Tag: probability distributions

Questions Related to probability distributions

If $X$ is a Poisson variate such that $P(X=1) = P(X=2)$ then $P(X=4)=$

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\dfrac{1}{2e^{2}}$

  2. $\dfrac{1}{3e^{2}}$

  3. $\dfrac{2}{3e^{2}}$

  4. $\dfrac{1}{e^{2}}$

Show answer
Correct Option: C
Explanation:

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


$\dfrac { { e }^{ -\mu }{ \mu }^{ 1 } }{ 1! } =\dfrac { { e }^{ -\mu }{ \mu }^{ 2 } }{ 2! } $

$ \mu = 2 $
$ P(X=4)= \dfrac { { e }^{ -2 }{ 2 }^{ 4 } }{ 4! }= \dfrac {2}{3{e}^{2}} $

If a random variable $X$ follows a P.D. such that $P(X=1)=P(X=2)$, then $P(X=0)=$

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $e^{2}$

  2. $\dfrac{1}{e^{2}}$

  3. $\dfrac{1}{e}$

  4. $e$

Show answer
Correct Option: B
Explanation:

$ 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 )]$ $= 0$

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

So, either $ { e }^{-0}$or ${ e }^{-2}$
that is 1 or $\dfrac {1}{{ e }^{2}} $

If the first two terms of a Poisson distribution are equal to $k$, find $k$.

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $e$

  2. $\displaystyle \frac{1}{e}$

  3. $1$

  4. $2$

Show answer
Correct Option: B
Explanation:

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 $P(X=0)=P(X=1)$
$=\dfrac{1}{e}$

In a binomial distribution $n = 200, p = 0.04$. Taking Poisson distribution as an approximation to the binomial distribution .
Assertion (A) :- Mean of the Poisson distribution $= 8$
Reason (R) : In a Poisson distribution, $\displaystyle P(X=4)=\frac{512}{3e^{8}}$

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. both A and R are true and R is the correct explanation of A

  2. both A and R are true and R is not correct explanation of A

  3. A is true but R is false

  4. A is false but R is true

Show answer
Correct Option: B
Explanation:

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.

If $X$ is a random poission variate such that $2P(X=0)+P(X=2)=2P(X=1)$ then $E(X)=$

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $4$

  2. $3$

  3. $2$

  4. $1$

Show answer
Correct Option: C
Explanation:

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


$2e^{-\lambda}+\dfrac{\lambda^{2}e^{-\lambda}}{2!}=2\lambda: e^{-\lambda}$

$2\lambda: e^{-\lambda}-2e^{-\lambda}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$
$4\lambda-4=\lambda^{2}$
$\lambda^{2}-4\lambda+4=0$
$(\lambda-2)^{2}=0$
$\lambda=2$
Hence mean=variance=$2$

If the probability of that a poisson variable $X$ takes a positive value $\geq 1$ is $1-e^{-1.5}$, then the varianceof the distribution is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $4$

  2. $3$

  3. $1.5$

  4. $0$

Show answer
Correct Option: C
Explanation:

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

In a town $10$ accidents take place in a span of $50$ days. Assuming that number of accidents follows Poisson distribution, the probability that there will be atleast one accident on a selected day at random is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\displaystyle \frac{e^{-0.02}.2^{1}}{1!}$

  2. $1-e^{-0.2}$

  3. $e^{-0.2}$

  4. $1-e^{1.2}$

Show answer
Correct Option: B
Explanation:

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

A car hire firm has $2$ cars which it hires out day by day. If the number of demands for a car on each day follows Poisson distribution with parameter $1.5$, then the probability that both the cars is used is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $1.12 \times e^{-1.5}$

  2. $1-2.5 \times e^{-1.5}$

  3. $1-3.625 \times e^{-1.5}$

  4. $3.625 \times e^{-1.5}$

Show answer
Correct Option: B
Explanation:

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

If $X$ is a Poisson variate with parameter $\displaystyle \frac{3}{2}$, find $P(X\geq 2)$

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\displaystyle \frac{5}{2}e^{\frac{-3}{2}}$

  2. $\displaystyle 1-\frac{5}{2}e^{\frac{-3}{2}}$

  3. $\displaystyle 1-e^{\frac{-3}{2}}$

  4. $\displaystyle e^{\frac{-3}{2}}$

Show answer
Correct Option: B
Explanation:

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

If $X$ is a random Poisson variate such that $P(X=0)=\displaystyle\frac{1}{e}$, then the variance of the same distribution is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:

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=$1$