In Poisson probability distribution if P(r=2)= 3P(r=3), and P(r=4) is given by

A
e/24
B
24/e

1 thought on “<br />In Poisson probability distribution if P(r=2)= 3P(r=3), and P(r=4) is given by<br /><br />A<br />e/24<br />B<br />24/e<br />”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   We know,

   Probability of any random variable ‘r’ having mean m using Poisson Distribution is given by

   [tex]\rm :\longmapsto\:P(r) = \dfrac{ {e}^{ – m} \: \: {m}^{r} }{r! } [/tex]

   According to statement,

   It is given that

   P(2) = 3 P(3)

   [tex]\rm :\longmapsto\:\dfrac{ \cancel{ {e}^{ – m} }\: {m}^{2} }{2!} = 3 \dfrac{ \cancel{{e}^{ – m} }\: {m}^{3} }{3!} [/tex]

   [tex]\rm :\longmapsto\:\dfrac{1}{2 \times 1} = 3 \times \dfrac{m}{3 \times 2 \times 1} [/tex]

   [tex]\bf\implies \:m \: = \: 1[/tex]

   Now,

   [tex]\rm :\longmapsto\:P(4)[/tex]

   [tex]\rm \: \: = \: \dfrac{ {e}^{ – m} \: {m}^{4} }{4!} [/tex]

   [tex]\rm \: \: = \: \dfrac{ {e}^{ – 1} {(1)}^{4} }{4!} [/tex]

   [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:m \: = \: 1 \bigg \}}[/tex]

   [tex]\rm \: \: = \: \dfrac{ {e}^{ – 1} }{4 \times 3 \times 2 \times 1} [/tex]

   [tex]\rm \: \: = \: \dfrac{1}{24e} [/tex]

   Hence,

   [tex]\bf :\longmapsto\:P(4) \: = \: \dfrac{1}{24e} [/tex]

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