Q14: If X is random Poisson variate such that P(X=1) =... - Statistical Methods & Experimental Designs - ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper

ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper

MCQ (Single Correct) +4 −1

If X is random Poisson variate such that P(X=1) = P(X=2). What is the P(X=4) ?

यदि X एक ऐसा पॉइसन यादृच्छिक चर (Poisson random variate) है कि P(X=1) = P(X=2), तो P(X=4) का मान क्या होगा?

(5e^-2)/6 (5e^-2)/6 (2e^-2)/2 (2e^-2)/2 (e^-2)/3 (e^-2)/3 (2e^-2)/3 (2e^-2)/3

Step-by-step mathematical evaluation:
For a Poisson variate with parameter $\lambda$:

$$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$$

Given $P(X=1) = P(X=2)$:

$$\frac{e^{-\lambda}\lambda^1}{1!} = \frac{e^{-\lambda}\lambda^2}{2!}$$ $$\lambda = \frac{\lambda^2}{2} \implies \lambda = 2$$

Now, evaluate $P(X=4)$ with $\lambda = 2$:

$$P(X=4) = \frac{e^{-2} \cdot 2^4}{4!} = \frac{e^{-2} \cdot 16}{24} = \frac{2 e^{-2}}{3}$$

पॉइसन बंटन (Poisson Distribution) की चरणबद्ध गणना:
पॉइसन चर हेतु प्रायिकता द्रव्यमान फलन:

$$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$$

दिया है $P(X=1) = P(X=2)$:

$$\frac{e^{-\lambda}\lambda^1}{1!} = \frac{e^{-\lambda}\lambda^2}{2!} \implies \lambda = \frac{\lambda^2}{2} \implies \lambda = 2$$

अब $\lambda = 2$ के लिए $P(X=4)$ की गणना:

$$P(X=4) = \frac{e^{-2} \cdot 2^4}{4!} = \frac{16 \cdot e^{-2}}{24} = \frac{2 e^{-2}}{3}$$

Explanation not matching this question? क्या व्याख्या इस प्रश्न से मेल नहीं खा रही? Report Mismatch त्रुटि बताएं

4 Exams

1 Exam

4 Exams

Previous Year Papers of Statistical Sciences