A random variable X has Poisson distribution with mean 2. Then P(X 1.5) equals (a) 2/e^2 (b) 0 (c)

Q: Wait, why do we change $X > 1.5$ to $X \ge 2$? Is that allowed?

Absolutely! This is the most critical step. A Poisson random variable represents counts of events, so it can only take non-negative integer values ($0, 1, 2, \dots$). Since $X$ cannot be $1.5$ or any other fraction, the first integer strictly greater than $1.5$ is $2$. Therefore, the event $X > 1.5$ is identical to the event $X \ge 2$.

Q: Why can't we just sum the probabilities from $2$ to infinity? Isn't that more direct?

While theoretically possible, it is practically impossible! Summing an infinite series is a recipe for disaster in an exam. The Complement Rule is your best friend here. By recognizing that the sum of all probabilities is $1$, we turn an infinite, impossible task into a simple calculation of just two terms ($X=0$ and $X=1$). Always look for the 'shortcut' that the Complement Rule provides.

Q: I always get confused with $0!$. Is it really $1$?

Yes, it is! By definition, $0! = 1$. It is a common point of confusion, but mathematically, it is essential for the Poisson formula to work. If $0!$ were $0$, the entire probability for $X=0$ would be undefined, which would break the distribution. Trust the definition: $0! = 1$.

Question

A random variable $X$ has Poisson distribution with mean 2. Then $P(X > 1.5)$ equals

Options:

(A) $2/e^2$

(B) 0

(C) 1 - 3/e^2

(D) 3/e^2

Sigma Percentile · Accepted Answer

### The Elegance of the Poisson Distribution Welcome, future engineer. Today, we are not just solving a probability problem; we are stepping into the shoes of a statistician modeling the randomness of the universe. The Poisson distribution is one of the most beautiful tools in our arsenal. It describes the number of events occurring in a fixed interval of time or space, provided these events happen with a known constant mean rate and independently of the time since the last event. Whether it is the number of emails you receive in an hour or the number of radioactive decays in a sample, the Poisson distribution is there. ### Phase 1: Decoding the Parameter $\lambda$ We are given that our random variable $X$ follows a Poisson distribution with a mean of $2$. In the language of statistics, we denote this mean as $\lambda = 2$. This $\lambda$ is the heartbeat of our distribution. It tells us the average intensity of the events we are observing. Our goal is to find $P(X > 1.5)$. ### Phase 2: The Discrete Trap Here is where many students stumble. They see $1.5$ and start thinking about continuous variables, perhaps even calculus. But stop! Take a breath. The Poisson distribution is discrete. It counts events. You cannot have $1.5$ emails or $1.5$ radioactive decays; you can only have $0, 1, 2, 3, \dots$ events. Therefore, when the problem asks for $P(X > 1.5)$, it is asking for the probability that $X$ takes any value strictly greater than $1.5$. On our discrete number line, the first integer that satisfies this is $2$. Thus, $P(X > 1.5)$ is exactly the same as $P(X \ge 2)$. ### Phase 3: The Power of the Complement Rule Now, we face a choice. We could calculate $P(X=2) + P(X=3) + P(X=4) + \dots$ all the way to infinity, but that is a trap! This is where we invoke the Complement Rule, a fundamental principle of probability. We know that the sum of all possible probabilities in a distribution must equal $1$. Therefore, the probability of our desired event is simply $1$ minus the probability of the unwanted events. Mathematically, we write this as: $$P(X \ge 2) = 1 - P(X < 2)$$ Since $X$ can only be $0$ or $1$ when it is less than $2$, our equation simplifies to: $$P(X > 1.5) = 1 - [P(X=0) + P(X=1)]$$ ### Phase 4: The Execution Now, we use the Probability Mass Function (PMF) for the Poisson distribution: $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ Let us calculate the unwanted terms. For $k=0$: $$P(X=0) = \frac{e^{-2} \cdot 2^0}{0!} = \frac{e^{-2} \cdot 1}{1} = e^{-2}$$ For $k=1$: $$P(X=1) = \frac{e^{-2} \cdot 2^1}{1!} = \frac{e^{-2} \cdot 2}{1} = 2e^{-2}$$ Summing these up, we get $P(X < 2) = e^{-2} + 2e^{-2} = 3e^{-2}$. Finally, we return to our complement equation: $$P(X > 1.5) = 1 - 3e^{-2}$$ To align with standard notation, we write this as **$1 - \frac{3}{e^2}$**. ### Final Thoughts Look at the elegance of that result. We started with a potentially infinite problem and reduced it to a simple, clean expression using the logic of discrete variables and the power of the complement rule. This is the essence of JEE Advanced physics and mathematics: it is not about brute force; it is about finding the most elegant path to the truth. Keep practicing, keep questioning, and keep falling in love with the logic behind the numbers.

Animated Solution for Mathematics - Probability: A random variable $X$ has Poisson distribution with mean 2. Then $P (X > 1.5)$ equals

Select Answer:

Visualized Solution

Identify the Distribution for $λ = 2$

Poisson Probability Mass Function

Analyze the Condition $X > 1.5$

Discrete Nature of $X$

The Complement Rule

Identify Unwanted Terms

Setup for $P (X = 0)$

Evaluate $P (X = 0)$

Setup for $P (X = 1)$

Evaluate $P (X = 1)$

Sum the Unwanted Probabilities

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

The Elegance of the Poisson Distribution

Phase 1

Phase 2

Phase 3

Phase 4

Final Thoughts

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Animated Solution for Mathematics - Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A random variable X has Poisson distribution with mean 2. Then P(X>1.5) equals