Tag: probability distributions

Quizzes (4) Questions (106)

Questions Related to probability distributions

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution 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

  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. $\displaystyle \frac{e^{-3}12^{2}}{ 4!}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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! }$

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

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

  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. $1-\displaystyle \sum _{{x}=0}^{\infty }\frac{{e}^{-12}(12)^{{x}}}{ x!}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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) $= 1 - (P(1)+ . . . . +P(11))$
$=1-\displaystyle \sum _{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$

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

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

  1. $1-e^{-2}$
  2. $1-2e^{-2}$
  3. $1-3e^{-2}$
  4. $1-4e^{-2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 
$P=1-e^{-\lambda}-\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$=1-e^{-2}-2e^{-2}$
$=1-3e^{-2}$.

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

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

  1. $\cfrac 38$
  2. $\cfrac 13$
  3. $\cfrac 23$
  4. $\cfrac 54$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

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

  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. $\dfrac { { e }^{ -0.3 }{ 3 }^{ -5 } }{5! }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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! } $

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

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

  1. $e^{\lambda}$
  2. $\log \dfrac{1}{k}$
  3. $\dfrac{1}{k}$
  4. $\log k$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In Poisson distribution, P(X=0) = e^(-lambda) = k. Taking the natural log, -lambda = ln(k), so lambda = -ln(k) = ln(1/k). Since mean = variance = lambda for Poisson, the variance is ln(1/k).

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

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

  1. $e^{-1}$
  2. $e^{-2}$
  3. $e^{3}$
  4. $e^{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }$
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

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

  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. $1-\displaystyle \sum _{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)$= _{x=0}^{5}\sum \frac{e^{-40}40^{x}}{x!}$