Two distinct numbers a and b are selected at random from {1, 2, 3, ..., 50}. The probability, that t

Q: Wait, why do we use the complement? Can't we just count the cases where one is a multiple of 3 and the other isn't, plus the case where both are?

You absolutely can! That is the 'direct' method. You would calculate $\binom{16}{1} \times \binom{34}{1} + \binom{16}{2}$. Both methods are mathematically sound, but the complement method is often faster and less prone to calculation errors in high-pressure exams like the JEE.

Q: How do I know how many multiples of 3 are in the set? Is there a trick?

The trick is the floor function! For any range from 1 to $N$, the number of multiples of $k$ is simply $\lfloor \frac{N}{k} \rfloor$. Here, $\lfloor \frac{50}{3} \rfloor = 16$. It works every time because it counts how many 'groups of 3' fit into 50.

Q: Why is the total number of ways $\binom{50}{2}$ and not $50 \times 49$?

Great question! $\binom{50}{2}$ represents selecting a 'set' of two numbers where order doesn't matter (choosing {3, 6} is the same as {6, 3}). If you used $50 \times 49$, you would be counting permutations, which would double-count every pair. As long as you are consistent—using combinations for both the numerator and denominator—you will get the correct probability.

Q: Does the fact that the numbers must be 'distinct' change the formula?

Yes, it is crucial! The word 'distinct' tells us we are sampling without replacement. If we could pick the same number twice, we would use $50^2$ as our sample space. Always look for keywords like 'distinct' or 'different' in probability problems!

Q: Is there any scenario where the product is divisible by 3 but neither number is a multiple of 3?

No, and that is the beauty of prime numbers! Since 3 is a prime number, if a product $ab$ is divisible by 3, then by Euclid's Lemma, at least one of the factors $a$ or $b$ must be divisible by 3. This is why our complement strategy is perfectly valid.

Question

Two distinct numbers $a$ and $b$ are selected at random from $\{1, 2, 3, \dots, 50\}$. The probability, that their product $ab$ is divisible by 3, is :

Options:

(A) $\frac{272}{1225}$

(B) $\frac{561}{1225}$

(C) $\frac{664}{1225}$

(D) $\frac{8}{25}$

Sigma Percentile · Accepted Answer

### The Elegance of the Complementary Path

Imagine you are standing before a vast collection of 50 numbered tiles, ranging from 1 to 50. You are tasked with picking two distinct tiles. The challenge is to find the probability that their product is divisible by 3.

At first glance, you might feel the urge to start listing pairs—$(3,1), (3,2), (6,1), (6,2) \dots$ but stop. In the world of JEE Advanced, brute force is rarely the path to victory. Instead, let us look for the 'hidden geometry' of the problem.

### The Power of the Complement

When a problem asks for the probability of an event occurring, we often find ourselves staring at a mountain of possibilities. But what if we looked at the valley instead?

The complement of our event—the scenario where the product $ab$ is **NOT** divisible by 3—is much easier to visualize. If the product of two numbers is not divisible by 3, then neither $a$ nor $b$ can contain a factor of 3.

Because 3 is prime, if it does not divide $a$ and it does not divide $b$, it cannot possibly divide their product. This is the fundamental beauty of prime numbers.

### Mapping the Territory

First, let us define our universe. We have 50 numbers. The total number of ways to choose any two distinct numbers is given by the combination formula:

$$
\binom{50}{2} = \frac{50 	imes 49}{2 	imes 1} = 1225
$$

This is our denominator, the total number of possible outcomes. Now, let us identify the 'multiples of 3' within our set.

By calculating $\lfloor \frac{50}{3} floor$, we find there are 16 such numbers: $\{3, 6, 9, \dots, 48\}$. Conversely, the numbers that are **NOT** multiples of 3 are $50 - 16 = 34$.

### The Execution

If we want the product to **NOT** be divisible by 3, we must choose both of our numbers from the pool of 34 non-multiples. The number of ways to do this is:

$$
\binom{34}{2} = \frac{34 	imes 33}{2 	imes 1} = 17 	imes 33 = 561
$$

So, the probability of the 'bad' event (the product not being divisible by 3) is simply the ratio of these favorable outcomes to the total:

$$
P(E^c) = \frac{561}{1225}
$$

### The Final Victory

We are almost there. We know that the sum of the probability of an event and its complement must equal 1. Therefore, the probability that the product **IS** divisible by 3 is:

$$
P(E) = 1 - P(E^c) = 1 - \frac{561}{1225}
$$

Performing the final subtraction, we arrive at:

$$
P(E) = \frac{1225 - 561}{1225} = \frac{664}{1225}
$$

Look at that result. It is not just a fraction; it is the culmination of logical deduction. By choosing to count what we didn't want, we saved ourselves from a labyrinth of complex cases.

Remember, in physics and mathematics, the most elegant solution is often the one that simplifies the problem into its most basic, fundamental truths. You have mastered the logic—now carry this confidence into your next challenge!

Animated Solution for Mathematics - Probability: Two distinct numbers $a$ and $b$ are selected at random from ${1, 2, 3, \dots, 50}$ . The probability, that their product $ab$ is divisible by 3, is :

Select Answer:

Visualized Solution

Sample Space $S = {1, 2, \dots, 50}$

Total Ways $(2 50)$

Evaluating $(2 50)$

Condition for $ab$ to be Divisible by $3$

Counting Multiples of $3$

Counting Non-multiples of $3$

Complementary Event Strategy

Ways for Neither to be Divisible by $3$

Evaluating $(2 34)$

Probability of Complement $P (E^{c})$

Final Probability $P (E)$

The Sigma Insight: Classical Definition of Probability

The Elegance of the Complementary Path

The Power of the Complement

Mapping the Territory

The Execution

The Final Victory

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Select Answer:

Visualized Solution

Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S={1,2,…,50}

Total Ways .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (250​)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (250​)

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ab to be Divisible by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Counting Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Counting Non-multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3