Tag: poisson distribution

Questions Related to poisson distribution

Suppose $2$% of the people on an average are left handed. The probability of 3 or more left handed among 100 people is

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

  1. $3e^{-2}$

  2. $4e^{-2}$

  3. $1-5e^{-2}$

  4. $5 e^{-2}$

Show answer
Correct Option: C
Explanation:

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

Suppose there is an average of $2$ suicides per year per $50,000$ population. In a city of population $1,00,000$, the probability that in a given year there are, zero suicides is

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

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

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

  3. $e^{-4}$

  4. $1-e^{-4}$

Show answer
Correct Option: C
Explanation:

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

Suppose on an average $5$ out of $2000$ houses get damaged due to fire accident during summer. Out of $10,000$ houses in a locality, the probability that exactly $10$ houses will get damaged during summer is

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

  1. $\displaystyle \frac{e^{-5}5^{10}}{ 10!}$

  2. $\displaystyle \frac{e^{-10}10^{10}}{ 10!}$

  3. $\displaystyle \frac{e^{-25}25^{10}}{10!}$

  4. $\displaystyle \frac{e^{-15}15^{10}}{10!}$

Show answer
Correct Option: C
Explanation:

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

A manufacturer who produces medicine bottles finds that $0.1$$\%$ of the bottles are defective. The bottles are packed in boxes containing $500$ bottles. A drug manufacturer buys $100$ boxes from the producer of bottles. Using Poisson distribution, the number of boxes with no defective bottle is

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

  1. $100\times e^{-0.1}$

  2. $100\times e^{-0.5}$

  3. $100\times e^{-0.05}$

  4. $100\times e^{-0.01}$

Show answer
Correct Option: B
Explanation:

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

A company knows on the basis of past experience that $2$% of its blades are defective. The probability of having $3$ defective blades in a sample of $100$ blades if $e^{-2}=0.1353$ is

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

  1. $0.1353$

  2. $0.1804$

  3. $0.2706$

  4. $0.3606$

Show answer
Correct Option: B
Explanation:

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

On the 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 $4$ ships in the next two hours is

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

  1. $\displaystyle \frac{e^{-12}12^{4}}{ 4!}$

  2. $\displaystyle \frac{e^{-4}12^{12}}{ 3!}$

  3. $\displaystyle \frac{e^{-6}12^{4}}{ 4!}$

  4. $\displaystyle \frac{e^{-3}12^{2}}{ 4!}$

Show answer
Correct Option: A
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(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! }$

Patients arrive randomly and independently at a Doctor's room from 8 AM at an average rate of one in 5 minutes. The waiting room can accommodate 12 persons. The probability that the room will be full when the doctor arrives at 9AM is

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

  1. $\displaystyle \frac{{e}^{-12}(12)^{12}}{ 12!}$

  2. $\displaystyle \sum _{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$

  3. $1-\displaystyle \sum _{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$

  4. $1-\displaystyle \sum _{{x}=0}^{\infty }\frac{{e}^{-12}(12)^{{x}}}{ x!}$

Show answer
Correct Option: C
Explanation:

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) $= 1 - (P(1)+ . . . . +P(11))$
$=1-\displaystyle \sum _{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$

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 at least two ships in the next $20$ minutes is

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

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

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

  3. $1-3e^{-2}$

  4. $1-4e^{-2}$

Show answer
Correct Option: C
Explanation:

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 
$P=1-e^{-\lambda}-\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$=1-e^{-2}-2e^{-2}$
$=1-3e^{-2}$.

For a poisson distribution with parameter $\lambda = 0.25$, the value of the $2^{nd}$ moment about the origin is

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

  1. $0.25$

  2. $0.3125$

  3. $0.0625$

  4. $0.025$

Show answer
Correct Option: B
Explanation:

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

If $X$ is a Poisson's variate such that $P(X=1)=3P(X=2)$, then find the variance of $X$.

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

  1. $\cfrac 38$

  2. $\cfrac 13$

  3. $\cfrac 23$

  4. $\cfrac 54$

Show answer
Correct Option: C
Explanation:
Fact:  Poisson distribution is $P(X=r)=\dfrac{e^{-\lambda}\lambda ^r}{r!}$

Now given $P(X=1)=3P(X=2)$

$\Rightarrow \dfrac{e^{-\lambda}\lambda}{1!}=3\cdot \dfrac{e^{-\lambda} \lambda^2}{2!}$

$\Rightarrow \lambda =\dfrac{2}{3}=$ Variance