Mathematics

Probability Distributions

488 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

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

If the first two terms of a Poisson distribution are equal to $k$, find $k$.

  1. $e$
  2. $\displaystyle \frac{1}{e}$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=P(x=1)$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$\lambda=1$
Hence $P(X=0)=P(X=1)$
$=\dfrac{1}{e}$

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

In a binomial distribution $n = 200, p = 0.04$. Taking Poisson distribution as an approximation to the binomial distribution .
Assertion (A) :- Mean of the Poisson distribution $= 8$
Reason (R) : In a Poisson distribution, $\displaystyle P(X=4)=\frac{512}{3e^{8}}$

  1. both A and R are true and R is the correct explanation of A

  2. both A and R are true and R is not correct explanation of A

  3. A is true but R is false

  4. A is false but R is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

mean of P.D. $=np = 8$
And $P(X = 4) = \cfrac{e^{-8}(8)^4}{4!}=\cfrac{512}{3e^8}$
Hence both statement are correct but they are not related to each other.

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

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

  1. $4$
  2. $3$
  3. $1.5$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $P(X\geq 1) = 1-e^{-1.5}$
$\Rightarrow 1-P(X=0)=1-e^{-1.5}\Rightarrow P(X=0)=e^{-1.5}=e^{-\lambda}\therefore \lambda = 1.5$

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

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

  1. $\displaystyle \frac{e^{-0.02}.2^{1}}{1!}$
  2. $1-e^{-0.2}$
  3. $e^{-0.2}$
  4. $1-e^{1.2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here Poisson parameter $\lambda = \cfrac{10}{50}=0.2$
$\therefore P(x \geq 1)=1-P(X=0)=1-\cfrac{e^{-0.2}.(.2)^0}{0!}=1-e^{-0.2}$ 

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

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

  1. $1.12 \times e^{-1.5}$
  2. $1-2.5 \times e^{-1.5}$
  3. $1-3.625 \times e^{-1.5}$
  4. $3.625 \times e^{-1.5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = 1$
Hence probability that both the cars are used $=1-P(X=0)-P(X=1)=1-\cfrac{e^{-1.5}(1.5)^0}{0!}-\cfrac{e^{-1.5}(1.5)^1}{1!}=1-2.5e^{-1.5}$

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

If $X$ is a random Poisson variate such that $P(X=0)=\displaystyle\frac{1}{e}$, then the variance of the same distribution is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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=$1$

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

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

  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. $\displaystyle e^{-10}\left [ 1-\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
$=\sum _{k=0} ^{k=4} \dfrac{e^{-10}.10^{k}}{k!}$

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

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

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

  1. $1.e^{-0.6}$
  2. $1-e^{-0.6}$
  3. $(0.6)e^{-0.6}$
  4. $(0.06)e^{-0.6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$

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

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 

  1. $e^{-2}$
  2. $1-e^{-2}$
  3. $\displaystyle \frac{e^{-2}2^{1}}{1!}$
  4. $e^{-0.02}$
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 $
$ 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}$
                             

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

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

  1. $e^{-1.5}$
  2. $1.5\times e^{-1.5}$
  3. $1-2.5\times e^{-1.5}$
  4. $1-1.5\times e^{-1.5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. 0
  2. $e^{-3}$
  3. $1-e^{-3}$
  4. $3e^{-3}$
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 = 100 \times 0.03 = 3 $    
$x = 0$ (No defective bulbs) 
$ P(0; 3)=\dfrac { { e }^{ -3 }{ 3 }^{ 0 } }{ 0! }  $
$P(0;3)={ e }^{ -3 } $

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 atleast one ship in the next $15$ minutes is

  1. $e^{-15}$
  2. $1-e^{-6}$
  3. $1-e^{-15}$
  4. $e^{-6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$=1-e^{-15}$.

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

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

  1. $e^{-2}$
  2. $1-e^{-2}$
  3. $2.e^{-2}$
  4. $3.e^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

A manufactured product on an average has $2$ defects per unit of product produced. If the number of defects follows P.D., the probability of finding zero defects is

  1. $e^{-2}$
  2. $1-e^{-2}$
  3. $\displaystyle \frac{e^{-2}2^{1}}{\angle 1}$
  4. $e^{-002}$
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! } $
: 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; ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is 
$ \mu = 2$ (defects per unit of product produced) 
x = 0 (zero defects)
$ 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

If the number of telephone calls coming into a telephone exchange between 10 AM and 11 AM follows Poisson distribution with parameter 2 then the probability of obtaining at least one call in that time interval is 

  1. $e^{-2}$
  2. $(1-e^{-2})$
  3. $2e^{-2}$
  4. $3e^{-2}$
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 $: The mean number of successes that occur in a specified region(Parameter)
x: The actual number of successes that occur in a specified region.
P(x;$ \mu $ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $ \mu $
$ P(x \ge 1; \mu) = 1 - P (x=0 ; \mu) $
$ \mu = 2 $    
$x = 0$ (No call comes) 
$ P(x \ge 1; 2) = 1 - P (x=0 ; 2) = 1 - \dfrac { { e }^{ -2 }{ 2 }^{ 0} }{ 0! } = 1 - {e}^{-2}$