Tag: business maths

Questions Related to business maths

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 } $

On an average, a submarine on patrol sights $6$ enemy ships per hour. Assuming the number of ships sighted in a given length of time is a poisson variate, the probability of sighting atleast one ship in the next $15$ minutes is

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

  1. $e^{-15}$

  2. $1-e^{-6}$

  3. $1-e^{-15}$

  4. $e^{-6}$

Show answer
Correct Option: C
Explanation:

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
$=1-e^{-15}$.

If the number of telephone calls coming into a telephone exchange between 10 AM and 11 AM follows P.D. with parameter 2, then the probability of obtaining zero calls in that time interval is

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

  1. $e^{-2}$

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

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

  4. $3.e^{-2}$

Show answer
Correct Option: A
Explanation:

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