The random variable X follows binomial distribution B(n, p), for which the difference of the mean an

Q: Wait, why do we reject the root $p=1$?

That is a fantastic catch! In a binomial distribution $B(n, p)$, $q$ represents the probability of failure, where $q = 1 - p$. If $p = 1$, then $q = 0$. This would mean the probability of failure is zero, effectively collapsing the binomial distribution into a trivial state where every trial is a guaranteed success. Furthermore, the problem asks us to evaluate $P(X=2)$, which implies that the random variable must be able to take the value 2, requiring $n \ge 2$. A distribution with $p=1$ and $n=1$ (which is what we get if we plug $p=1$ into our derived relation) fails these logical constraints. Always check your boundary conditions!

Q: Why can't we just calculate $P(X=2) + P(X=3) + P(X=4)$ for the final step?

You absolutely can! There is no mathematical law forbidding that path. However, as an engineer and a mathematician, we always look for the path of least resistance. Calculating $P(X=2)$, $P(X=3)$, and $P(X=4)$ involves three separate binomial calculations. Using the complement rule, $P(X > 1) = 1 - P(X=0) - P(X=1)$, requires only two. In the heat of a JEE Advanced exam, saving time and reducing the number of places where a calculation error can creep in is the hallmark of a top-tier student.

Q: Is there a faster way to handle the mean-variance relation?

Yes, there is! You can memorize the identity $\mu - \sigma^2 = np - npq = np(1-q) = np^2$. If you remember that the difference between the mean and variance of a binomial distribution is always $np^2$, you can skip the derivation steps entirely. However, I always encourage deriving it on the fly during the exam to ensure you don't misremember the formula under pressure.

Question

The random variable $X$ follows binomial distribution $B(n, p)$, for which the difference of the mean and the variance is 1. If $2 P(X = 2) = 3 P(X = 1)$, then $n^2 P(X > 1)$ is equal to

Options:

(A) 15

(B) 11

(C) 12

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The Dance of Probability: Unlocking the Binomial Mystery. Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a binomial distribution to reveal the elegant structure hidden beneath.

This problem is a classic JEE Advanced favorite because it tests your algebraic stamina and your conceptual grasp of probability. Let us begin.

### The Mean-Variance Trap

We start with a random variable $X \sim B(n, p)$. The problem gives us a gift: the difference between the mean and the variance is $1$.

We know the mean is $\mu = np$ and the variance is $\sigma^2 = npq$. The condition is $\mu - \sigma^2 = 1$. Substituting our formulas, we get:

$$np - npq = 1$$

Now, factor out the $np$: $np(1 - q) = 1$. Since $p + q = 1$, we know that $1 - q = p$. Suddenly, the equation simplifies to:

$$np^2 = 1$$

Keep this relationship, $n = \frac{1}{p^2}$, safe in your mind; we will need it later.

### The Ratio Dance

Next, we are given the condition $2 P(X = 2) = 3 P(X = 1)$. This looks intimidating, but let us break it down using the binomial probability formula: $P(X = k) = \binom{n}{k} p^k q^{n-k}$.

Expanding both sides, we get:

$$2 \left[ \binom{n}{2} p^2 q^{n-2} ight] = 3 \left[ \binom{n}{1} p^1 q^{n-1} ight]$$

Do not let the combinations scare you. Remember that $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n}{1} = n$. Substituting these, the equation becomes:

$$2 \cdot \frac{n(n-1)}{2} p^2 q^{n-2} = 3 \cdot n p q^{n-1}$$

Notice how the $2$s cancel out on the left? It is almost as if the problem was designed to be solved. We are left with:

$$n(n-1) p^2 q^{n-2} = 3 n p q^{n-1}$$

### The Algebraic Cleanup

Now, we perform the great cancellation. Divide both sides by $n p q^{n-2}$. Assuming $n 
eq 0$ and $p, q 
eq 0$, we are left with:

$$(n-1)p = 3q$$

This is beautiful. We have reduced a complex probability equation into a simple linear relationship. Substitute $q = 1 - p$ to get $(n-1)p = 3(1 - p)$.

Expanding this, we get $np - p = 3 - 3p$, which rearranges to:

$$np + 2p = 3$$

### The Final Convergence

We have two equations now: $np^2 = 1$ and $np + 2p = 3$. From the first, $np = \frac{1}{p}$. Substitute this into the second:

$$\frac{1}{p} + 2p = 3$$

Multiply by $p$ to clear the fraction: $1 + 2p^2 = 3p$, or $2p^2 - 3p + 1 = 0$. Factoring this quadratic gives:

$$(2p - 1)(p - 1) = 0$$

We reject $p=1$ (as discussed in our FAQs) and accept $p = \frac{1}{2}$. Plugging this back into $n = \frac{1}{p^2}$, we find:

$$n = \frac{1}{(1/2)^2} = 4$$

### The Victory Lap

We have our parameters: $n=4$ and $p=1/2$. The question asks for $n^2 P(X > 1)$. Since $n=4$, $n^2 = 16$.

For $P(X > 1)$, we use the complement rule: $1 - P(X=0) - P(X=1)$. Calculating these:

$$P(X=0) = \binom{4}{0} (1/2)^0 (1/2)^4 = \frac{1}{16}$$

$$P(X=1) = \binom{4}{1} (1/2)^1 (1/2)^3 = \frac{4}{16}$$

Thus, $P(X > 1) = 1 - \frac{1}{16} - \frac{4}{16} = \frac{11}{16}$. Finally:

$$16 	imes \frac{11}{16} = 11$$

You have navigated the complexity and arrived at the truth. The final answer is **11**. Well done.

Animated Solution for Mathematics - Probability: The random variable $X$ follows binomial distribution $B (n, p)$ , for which the difference of the mean and the variance is 1. If $2 P (X = 2) = 3 P (X = 1)$ , then $n^{2} P (X > 1)$ is equal to

Select Answer:

Visualized Solution

Binomial Distribution Parameters

Simplifying Mean and Variance

The Probability Condition

Expanding the Probabilities

Simplifying Combinations

Canceling Common Terms

Expressing in terms of p

Substituting n

Solving the Quadratic

Finding n and p

Target Expression

Calculating Probabilities

Final Answer

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Mean-Variance Trap

The Ratio Dance

The Algebraic Cleanup

The Final Convergence

The Victory Lap

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