Two numbers are selected randomly from the set S = {1, 2, 3, 4, 5, 6} without replacement one by one

Q: Wait, why do we use $6 \times 5$ for the total outcomes instead of $6 \times 6$?

That is a fantastic question! The problem specifies 'without replacement'. This means once you pick the first number, it cannot be picked again. For the first selection, you have 6 options. For the second, you have only 5 remaining. If it were 'with replacement', then yes, it would be $6 \times 6 = 36$.

Q: Why can't we just list all the pairs where the minimum is less than 4?

You absolutely can! But in JEE Advanced, time is your most precious resource. Listing pairs like $(1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3)...$ is prone to human error—you might miss one or double-count. The complementary method is a 'trap-catcher' that simplifies the logic, reducing the number of cases you have to manually track.

Q: Does the order of selection matter here?

It actually doesn't! Whether you treat this as permutations (where order matters) or combinations (where order doesn't), the probability remains the same. We used permutations here ($6 \times 5$ and $3 \times 2$), but if you used combinations ($^{6}C_{2}$ and $^{3}C_{2}$), you would get the exact same result. The key is to be consistent in both the numerator and the denominator.

Question

Two numbers are selected randomly from the set $S = \{1, 2, 3, 4, 5, 6\}$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is

Options:

(A) 1/15

(B) 14/15

(C) 1/5

(D) 4/5

Sigma Percentile · Accepted Answer

### The Art of the Complement: A Probability Masterclass

Welcome, future engineer. Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, elegant shortcut. We are dealing with a set $S = \{1, 2, 3, 4, 5, 6\}$ and we are selecting two numbers without replacement.

The goal is to find the probability that the minimum of these two numbers is less than $4$. Let's embark on this journey.

### Phase 1: Visualizing the Sample Space

Imagine you have a bag containing six balls, numbered $1$ through $6$. You reach in and pull out one ball, then another, without putting the first one back. This is the definition of "without replacement."

To find the total number of possible outcomes, we use the fundamental counting principle. For the first pick, you have $6$ choices, and for the second pick, you have $5$ choices. Thus, the total number of outcomes $N$ is:

$$N = 6 	imes 5 = 30$$

This is our universe. Every possible pair of numbers we could pick exists within these $30$ outcomes.

### Phase 2: The Trap of Direct Calculation

Now, consider the condition: we want the minimum of the two numbers to be less than $4$. If you were to calculate this directly, you would have to consider cases where the minimum is $1$, $2$, or $3$.

You would be listing pairs like $(1,2), (1,3), (2,1), (3,1)$, and so on. While this is mathematically sound, it is a path filled with potential pitfalls. In the high-pressure environment of the JEE, we want to minimize the number of steps where a simple arithmetic error could cost us marks.

### Phase 3: The Power of the Complement

This is where we pivot to a more sophisticated strategy: the complementary event. In probability, the sum of the probability of an event $A$ and its complement $A'$ is always $1$. Mathematically, we express this as:

$$P(A) = 1 - P(A')$$

Here, our event $A$ is $\min < 4$. The complement $A'$ is the opposite: $\min \ge 4$.

Think about what this means. If the minimum of two numbers is $4$ or greater, it implies that **both** numbers must be $4$ or greater. If even one number were less than $4$, the minimum would be less than $4$, violating our condition for the complement.

### Phase 4: Executing the Logic

Our condition for the complement $A'$ is that both numbers must be chosen from the subset $\{4, 5, 6\}$. This subset has $3$ elements. We need to choose two numbers from this subset without replacement.

Following the same logic as before, the number of favorable outcomes for our complement $n(A')$ is:

$$n(A') = 3 	imes 2 = 6$$

Now, we calculate the probability of this complementary event:

$$P(A') = \frac{n(A')}{N} = \frac{6}{30} = \frac{1}{5}$$

### Phase 5: The Final Victory

We are almost there. We have the probability of the complement. To find the probability of our original event, we simply subtract this from $1$:

$$P(\min < 4) = 1 - P(\min \ge 4) = 1 - \frac{1}{5} = \frac{4}{5}$$

And there it is. By choosing the path of the complement, we turned a tedious counting exercise into a swift, elegant calculation.

Remember, in physics and mathematics, the most complex problems often yield to the simplest, most elegant logic. The final answer is **$\frac{4}{5}$**. Keep practicing this mindset, and you will master the JEE.

Animated Solution for Mathematics - Probability: Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is

Select Answer:

Visualized Solution

Understanding the Set $S$

Calculating Total Outcomes

The Complementary Event Strategy

Defining the Condition $min \geq 4$

Calculating Favorable Outcomes

Probability of Complementary Event

Final Calculation

Key Takeaway

The Sigma Insight: Classical Definition of Probability

The Art of the Complement

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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