Tag: probability distributions

Questions Related to probability distributions

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 

If X is a random poisson variate such that $E(X^2)=6$, then $E(x)=$?

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

  1. $3$

  2. $2$

  3. $-3$ & $2$

  4. $-2$

Show answer
Correct Option: C
Explanation:

In poison distribution mean=variance=x

E(X)=x
variance=$E({ X }^{ 2 })$-${ E({ X }) }^{ 2 }$
$x$=6-${ x }^{ 2 }$
on solving we get x=-3 and 2

If $3 percent $ bulb manufactured by a company are defective; the probability that in a sample of $100$ bulbs exactly five defective is

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

  1. $\dfrac { { { e }^{ -0.003 } }\left( 0.03 \right) ^{ 5 } }{  5! }$

  2. $\dfrac { { { e }^{ -0.3 } }0.03^{ 5 } }{ 5!}$

  3. $\dfrac { { { e }^{ -3 } }3^{ 5 } }{5! }$

  4. $\dfrac { { e }^{ -0.3 }{ 3 }^{ -5 } }{5! }$

Show answer
Correct Option: C
Explanation:

$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! } $

If, in a Poisson distribution $P(X= 0)=k$ then the variance is: 

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

  1. $e^{\lambda}$

  2. $\log \dfrac{1}{k}$

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

  4. $\log k$

Show answer
Correct Option: B

The incidence of an occupational disease to the workers of a factory is found to be $\displaystyle \frac{1}{5000}$ . If there are $10,000$ workers in a factory then the probability that none of them will get the disease is

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

  1. $e^{-1}$

  2. $e^{-2}$

  3. $e^{3}$

  4. $e^{4}$

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 = 10000 \times \dfrac{1}{5000} = 2 $
$x = 0$
$ P(0;2)=\dfrac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } =  { e }^{ -2 }$

The probability that atmost $5$ defective fuses will be found in a box of $200$ fuses, if experience shows that $20 \%$ of such fuses are defective,  is

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

  1. $\displaystyle \frac{e^{-40}40^{5}}{ 5!}$

  2. $\displaystyle \sum _{x=0}^{5}\frac{e^{-40}40^{x}}{ x!}$

  3. $\displaystyle \sum _{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$

  4. $1-\displaystyle \sum _{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$

Show answer
Correct Option: B
Explanation:

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

$p=\frac{20}{100}$
$n=200$
mean$ = \lambda  = np = 200\times \frac{20}{100} = 40$
$\displaystyle => P($atmost $5$ defective fuses)$= _{x=0}^{5}\sum \frac{e^{-40}40^{x}}{x!}$