Algebra

Probability

Medium

applied

If a random variable y follows Poisson's distribution such that $P(X = 2) = 9 P(X = 4) + 90 P(X = 6)$ , then sum of the mean and variance of X is

✅ Correct Option: 4

For a Poisson distribution with parameter $\lambda$ :

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

Mean $=$ Variance $= \lambda$

Given: $P(X = 2) = 9\,P(X = 4) + 90\,P(X = 6)$

$\dfrac{e^{-\lambda}\,\lambda^2}{2!} = 9 \cdot \dfrac{e^{-\lambda}\,\lambda^4}{4!} + 90 \cdot \dfrac{e^{-\lambda}\,\lambda^6}{6!}$

$\dfrac{e^{-\lambda}\,\lambda^2}{2} = 9 \cdot \dfrac{e^{-\lambda}\,\lambda^4}{24} + 90 \cdot \dfrac{e^{-\lambda}\,\lambda^6}{720}$

Cancelling $e^{-\lambda}$ from every term:

$\dfrac{\lambda^2}{2} = \dfrac{9\lambda^4}{24} + \dfrac{90\lambda^6}{720}$

$\dfrac{\lambda^2}{2} = \dfrac{3\lambda^4}{8} + \dfrac{\lambda^6}{8}$

$4\lambda^2 = 3\lambda^4 + \lambda^6$

$4 = 3\lambda^2 + \lambda^4$

$\lambda^4 + 3\lambda^2 - 4 = 0$

Let $u = \lambda^2$ :

$u^2 + 3u - 4 = 0$

$(u + 4)(u - 1) = 0$

$u = -4 \quad \text{or} \quad u = 1$

Since $u = \lambda^2$ cannot be negative:

$\lambda^2 = 1$

$\lambda = 1$

Mean $+$ Variance $= \lambda + \lambda$

$= 1 + 1$

$= 2$

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If a random variable y follows Poisson's distribution such that P(X=2)=9P(X=4)+90P(X=6)P(X = 2) = 9 P(X = 4) + 90 P(X = 6)P(X=2)=9P(X=4)+90P(X=6), then sum of the mean and variance of X is

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CUET Mathematics Past Year Questions (Topic-Wise)

If a random variable y follows Poisson's distribution such that $P(X = 2) = 9 P(X = 4) + 90 P(X = 6)$ , then sum of the mean and variance of X is