Tag: random variables and probability distribution

Questions Related to random variables and probability distribution

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. $0$

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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=$1$

If on an average ,5 percent of the output in a factory making certain parts, is defective and that 200 units are in a package then the probability that atmost 4 defective parts may be found in that package is

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

  1. $\displaystyle e^{-10}\left [ 1+\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$

  2. $\displaystyle e^{-10}\left [ 1+\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$

  3. $\displaystyle e^{-10}\left [ 1-\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$

  4. $\displaystyle e^{-10}\left [ 1-\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$

Show answer
Correct Option: B
Explanation:

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 
$=\sum _{k=0} ^{k=4} \dfrac{e^{-10}.10^{k}}{k!}$

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

Suppose $300$ misprints are distributed randomly throughout a book of $500$ pages. The probability that a given page contains, at least one misprint is 

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

  1. $1.e^{-0.6}$

  2. $1-e^{-0.6}$

  3. $(0.6)e^{-0.6}$

  4. $(0.06)e^{-0.6}$

Show answer
Correct Option: B
Explanation:

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}$

A manufactured product on an average has 2 defects per unit of product produced. If the number of defects follows Poisson distribution, the probability of finding at least one defect is 

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

  1. $e^{-2}$

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

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

  4. $e^{-0.02}$

Show answer
Correct Option: B
Explanation:

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}$
                             

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 only one car is used is

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

  1. $e^{-1.5}$

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

  3. $1-2.5\times e^{-1.5}$

  4. $1-1.5\times e^{-1.5}$

Show answer
Correct Option: B
Explanation:

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

If $3$% of electric bulbs manufactured by a company are defective, the probability that a sample of $100$ bulbs has no defective bulbs is

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

  1. 0

  2. $e^{-3}$

  3. $1-e^{-3}$

  4. $3e^{-3}$

Show answer
Correct Option: B
Explanation:

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 } $