Fifteen football players of a club-team are given 15 T-shirts with their names written on the backsi

Q: Wait, why do we use the complement rule here? Can't we just calculate $P(X \ge 3)$ directly?

You absolutely could, but it would be a nightmare! Calculating $P(X \ge 3)$ requires summing the probabilities for $X=3, 4, 5, \dots, 15$. That is 13 separate, complex calculations. By using the complement rule, $P(X \ge 3) = 1 - P(X < 3)$, you only need to calculate 3 cases ($X=0, 1, 2$). In competitive exams, choosing the path of least resistance is a skill just as important as knowing the formulas.

Q: Why does the term $e$ appear in the probability formula? I thought this was just combinatorics.

That is a brilliant observation! The number of derangements $D_n$ is approximately $\frac{n!}{e}$. When we divide by $n!$ to get the probability, the $n!$ cancels out, leaving us with $\frac{1}{e}$. This is a beautiful connection between discrete combinatorics and the transcendental number $e$. For large $n$, the distribution of matches follows a Poisson distribution with a mean of 1, which is why $e^{-1}$ appears so naturally.

Q: Why are we approximating $e \approx 3$? Isn't $e \approx 2.718$?

In a real-world scenario, you would use $2.718$. However, this is a specific type of JEE problem where the options are simple fractions. The examiners have designed the problem such that using the integer approximation $e \approx 3$ leads you directly to one of the given options. It is a common 'meta-game' in competitive exams to recognize when an approximation is intended to simplify the arithmetic.

Question

Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is

Options:

(A) $\frac{5}{24}$

(B) $\frac{2}{15}$

(C) $\frac{1}{6}$

(D) $\frac{5}{36}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a football locker room. There are fifteen players and fifteen T-shirts, each bearing a player's name on the back. Chaos ensues as everyone grabs a shirt at random.

You are asked: what is the probability that at least three players end up with their own shirt? This is a classic encounter with the concept of **derangements**, and it holds a beautiful mathematical secret.

### The Strategy: Working Smarter, Not Harder

When faced with the prospect of calculating the probability of at least three matches, your first instinct might be to calculate the probability for exactly three, then four, then five, and so on, all the way to fifteen. Stop! That is the trap.

In the world of probability, whenever you see "at least," your mind should immediately jump to the **complement rule**:

$$P(A) = 1 - P(A^c)$$

Here, the complement of "at least 3" is "fewer than 3." This means we only need to calculate the probabilities for $X=0$, $X=1$, and $X=2$. By finding these three values and subtracting their sum from 1, we bypass the tedious work of calculating the higher-order matches.

### The Derangement Magic

To solve this, we need the derangement formula. A derangement is a permutation where no element appears in its original position. The probability of exactly $r$ matches in a set of $n$ items is given by:

$$P(X=r) = \frac{\binom{n}{r} D_{n-r}}{n!}$$

When we expand the derangement term $D_{n-r}$ as $(n-r)! \sum_{i=0}^{n-r} \frac{(-1)^i}{i!}$, something magical happens. The $n!$ and $(n-r)!$ terms cancel out, leaving us with a remarkably clean expression:

$$P(X=r) = \frac{1}{r!} \sum_{i=0}^{n-r} \frac{(-1)^i}{i!}$$

### The Poisson Connection

Now, look at that alternating sum. For a large $n$ (and 15 is large enough for this approximation), the sum $\sum_{i=0}^{n-r} \frac{(-1)^i}{i!}$ is the Taylor series expansion for $e^{-1}$.

Thus, the probability of exactly $r$ matches simplifies to:

$$P(X=r) \approx \frac{1}{r!e}$$

This is the heart of the problem. It tells us that the probability of having $r$ matches is inversely proportional to $r!$ and the constant $e$.

### The Final Calculation

Let's apply this to our cases:

For $r=0$: 
$$P(X=0) \approx \frac{1}{0!e} = \frac{1}{e}$$

For $r=1$: 
$$P(X=1) \approx \frac{1}{1!e} = \frac{1}{e}$$

For $r=2$: 
$$P(X=2) \approx \frac{1}{2!e} = \frac{1}{2e}$$

Summing these up, we get:

$$P(X < 3) \approx \frac{1}{e} + \frac{1}{e} + \frac{1}{2e} = \frac{5}{2e}$$

### The JEE Twist

Finally, we return to our complement rule:

$$P(X \ge 3) = 1 - P(X < 3) \approx 1 - \frac{5}{2e}$$

If you look at the options provided, you will notice they are all simple fractions. This is the final hurdle. The exam setters expect you to use the approximation $e \approx 3$.

Substituting this into our result, we get:

$$1 - \frac{5}{2(3)} = 1 - \frac{5}{6} = \frac{1}{6}$$

And there it is—the elegance of the solution, hidden behind a clever approximation. You have navigated the trap, used the derangement formula, and arrived at the **final answer of $\frac{1}{6}$**.

Animated Solution for Mathematics - Probability: Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is

Select Answer:

Visualized Solution

Problem Identification: $n = 15$

Complementary Event Strategy

Defining the Complement Cases

The Derangement Formula

Simplifying the Probability Expression

Large $n$ Approximation

Calculating $P (X = 0)$

Calculating $P (X = 1)$

Calculating $P (X = 2)$

Summing the Probabilities

Final Substitution

Applying the $e \approx 3$ Approximation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Strategy

The Derangement Magic

The Poisson Connection

The Final Calculation

The JEE Twist

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