$e^{-m}\cos h(m)$

For a Poisson distribution, mean and variance are equal .
i.e.$\lambda = \sigma^2$

The mean of P.D. is $5$. 
In P.D,  the variance and mean of the same distribution is same.
So, variance = $5$

For a Poisson distribution, mean ($\mu$) and variance ($\sigma^2$ )are equal .
i.e.$\mu = \sigma^2$

Mean = $ \lambda$
In Poisson distribution, the expected value of a Poisson-distributed random variable is equal to that is mean and is equal to its variance. 
Variance =  $ \lambda$
S.D = $ \sqrt {variance} =\sqrt { \lambda  } $

$P(X=3)/P(X=2)$
$=\dfrac{\lambda^{3}(2!)}{3!(\lambda^{2})}$
$=\dfrac{\lambda}{3}$
$=\dfrac{1}{2(3)}$
$=\dfrac{1}{6}$
The ratio is $1:6$

Given $P(X=0)=.1$
$\Rightarrow e^{-\lambda}=\cfrac{1}{10}\therefore \lambda = \log _e 10$

In Poisson distribution the  parameter $(  > 0)$, it is always greater than zero or a finite positive value.

X is a Poisson variable with parameter 0.09.
In Poisson distribution,  Poisson variable(X) is equal to  expected value of X and also to its variance. 
 S.D =  $\sqrt {variance}  =  \sqrt { X } =\sqrt { 0.09 } =0.3$

Given $\sigma =1.5$
We know that for a Poisson's distribution,mean and variance are equal.
$\mu=\sigma^2$
$\Rightarrow \mu =2.25$

$S.D= \sqrt { \lambda  }$

Therefore, the probability of getting $6$ heads with $6$ coins $ = {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{64}} = P\left( { > {a _y}} \right)$
Then, 
$\eta P = 6400 \times \frac{1}{{64}} = 100 = m\left( {{a _y}} \right)$
So, by poison's law, $P\left( {x = n} \right) = \frac{{{e^{ - m}}{m^x}}}{{x!}}$
$ = \frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$

According to Poisson distribution
$\displaystyle P(X=r)=\frac{e^{-m}m^{r}}{r!}$
$\displaystyle \therefore P(X\leq 1)=P(X=0)+P(X=1)$
$\displaystyle =e^{-m}+\frac{e^{-m} m}{1!}$
Given m $\displaystyle =$mean$ = 5 $
$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$

in Poisson distribution, probability $ P(x; m  )=  \dfrac { { e }^{ - m  }{ m }^{ x } }{ \angle x } $
$m$ = mean  ;   $x$= number of success     $ \angle x =$ factorial $ x $
 For $r$ success  $x= r$;        
  $ P(r; m )=  \dfrac { { e }^{ -m  }{ m  }^{ r } }{ \angle r } $ 

For Poisson distribution of mean = $\mu = 2 $ 
 P$(x ; \mu) = \frac { { e }^{ -\mu  }{ \mu  }^{ x } }{ x! }$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)]$ 

In  Poisson distribution such that  $ \alpha = p(X=1)=p(X=2) $ 
        $   p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $  
       => $ \alpha = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}  $       =>   $ \alpha =  { e }^{-\mu}{\mu}  $
                 =>  $ { e }^{-\mu}{\mu} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2} $
                 =>   $ \mu = 2 $ 
  $   p(4; \mu) = \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} =  \dfrac { { e }^{ -\mu  }{ \mu \times { \mu  }^{ 3 } } }{ 24 }  = \dfrac { \alpha {\times { 2 }^{ 3 } } }{ 24 }  = \dfrac { \alpha  }{ 3 } $  

$E[{ X }^{ 2 }]= \sum _{ k=0 }^{ \infty  }{ k^{ 2 } } \sum _{  }^{  }{  } \dfrac { 1 }{ k! } \lambda ^{ k }e^{ -\lambda  }\ \ = ^{ k }e^{ -\lambda  }\sum _{ 1 }^{ \infty  }{ k\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } \= ^{ k }e^{ -\lambda  }(\sum _{ 1 }^{ \infty  }{ (k-1)\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } ) +\sum _{ 1 }^{ \infty  }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ =^{ k }e^{ -\lambda  }(\sum _{ 2 }^{ \infty  }{ (\lambda \dfrac { 1 }{ (k2)! } \lambda ^{ k-2 } } )+ \sum _{ 1 }^{ \infty  }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ = ^{ k }e^{ -\lambda  }(\sum _{ i=0 }^{ \infty  }{ (\lambda \dfrac { 1 }{ i! } \lambda ^{ i } } )+ \sum _{ j=0 }^{ \infty  }{ \dfrac { 1 }{ j! } \lambda ^{ j } } )\ = ^{ k }e^{ -\lambda  }(\lambda e^{ \lambda  }+e^{ \lambda  })\  \ = { \lambda  }^{ 2 }+ \lambda \ \ Variance= E[{ X }^{ 2 }]- { E[X] }^{ 2 }\ = { \lambda  }^{ 2 }+ \lambda - { \lambda  }^{ 2 }\ = \lambda \ \ \ $

In  Poisson distribution such that $P(X=1)=P(X=2)$
    $ P(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!}$
          =>  $ P(1; \mu)  =  P(2; \mu) $
         =>  $ \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}  $ 
          => $ 1 =$ $ \dfrac {\mu}{2} $ 
         => $ \mu = 2 $
 In Poisson distribution Variance $(m)$ is equal to mean. 
       Mean = Variance  = $2 $

Sum of the terms in Even places will be 
$=e^{-m}[m+\dfrac{m^{3}}{3!}+\dfrac{m^{5}}{5!}+...]$

$=e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]$ ...(i)
Sum of the terms in Odd places will be 
$=e^{-m}[1+\dfrac{m^{2}}{2!}+\dfrac{m^{4}}{4!}+...]$

$=e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]$ ...(ii)
Taking the ratio of i and ii, we get 
$\dfrac{i}{ii}$

$=\dfrac{e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]}{e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]}$

$=\dfrac{e^{m}-e^{-m}}{e^{m}+e^{-m}}$

$=\dfrac{2sinhm}{2coshm}$

$=tanh(m)$.

If m is the variance of P. D, then  P$(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in even places = [ P$(1; \mu) + P(3; \mu)  + P(5; \mu) + .........$ ]

If m is the variance of P. D, then  P$(x; \mu) = \frac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in odd places = [ P$(0; \mu) + P(2; \mu)  + P(4; \mu) + .........$ ]

Considering all odd places in Poisson's Distribution, we get 
$e^{-\lambda}+\dfrac{e^{-\lambda}.\lambda^{2}}{2!}+\dfrac{e^{-\lambda}.\lambda^{4}}{4!}....$

$=e^{-\lambda}[1+\dfrac{\lambda^{2}}{2!}+\dfrac{\lambda^{4}}{4!}+....]$

$=e^{-\lambda}[\dfrac{e^{-\lambda}+e^{\lambda}}{2}]$

$=e^{-\lambda}(\cosh\lambda)$.
Hence reason is false.

Here,  $P(X=0)=P(X=1)$
$\Rightarrow \displaystyle \frac{e^{-\lambda}\lambda^0}{0!}=\frac{e^{-\lambda}\lambda^1}{1!}\Rightarrow \lambda = 1$
$\displaystyle \therefore P(X = 2) = \frac{e^{-\lambda}1^2}{2!}=\frac{1}{2e}$

Variance  = $E(X^{ 2 }) - E[X]^{ 2 }$
Mean = Variance for P.D.
Let mean = $m$
Therefore
$m = $ $ 6 - m*m $
Hence
$m = 2$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=2)=3P(x=3)$
$3\dfrac{\lambda^{3}e^{-\lambda}}{3!}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$
$3\lambda=3$
$\lambda=1$
Hence mean=variance=$1$

Poisson's distribution implies
$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence substituting $x=0$ 

According to question
$P(X=3; \mu )= \dfrac {2}{3}P(X=4; \mu) $
$\dfrac { { e }^{-\mu}{\mu}^{3}}{3!}  = \dfrac {2}{3} \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} $
=> $\dfrac { { e }^{-\mu}{\mu}^{3}}{6}  = \dfrac { { e }^{-\mu}{\mu}^{4}}{36} $
=> ${ e }^{ -\mu  }{ \mu  }^{ 3 }\left[ \dfrac { 1 }{ 6 } -\dfrac { \mu  }{ 36 }  \right] $ 
=> $ \mu = 0 \  or\   \mu = \dfrac{36}{6} = 6 $

For Poisson's distribution, $P(X)$ is given by 
$\displaystyle P(X)=\frac { { e }^{ -\lambda  }.{ \lambda  }^{ x } }{ x! } $
It is given that $P(X=2)=9P(X=4)+  90P(X=6)$
$\Rightarrow \displaystyle \frac { { e }^{ -\lambda  }.{ \lambda  }^{ 2 } }{ 2! } =9.\frac { { e }^{ -\lambda  }.{ \lambda  }^{ 4 } }{ 4! } +90.\frac { { e }^{ -\lambda  }.{ \lambda  }^{ 6 } }{ 6! } $
Cancelling ${ e }^{ -\lambda  }$ from both sides, we get
$\displaystyle \frac { { \lambda  }^{ 2 } }{ 2! } =9.\frac { { \lambda  }^{ 4 } }{ 4! } +90.\frac { { \lambda  }^{ 6 } }{ 6! } $
Since $\lambda$ can not be zero, cancel out $\lambda^2$ from both sides to obtain
${ \lambda  }^{ 4 }+3{ \lambda  }^{ 2 }-4=0$.
Solving , we get $\lambda = 1$ as a valid solution.

In Poisson distribution,   $P(X=0) = 0.8,$
 $ p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $ 
=>  $ p(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!} $
=>    0.8 = $ { e }^{-\mu} $ 
=>    $ \dfrac {4}{5} = { e }^{-\mu} $
=>    ${ e }^{\mu}  = \dfrac {5}{4} $ 
taking log to both sides 
 =>  $  \log _{ e }{ { e }^{ \mu  } }  = \log _{e} (\dfrac {5}{4}) $ 
 =>  $ \mu =  \log _{e} (\dfrac {5}{4}) $

Poisson's distribution implies
$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=1)=P(x=2)$
$\frac{\lambda^{1}e^{-\lambda}}{1!}=\frac{\lambda^{2}e^{-\lambda}}{2!}$
$\lambda=2$
Hence mean=variance=$2$

Poisson's distribution implies
$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence substituting $x=0$ we get
$e^{-\lambda}=0.2$
$e^{\lambda}=5$
$\lambda=ln(5)$
Hence mean=variance=$log _{e}(5)$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=2P(x=1)$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=2\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$2\lambda=1$
$\lambda=\dfrac{1}{2}$
Hence mean=variance=$0.5$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$

$P(X=2)= \cfrac {2}{3}P(X=1) $

$P$$(2; \mu) = \cfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \cfrac {2}{3} P(1; \mu) $=$ \cfrac {2}{3} \cfrac { { e }^{-\mu}{\mu}^{1}}{1!}$

$ { e }^{-\mu} \mu (\cfrac {\mu}{2} - \cfrac {2}{3})=0$

either $\mu = 0\  or \mu = \cfrac{4}{3}$

but here $ \mu = \cfrac{4}{3}$$P(X=3) =$ $ \cfrac { { e }^{-\mu}{\mu}^{3}}{3!} = \cfrac { { e }^{ \frac { -4 }{ 3 }  }{( \frac { 4 }{ 3 }  })^{ 3 } }{ 6 } = \cfrac { 64 }{ 162 } { e }^{ \frac{-4}{3} } $

$P(X>1)$
$=1-[P(X=0)+P(X=1)]$
$=1-[e^{-1.5}+1.5e^{-1.5}]$
$=1-e^{1.5}(1+1.5)$
$=1-e^{1.5}(2.5)$

$P(X=0)=P(X=1) $
P$(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!} =  P(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!}$
$ { e }^{-\mu}  (1 - \mu )$ = 0 
 $\mu = 1 $
P(X=2) = $ \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \dfrac{{ e }^{-1}}{2} = \dfrac {1}{2e} $

$P(X=k)=P(X=k+1)$
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

For Poisson's distribution, $P(X)$ is given by 
$\displaystyle P(X)=\frac { { e }^{ -\lambda  }.{ \lambda  }^{ x } }{ x! } $
Hence $\displaystyle P(X=1) =\displaystyle  \frac { { e }^{ -\lambda }.{ \lambda  }^{ 1 } }{ 1! } $.
Similarly $\displaystyle P(X=2)=\frac { { e }^{ -\lambda }.{ \lambda  }^{ 2 } }{ 2! } $.
Given that, $P(X=1) = 2 P(X=2)$
$\Rightarrow \displaystyle \frac { { e }^{ -\lambda }.{ \lambda  }^{ 1 } }{ 1! } =2.\frac { { e }^{ -\lambda }.{ \lambda  }^{ 2 } }{ 2! } $
Simplifying we get $\lambda = 1$

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$
$P(X=1) = P(X=2)$

$ P(X=1)=P(X=2)$
 $P$$(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} =  P(2; \mu) = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}$
$ { e }^{-\mu} [\mu (2 - \mu )]$ $= 0$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=P(x=1)$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$\lambda=1$
Hence $P(X=0)=P(X=1)$
$=\dfrac{1}{e}$

mean of P.D. $=np = 8$
And $P(X = 4) = \cfrac{e^{-8}(8)^4}{4!}=\cfrac{512}{3e^8}$
Hence both statement are correct but they are not related to each other.

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $2.P(x=0)+P(x=2)=2P(x=1)$
$2\dfrac{\lambda^{0}e^{-\lambda}}{0!}+\dfrac{\lambda^{2}e^{-\lambda}}{2!}=2\dfrac{\lambda^{1}e^{-\lambda}}{1!}$

Given $P(X\geq 1) = 1-e^{-1.5}$
$\Rightarrow 1-P(X=0)=1-e^{-1.5}\Rightarrow P(X=0)=e^{-1.5}=e^{-\lambda}\therefore \lambda = 1.5$

Here Poisson parameter $\lambda = \cfrac{10}{50}=0.2$
$\therefore P(x \geq 1)=1-P(X=0)=1-\cfrac{e^{-0.2}.(.2)^0}{0!}=1-e^{-0.2}$ 

Here $\lambda = 1$
Hence probability that both the cars are used $=1-P(X=0)-P(X=1)=1-\cfrac{e^{-1.5}(1.5)^0}{0!}-\cfrac{e^{-1.5}(1.5)^1}{1!}=1-2.5e^{-1.5}$

Here $\lambda = \cfrac{3}{2}$
$\therefore P(X\geq 2) = 1-P(X=0)-P(X=1)$
$\displaystyle =1-\cfrac{e^{-3/2}(3/2)^0}{0!}-\cfrac{e^{-3/2}(3/2)^1}{1!}=1-\cfrac{5}{2}e^{-3/2}$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=\frac{1}{e}$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=\dfrac{1}{e}$
$\lambda=1$
Hence mean=variance=$1$

Here $\lambda$
$=\dfrac{5}{100}.200$
$=10$.
Hence by applying Poisson distribution, we get that the probability that atmost 4 defective part are found is 
$=\sum _{k=0} ^{k=4} \dfrac{e^{-10}.10^{k}}{k!}$

$=e^{-10}[1+\dfrac{10}{1!}+\dfrac{10^{2}}{2!}+\dfrac{10^{3}}{3!}+\dfrac{10^{4}}{4!}]$.

Here $\lambda = \cfrac{300}{500}=0.6$
Hence $ P(X\geq 1) = 1-P(X=0)=1-\cfrac{e^{-0.6}(0.6)^0}{0!}=1-e^{-0.6}$

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region.
$x$: The actual number of successes that occur in a specified region.
$P(x$; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
$ P(x \ge 1; \mu) = 1 - P (x=0 ; \mu) $
$ \mu = 2 $    
$x = 0$ (No defective product) 
$ P(x \ge 1; 2) = 1 - P (x=0 ; 2) = 1 - \dfrac { { e }^{ -2 }{ 2 }^{ 0} }{ 0! } = 1 - {e}^{-2}$
                             

Here $\lambda = 1.5$
Hence probability that only one car is used is $=P(X=1) = \cfrac{e^{-1.5}(1.5)}{1!}=1.5e^{-1.5}$

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region.
x: The actual number of successes that occur in a specified region.
P(x;$ \mu $ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
$ \mu = 100 \times 0.03 = 3 $    
$x = 0$ (No defective bulbs) 
$ P(0; 3)=\dfrac { { e }^{ -3 }{ 3 }^{ 0 } }{ 0! }  $
$P(0;3)={ e }^{ -3 } $

The probability of seeing atleast one ship
=1-(probability of seeing no ship)
$=1-\dfrac{\lambda^{0}.e^{-\lambda}}{0!}$
$=1-e^{-\lambda}$
It is given the Poisson's variate is the number of ships passing per unit time.
Hence in the above case $\lambda=15$
Thus the required probability is
$=1-e^{-15}$.

Here P.D parameter $\lambda = 2$
Hence probability of obtaining zero calls during 10 AM to 11 AM is $=(P(X=0)=e^{-2}$ 

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
: The mean number of successes that occur in a specified region.
x: The actual number of successes that occur in a specified region.
P(x; ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is 
$ \mu = 2$ (defects per unit of product produced) 
x = 0 (zero defects)
$ P(0; 2)=\dfrac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } = {e}^{-2}$

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(Parameter)
x: The actual number of successes that occur in a specified region.
P(x;$ \mu $ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
$ P(x \ge 1; \mu) = 1 - P (x=0 ; \mu) $
$ \mu = 2 $    
$x = 0$ (No call comes) 
$ P(x \ge 1; 2) = 1 - P (x=0 ; 2) = 1 - \dfrac { { e }^{ -2 }{ 2 }^{ 0} }{ 0! } = 1 - {e}^{-2}$
                             

Here $\lambda = \cfrac{1}{500}\times 10 = 0.02$
Thus probability that  lot is not defective is $=P(X=0)=\cfrac{e^{-0.002}(.0020^0}{0!}=e^{-.002}=0.9802$
Hence number of no defective lots out of $10,000$ is $=.9802\times 10,000=9802$

Here $\lambda = 0.001$
Hence number of day out of 1000 days without accident is $1000\times P(X=0)=1000\times e^{-0.001}$

we are given $n=100$
let $p=$ probability of a defective bulb $=5$%$=0.05$
$\therefore m=$ mean number of defective bulbs in a box of $100=np=100\times 0.05=5$
Since p is small , we can use poison's distribution.
Probability of $x$ defective bulbs in a box of $100$ is
$\displaystyle P\left( X=x \right) =\frac { { e }^{ -m }{ m }^{ x } }{ x! } =\frac { { e }^{ -5 }{ 5 }^{ x } }{ x! } ,x=0,1,2...$
Probability that is box will fail to meet the guarented quality is $\displaystyle P\left( X>10 \right) =1-P\left( X\le 10 \right) =1-\sum _{ x=0 }^{ 10 }{ \frac { { e }^{ -5 }{ 5 }^{ x } }{ x! }  } =1-{ e }^{ -5 }\sum _{ x=0 }^{ 10 }{ \frac { { 5 }^{ x } }{ x! }  } $

Here $\lambda = 3$
Hence Number of drivers with no accident out of 1000 is $=1000\times P(X=0)=1000\times e^{-3}=1000\times 0.0498=49.8$

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
Here,$ \mu $  = 2 
x = 2 (exact 2 workers)
$ P(2;2)=\dfrac { { e }^{ -2 }{ 2 }^{2 } }{ 2! } $

Here $\lambda = \cfrac{0.1}{100}\times 500=0.5 $
Hence number of boxes out of 100 which contain at least one defective bottle is,
$=100\left(1-P(X=0)\right)=100\left(1-e^{-0.5}\right)$ 

Here P.D parameter $\lambda = \cfrac{2}{100}\times 100=2$
Hence probability that 3 or more people are left handed is $=1-P(X=0)-P(X=1)-P(X=2)$
$=1-e^{-2}-\cfrac{e^{-2}2}{1!}-\cfrac{e^{-2}2^2}{2!}=1-5e^{-2}$

P.D parameter $\lambda = \cfrac{2}{50000}\times 100000 = 4$
Thus probability that in a year there is no suicide is $=P(X=0)=e^{-4}$

Here P.D parameter $\lambda = \cfrac{5}{2000}\times 10000=25$
Hence probability that exactly 10 houses will get damaged $=P(X = 10) = \cfrac{e^{-25}(25)^{10}}{10!}$

$\displaystyle p=\frac { 0.1 }{ 100 } =0.001,n=500\ \lambda =np=500\times 0.001=0.5\ N=100$
We know that $\displaystyle p\left( r \right) =e^{ -1 }\frac { { \lambda  }^{ r } }{ r! } $
Number of boxes containing no defective bottle 
$\displaystyle =N.P.\left( r=0 \right) =1000\times { e }^{ 0.5 }\frac { \left( 0.5 \right) ^{ 0 } }{ 0! } =1000\times { e }^{ -0.5 }$

Here P.D parameter $\lambda = \cfrac{2}{100}\times 100=2$
Hence number of probability that 3 blade are defective is $=P(X=3)=\cfrac{e^{-2}(2)^3}{3!}=\cfrac{4}{3}e^{-2}=\cfrac{4}{3}\times .1353=.1804$

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
the average a submarine on patrol sights 6 enemy ships per hour, so for 2 hours
Here, $ \mu =  6 \times 2 = 12 $
x = 4 ( ships)
$ P(4;12)=\dfrac { { e }^{ -12 }{12 }^{4 } }{ 4! }$

Random Arrival process is a Poisson process
Given by
$P(k) = \dfrac{e^{-\lambda} \lambda^x}{x!}$
$given, \lambda = 1/5 min^{-1}= 12 s^{-1}$
Room is not full, if No.of patients $< 12$
Hence 
$P$(room is full) $= 1 - (P(1)+ . . . . +P(11))$
$=1-\displaystyle \sum _{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$

The probability of sighting at least two ships
=1-(Probability of sighting atmost 1 ships)
$=1-e^{-\lambda}-\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
Here $\lambda=2$
Hence 
$P=1-e^{-\lambda}-\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$=1-e^{-2}-2e^{-2}$
$=1-3e^{-2}$.

Second moment about the origin is $\lambda +{ \lambda  }^{ 2 } = 0.25+{(0.25)}^{2} = 0.3125$
Therefore the correct option is $B$.

In poison distribution mean=variance=x

$In\quad poison\quad distribution\quad \frac { { e }^{ -u }{ u }^{ x } }{ x! } \ Here\quad 3\quad are\quad defective\quad in\quad 100\quad so\quad u=3\ Probaility\quad that\quad exactly\quad 5\quad are\quad defective\quad (x=5)=\quad \frac { { e }^{ -3 }{ (3) }^{ 5 } }{ 5! } $

The poisson distribution is
$\displaystyle P(X=x)=\frac{e^{-\lambda }\lambda ^{x}}{x!}$ , $x=0,1,2,3,..$
Let, $X$ denote the defective fuse

Here $\lambda = .002\times 1000=2$
Thus probability that a box contain at least one defective ballot $=1-P(X=0)=1-e^{-2}=1-0.1353=.8647$   
Hence   number of box out of $500$ which contain at least one defective ballot is $=.8647\times 500 =432$

Here $\lambda = \left(\cfrac{1}{2}\right)^6\times 64100=100$
Hence probability of getting six heads two times $=P(X=2)=\cfrac{e^{-100}(100)^2}{2!}$

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
Here,$ \mu  =  100 \times 0.02 = 2 $
$x = 3$(defective blades)
$ P(3;2)=\dfrac { { e }^{ -2 }{ 2 }^{3 } }{ 3! } $

By using Poisson distribution, 
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $
$ \mu $: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $ \mu $): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
Here, $ \mu  = 1.5$ and $x = 0$ (neither car is used)
$ P(0;1.5)=\dfrac { { e }^{ -1.5 }{ 1.5 }^{0 } }{ 0! } = {e}^{-1.5} $

$\mu _2=E(X^2)-[E(X)]^2......(i)$