Question

If random variables has Poisson distribution such that p(2) =p(3), then p(5) =​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

POISSON DISTRIBUTION

If X is a Poisson Variate then its Probability Density function is given by

$\displaystyle \sf{ P(r) = \: {e}^{ - m} \: \frac{ {m}^{r} }{r \: !\: } \: \: } \: \:$

Where m > 0 and r = 0,1,2,3,4,5,........

GIVEN

A random variables has Poisson distribution such that P(2) = P(3)

TO DETERMINE

P (5)

CALCULATION

Here

$\displaystyle \sf{ P(r) = \: {e}^{ - m} \: \frac{ {m}^{r} }{r \: !\: } \: \: } \: \:$

So

$\displaystyle \sf{ P(2) = \: {e}^{ - m} \: \frac{ {m}^{2} }{2 \: !\: } \: \: } \: \:$

$\displaystyle \sf{ P(3) = \: {e}^{ - m} \: \frac{ {m}^{3} }{3\: !\: } \: \: } \: \:$

Now P(2) = P(3) gives

$\displaystyle \sf{ \: {e}^{ - m} \: \frac{ {m}^{2} }{2 \: !\: } \: = {e}^{ - m} \: \frac{ {m}^{3} }{3 \: !\: } \: } \: \:$

$\implies \: \displaystyle \sf{ \frac{ {m}^{2} }{2 \: !\: } \: = \frac{ {m}^{3} }{3 \: !\: } \: } \: \:$

$\implies \: \displaystyle \sf{ \: m = \frac{ 3 \: !\: }{2 \: !\: } \: } \: \:$

$\implies \: \displaystyle \sf{ \: m = 3 \: } \: \:$

Hence

$\sf{P(5)}$

$\displaystyle \sf{= \: {e}^{ - m} \: \frac{ {m}^{5} }{5\: !\: } \: \: }$

$\displaystyle \sf{= \: {e}^{ - 3} \: \frac{ {3}^{5} }{5\: !\: } \: \: }$

$\sf{ = 0.1008}$

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LEARN MORE FROM BRAINLY

The probability that the root of the equation

x^2+2nx+4n+(5/n) =0 are not real numbers

https://brainly.in/question/20849210

Answer 2

Answer:

if a random variable has a poisson distribution such that p(2)=p(3)find p(5)