Probability is about the final outcome. Whether student A solves it first or last, the event 'problem solved' remains the same. The independence of the students ensures that their individual successes don't influence each other.

Q: What if the students were not independent?

If they were not independent, we could not simply multiply their failure probabilities. We would need conditional probabilities, like $P(\bar{B}|\bar{A})$. But in JEE problems, unless stated otherwise, assume individual efforts are independent.

Q: Is there any other way to solve this without the complement?

Yes, you could use the Inclusion-Exclusion Principle: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(A \cap C)] + P(A \cap B \cap C)$. While mathematically sound, it is much more prone to calculation errors than the complement method!

Question

A problem in mathematics is given to three students $A, B, C$ and their respective probability of solving the problem is $1/2, 1/3$ and $1/4$. Probability that the problem is solved is

Options:

(A) $3/4$

(B) 1/2

(C) 2/3

(D) 1/3

Sigma Percentile · Accepted Answer

### The Art of Thinking Backward: Mastering Probability

Welcome, future engineer! Today, we are going to tackle a classic problem that appears simple on the surface but hides a profound lesson in logical strategy.

We have three students, $A$, $B$, and $C$, with individual probabilities of solving a problem: $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, and $P(C) = \frac{1}{4}$. Our goal is to find the probability that the problem is solved.

### The Trap of Direct Calculation

When you first look at this, your instinct might be to calculate the probability that exactly one student solves it, then exactly two, then all three, and add them up. Stop! Take a deep breath.

While that path is mathematically valid, it is a labyrinth of complexity. In the JEE, time is your most precious resource. We need to find the 'elegant' path.

The problem asks for the probability that the problem is solved, which is equivalent to saying 'at least one student solves it.' Mathematically, we are looking for $P(A \cup B \cup C)$.

### The Power of the Complement

Instead of looking at the light, let's look at the shadow. What is the only scenario where the problem is NOT solved? It is the scenario where everyone fails.

If $P(	ext{Solved})$ is our target, then $P(	ext{Not Solved})$ is the probability that $A$ fails, AND $B$ fails, AND $C$ fails. The beauty of the Complementary Law is that:

$$P(	ext{Solved}) = 1 - P(	ext{None solve})$$

This transforms a massive summation problem into a simple subtraction.

### Calculating the Failure

If the probability of success for student $A$ is $P(A) = \frac{1}{2}$, then the probability of failure, $P(\bar{A})$, is simply $1 - \frac{1}{2} = \frac{1}{2}$.

Similarly, for student $B$, $P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$, and for student $C$, $P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4}$.

Because these students are working independently, the probability that they all fail simultaneously is the product of their individual failure probabilities:

$$P(	ext{None solve}) = P(\bar{A}) 	imes P(\bar{B}) 	imes P(\bar{C})$$

Substituting our values, we get:

$$P(	ext{None solve}) = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4}$$

### The Elegance of Cancellation

Look at this expression. Do you see the beauty? The $2$ in the numerator and denominator cancels out, and the $3$ in the numerator and denominator cancels out.

We are left with:

$$P(	ext{None solve}) = \frac{1}{4}$$

This is the probability of total failure. But we want the probability of success! We return to our complement equation:

$$P(	ext{Solved}) = 1 - P(	ext{None solve}) = 1 - \frac{1}{4} = \frac{3}{4}$$

### Final Reflections

There it is—the answer is **$\frac{3}{4}$**. By choosing the path of the complement, we avoided a mountain of tedious arithmetic and arrived at the solution with clarity and confidence.

Remember, in physics and mathematics, the most complex problems often yield to the simplest logical shifts. Keep practicing this 'backward' thinking; it will serve you well in the most challenging exams of your life.

Animated Solution for Mathematics - Probability: A problem in mathematics is given to three students $A, B, C$ and their respective probability of solving the problem is $1/2, 1/3$ and $1/4$ . Probability that the problem is solved is

Select Answer:

Visualized Solution

Given Probabilities

Defining "Problem Solved"

The Complementary Law

Probability of Failure

Calculating $P (\overset{ˉ}{A})$

Calculating $P (\overset{ˉ}{B})$ and $P (\overset{ˉ}{C})$

Independent Events

Probability of Total Failure

Calculating Total Failure

Final Substitution

The Final Result

The Sigma Insight: Addition and Multiplication Theorems

The Art of Thinking Backward

The Trap of Direct Calculation

The Power of the Complement

Calculating the Failure

The Elegance of Cancellation

Final Reflections

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