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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Probability

BITSAT 2010 · Mathematics · Probability

If mean of a poisson distribution of a random variable X is 2 ,...

If mean of a poisson distribution of a random variable XX is 2 , then the value of P(X>1.5)P(X > 1.5) is

Options

  1. A(\frac{3}{e^2})
  2. B(\frac{3}{e})
  3. C(1-\frac{3}{e})
  4. D(1-\frac{3}{e^2})

Correct answer

D. (1-\frac{3}{e^2})

Step-by-step solution

Since, (\mathrm{P}(X=r\mathrm{X}=\mathrm{r})=\frac{e^{-\lambda} \lambda^r}{r !}) (where (\lambda=) mean) (∴P(X=r>1.5)=P(2)+P(3)+…∞=1−P(X=r≤1)=1−P(0)−P(1)=1−(e−2+e−2×21!)=1−3e2\begin{aligned} \therefore \quad & \mathrm{P}(\mathrm{X}=\mathrm{r} > 1.5)=\mathrm{P}(2)+\mathrm{P}(3)+\ldots \infty \\ & =1-\mathrm{P}(\mathrm{X}=\mathrm{r} \leq 1)=1-\mathrm{P}(0)-\mathrm{P}(1) \\ & =1-\left(e^{-2}+\frac{e^{-2} \times 2}{1 !}\right)=1-\frac{3}{e^2} \end{aligned})
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