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

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

There are $500$ boxes each containing $1000$ ballot papers for election. The chance that a ballot paper is defective is $0.002$. Assuming that the number of defective ballot papers follow Poisson distribution, the number of boxes containing at least one defective ballot paper given that $e^{-2}=0.1353$ is

  1. $216$
  2. $432$
  3. $648$
  4. $234$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = .002\times 1000=2$
Thus probability that a box contain at least one defective ballot $=1-P(X=0)=1-e^{-2}=1-0.1353=.8647$   
Hence   number of box out of $500$ which contain at least one defective ballot is $=.8647\times 500 =432$

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

Six unbiased coins are tossed $6400$ times. Using Poisson distribution, the approximate probability of getting six heads $2$ times is

  1. $\displaystyle \frac{e^{-64}(64)^{2}}{ 2!}$
  2. $\displaystyle \frac{e^{-100}(100)^{2}}{ 2!}$
  3. $1-\displaystyle \frac{e^{-100}(100)^{x}}{ x!}$
  4. $\displaystyle \frac{e^{-100}(100)^{x}}{x!}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $\lambda = \left(\cfrac{1}{2}\right)^6\times 64100=100$
Hence probability of getting six heads two times $=P(X=2)=\cfrac{e^{-100}(100)^2}{2!}$

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

A company knows on the basis of past experience that $2$% of the blades are defective. The probability of having 3 defective blades in a sample of $100$ blades is

  1. $e^{-2}2^{2}$
  2. $\displaystyle \frac{e^{-2}2^{3}}{3!}$
  3. $\displaystyle \frac{e^{-2}2^{3}}{ 2!}$
  4. $\displaystyle \frac{e^{-4}2^{-1}}{ 2!}$
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 $
Here,$ \mu  =  100 \times 0.02 = 2 $
$x = 3$(defective blades)
$ P(3;2)=\dfrac { { e }^{ -2 }{ 2 }^{3 } }{ 3! } $

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 neither 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
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 $
Here, $ \mu  = 1.5$ and $x = 0$ (neither car is used)
$ P(0;1.5)=\dfrac { { e }^{ -1.5 }{ 1.5 }^{0 } }{ 0! } = {e}^{-1.5} $

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

In a big city, $5$ accidents take place over a period of $100$ days. If the numebr of accidents follows P.D., the probability that there will be $2$ accidents in a day is

  1. $\displaystyle \frac{e^{-5}5^{2}}{ 2!}$
  2. $\displaystyle \frac{e^{-05}5^{2}}{ 2!}$
  3. $\displaystyle \frac{e^{-005}(0.05)^{2}}{ 2!}$
  4. $\displaystyle \frac{e^{5}5^{2}}{ 2!}$
Reveal answer Fill a bubble to check yourself
C 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 = \dfrac {5}{100} = 0.05 $
$x = 2$ 
$ P(2; 005)=\dfrac { { e }^{ -0.05 }{ 0.05 }^{ 2 } }{ 2! } $
Multiple choice

The probability of an event occurring is represented by:

  1. P(A)

  2. P(not A)

  3. P(A or B)

  4. P(A and B)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P(A) represents the probability of event A occurring.

Multiple choice

The probability of obtaining at least one success in a sequence of independent trials is calculated using the:

  1. Binomial distribution

  2. Poisson distribution

  3. Normal distribution

  4. Uniform distribution
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Binomial distribution is used to calculate the probability of obtaining at least one success in a sequence of independent trials.

Multiple choice

The probability of obtaining exactly k successes in a sequence of n independent trials with probability of success p is given by the:

  1. Binomial distribution

  2. Poisson distribution

  3. Normal distribution

  4. Uniform distribution
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Binomial distribution is used to calculate the probability of obtaining exactly k successes in a sequence of n independent trials with probability of success p.

Multiple choice

What is the probability mass function of a discrete random variable?

  1. A function that gives the probability of each possible value of the random variable

  2. A function that gives the cumulative probability of each possible value of the random variable

  3. A function that gives the expected value of the random variable

  4. A function that gives the variance of the random variable
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability mass function of a discrete random variable is a function that gives the probability of each possible value of the random variable.

Multiple choice

What is the moment generating function of a random variable?

  1. A function that gives the probability of each possible value of the random variable

  2. A function that gives the cumulative probability of each possible value of the random variable

  3. A function that gives the expected value of the random variable

  4. A function that gives the variance of the random variable
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The moment generating function of a random variable is a function that gives the expected value of the random variable raised to the power of t.