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

A examinations consists of $8$ questions in each of which  one of the $5$ alternatives is the correct one. On the assumption that a candidate who has done no preparatory  work, chooses for each questions any one of the five alternatives with equal probability, then the probability that he gets more than one correct answer is equal to:

  1. ${\left( {0.8} \right)^8}$
  2. $3{\left( {0.8} \right)^8}$
  3. $1-{\left( {0.8} \right)^8}$
  4. $1-3{\left( {0.8} \right)^8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Probability of an answers to be correct $=\dfrac{1}{5}=0.2$

Probability of an answers not to be correct $=1-0.2$
$=0.8$

Probability $\left(more\ than\ 1\ correct\right)=1-P\left(0\ correct\right)-P\left(1\ correct\right)$

$=1-^{8}{C} _{0}{\left(0.8\right)}^{8}-^{8}{C} _{1}\left(0.2\right) \left(0.8\right)$

$=1-{\left(0.8\right)}^{8}-1.6{\left(0.8\right)}^{7}$

$=1-{\left(0.8\right)}^{8}-2{\left(0.8\right)}^{8}$

$=1-3{\left(0.8\right)}^{8}$

$D$ is coorect.
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with a average of 5 phone calls during IO-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

  1. $\displaystyle \frac{6}{5^{e}}$
  2. $\displaystyle \frac{5}{6}$
  3. $\displaystyle \frac{6}{55}$
  4. $\displaystyle \frac{6}{e^{5}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to Poisson distribution
$\displaystyle P(X=r)=\frac{e^{-m}m^{r}}{r!}$
$\displaystyle \therefore P(X\leq 1)=P(X=0)+P(X=1)$
$\displaystyle =e^{-m}+\frac{e^{-m} m}{1!}$
Given m $\displaystyle =$mean$ = 5 $
$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$

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

The probability of r successes in case of poissons distrbution is

  1. $\dfrac{e^{\gamma }m}{\angle \gamma }$
  2. $\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$
  3. $\dfrac{e^{m}\gamma }{\angle \gamma }$
  4. $\dfrac{e^{-m}m^{r}}{\angle \gamma }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

in Poisson distribution, probability $ P(x; m  )=  \dfrac { { e }^{ - m  }{ m }^{ x } }{ \angle x } $
$m$ = mean  ;   $x$= number of success     $ \angle x =$ factorial $ x $
 For $r$ success  $x= r$;        
  $ P(r; m )=  \dfrac { { e }^{ -m  }{ m  }^{ r } }{ \angle r } $ 

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

A random variable $X$ has Poisson distribution with mean $2$. Then $P(X > 1.5)$ equals

  1. $2/e^{2}$
  2. $0$
  3. $1-\dfrac{3}{e^{2}}$
  4. $\dfrac{3}{e^{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For Poisson distribution of mean = $\mu = 2 $ 
 P$(x ; \mu) = \frac { { e }^{ -\mu  }{ \mu  }^{ x } }{ x! }$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)]$ 

P$(0; 2) = \dfrac { { e }^{-2}{2}^{0}}{0!}={e}^{-2}$
 P$(1; 2) = \dfrac { { e }^{-2}{2}^{} } {1!}=2{e}^{-2}$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)] $
$ = 1 - [{e}^{-2}+ 2{e}^{-2} ] = 1 - 3  {e}^{-2} = 1 - \frac { 3 }{ { e }^{ 2 } } $

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

If $X$ is a random poisson variate such that $\alpha =p(X=1)=p(X=2)$, then $p(X=4)=$

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

In  Poisson distribution such that  $ \alpha = p(X=1)=p(X=2) $ 
        $   p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $  
       => $ \alpha = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}  $       =>   $ \alpha =  { e }^{-\mu}{\mu}  $
                 =>  $ { e }^{-\mu}{\mu} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2} $
                 =>   $ \mu = 2 $ 
  $   p(4; \mu) = \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} =  \dfrac { { e }^{ -\mu  }{ \mu \times { \mu  }^{ 3 } } }{ 24 }  = \dfrac { \alpha {\times { 2 }^{ 3 } } }{ 24 }  = \dfrac { \alpha  }{ 3 } $  

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

If for a poisson distribution $P(X=0)=0.2$, then the variance of the distribution is

  1. $5$
  2. $log _{10}5$
  3. $log _{e}5$
  4. $log _{5}e$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Poisson's distribution implies
$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence substituting $x=0$ we get
$e^{-\lambda}=0.2$
$e^{\lambda}=5$
$\lambda=ln(5)$
Hence mean=variance=$log _{e}(5)$

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

In a Poisson distribution, the probability $P(X=0)$ is twice the probability $P(X=1)$. The mean of the distribution is

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

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=2P(x=1)$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=2\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$2\lambda=1$
$\lambda=\dfrac{1}{2}$
Hence mean=variance=$0.5$

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

A random variable $X$ follows poisson distribution such that $P(X=k)=P(X=k+1)$ then the parameter of the distribution $\lambda =$

  1. $K$
  2. $K+1$
  3. $\dfrac{K}{2}$
  4. $\dfrac{K+1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$P(X=k)=P(X=k+1)$
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $


$ P(k;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ k } }{ k! } =P(k+1;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ k+1 } }{ (k+1)! }$

$ 1 = \dfrac {\mu}{k+1} $

$ \mu = k+1 $

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

In a poisson distribution $P(X=0)=P(X=1)=k$, then the value of $k$ is

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

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

Given,$P(X=0)=P(X=1)=k $

$ \dfrac { { e }^{ -\mu }{ \mu }^{ 0 } }{ 0! } =\dfrac { { e }^{ -\mu }{ \mu }^{ 1 } }{ 1! }  $
$ { e }^{ -\mu } ={ e }^{ -\mu } \mu $
$ { e }^{ -\mu } (1 - \mu) = 0 $ 
Since $ { e }^{ -\mu } \neq 0$ , => $ \mu = 1 $
$ P(X=0) = \dfrac { { e }^{ -\mu }{ \mu }^{ 0 } }{ 0! } = { e }^{ -\mu } ={ e }^{ - 1} = \dfrac {1}{e} $