MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 5

If for a poisson distribution

$P(X=0)=0.2$ , then the variance of the distribution is

0%
$5$
0%
$log_{10}5$
0%
$log_{e}5$
0%
$log_{5}e$

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=

$log_{e}(5)$

$X$ is a poisson variate with

$P(X=0)=P(X=1)$ , then

$P(X=2)$ is

0%
$\dfrac{e}{2}$
0%
$\dfrac{e}{6}$
0%
$\dfrac{1}{6e}$
0%
$\dfrac{1}{2e}$

Explanation

Here,

$P(X=0)=P(X=1)$

$\Rightarrow \displaystyle \frac{e^{-\lambda}\lambda^0}{0!}=\frac{e^{-\lambda}\lambda^1}{1!}\Rightarrow \lambda = 1$

$\displaystyle \therefore P(X = 2) = \frac{e^{-\lambda}1^2}{2!}=\frac{1}{2e}$

A random variable

$X$ has Poisson distribution with mean

$2$ . Then

$P(X > 1.5)$ equals

0%
$2/e^{2}$
0%
$0$
0%
$1-\dfrac{3}{e^{2}}$
0%
$\dfrac{3}{e^{2}}$

Explanation

For Poisson distribution of mean =

$\mu = 2$
P

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

$P(X>1.5) = 1 - [P(X=0) + P(X=1)]$

$(0; 2) = \dfrac { { e }^{-2}{2}^{0}}{0!}={e}^{-2}$
P

$(1; 2) = \dfrac { { e }^{-2}{2}^{} } {1!}=2{e}^{-2}$

$P(X>1.5) = 1 - [P(X=0) + P(X=1)]$

$= 1 - [{e}^{-2}+ 2{e}^{-2} ] = 1 - 3 {e}^{-2} = 1 - \frac { 3 }{ { e }^{ 2 } }$

$X$ is a random poisson variate such that

$\alpha =p(X=1)=p(X=2)$ , then

$p(X=4)=$

0%
$\alpha e^{-2}$
0%
$\alpha e^{2}$
0%
$2\alpha$
0%
$\dfrac{\alpha }{3}$

Explanation

In Poisson distribution such that

$\alpha = p(X=1)=p(X=2)$

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

$\alpha = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}$ =>

$\alpha = { e }^{-\mu}{\mu}$
=>

${ e }^{-\mu}{\mu} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2}$
=>

$\mu = 2$

$p(4; \mu) = \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} = \dfrac { { e }^{ -\mu }{ \mu \times { \mu }^{ 3 } } }{ 24 } = \dfrac { \alpha {\times { 2 }^{ 3 } } }{ 24 } = \dfrac { \alpha }{ 3 }$

$X$ is a random poisson variate such that

$E(X^{2})=6$ , then

$E(X)=$

0%
$-3$
0%
$2$
0%
$-3\&2$
0%
$-2$

Explanation

Variance =

$E(X^{ 2 }) - E[X]^{ 2 }$
Mean = Variance for P.D.
Let mean =

$m$
Therefore

$m =$

$6 - m*m$
Hence

$m = 2$

$X$ is a poisson variate with

$P(X=0) = 0.8,$ then the variance of

$X$ is

0%
$log_{e}20$
0%
$log_{10}20$
0%
$log_{e}(5/4)$
0%
$0$

Explanation

In Poisson distribution,

$P(X=0) = 0.8,$

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

$p(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!}$
=> 0.8 =

${ e }^{-\mu}$
=>

$\dfrac {4}{5} = { e }^{-\mu}$
=>

${ e }^{\mu} = \dfrac {5}{4}$
taking log to both sides
=>

$\log _{ e }{ { e }^{ \mu } } = \log_{e} (\dfrac {5}{4})$
=>

$\mu = \log_{e} (\dfrac {5}{4})$

$X$ is a poisson variate such that

$P(X=0)=0.1,P(X=2)=0.2$ , then the parameter

$\lambda$

0%
$2$
0%
$4$
0%
$5$
0%
$3$

Explanation

$P(X=0)=0.1$ and

$P(X=2)=0.2$

$P$

$(0; \mu) = \dfrac{ { e }^{-\mu}{\mu}^{0}}{0!} = { e }^{-\mu}$ =

$0.1$

$P$

$(2; \mu) = \dfrac{ { e }^{-\mu}{\mu}^{2}}{2!}$

$= 0.2$

$\dfrac {{0.1}{\mu}^{2}}{2}$

$= 0.2$

$\mu$

$= 2$

The parameter is mean (

$\mu$ ) in Poisson distribution, so

$=$

$2$

$X$ is a poisson variate such that

$P(X=0)=P(X=1)$ ,then

$P(X=2)=$

0%
$\dfrac{e}{2}$
0%
$\dfrac{e}{6}$
0%
$\dfrac{1}{6e}$
0%
$\dfrac{1}{2e}$

Explanation

$P(X=0)=P(X=1)$
P

$(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!} = P(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!}$

${ e }^{-\mu} (1 - \mu )$ = 0

$\mu = 1$
P(X=2) =

$\dfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \dfrac{{ e }^{-1}}{2} = \dfrac {1}{2e}$

$X$ is a poisson variate such that

$P(X=0)=\dfrac{1}{2}$ , the variance of

$X$ is

0%
$\dfrac{1}{2}$
0%
$2$
0%
$\log_{e}2$
0%
$3$

Explanation

Poisson's distribution implies

$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=

$\lambda$
Hence substituting

$x=0$

we get

$e^{-\lambda}=0.5$

$e^{\lambda}=2$

$\lambda=ln(2)$
Hence mean=variance=

$log_{e}(5)$

$X$ is a Poisson variate such that

$P(X = 2) = 9P(X = 4)$ then mean and variance of

$X$ are

0%
$1, 1$
0%
$2, 2$
0%
$\displaystyle \frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$
0%
$\displaystyle\frac{2}{3},\frac{2}{3}$

Explanation

Poisson's distribution implies

$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=

$\lambda$
Hence

$P(x=2)=9P(x=4)$

$9\dfrac{\lambda^{4}e^{-\lambda}}{4!}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$

$9\lambda^{2}=12$

$\lambda^{2}=\dfrac{4}{3}$

$\lambda=\dfrac{2}{\sqrt{3}}$
Hence mean

$=$ variance

$=$

$\dfrac{2}{\sqrt{3}}$

$X$ is a Poisson variate such that

$P(X=1) = P(X=2)$ then

$P(X=4)=$

0%
$\dfrac{1}{2e^{2}}$
0%
$\dfrac{1}{3e^{2}}$
0%
$\dfrac{2}{3e^{2}}$
0%
$\dfrac{1}{e^{2}}$

Explanation

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

$P(X=1) = P(X=2)$

$\dfrac { { e }^{ -\mu }{ \mu }^{ 1 } }{ 1! } =\dfrac { { e }^{ -\mu }{ \mu }^{ 2 } }{ 2! }$

$\mu = 2$

$P(X=4)= \dfrac { { e }^{ -2 }{ 2 }^{ 4 } }{ 4! }= \dfrac {2}{3{e}^{2}}$

If a random variable

$X$ follows a P.D. such that

$P(X=1)=P(X=2)$ , then

$P(X=0)=$

0%
$e^{2}$
0%
$\dfrac{1}{e^{2}}$
0%
$\dfrac{1}{e}$
0%
$e$

Explanation

$P(X=1)=P(X=2)$

$P$

$(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = P(2; \mu) = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}$

${ e }^{-\mu} [\mu (2 - \mu )]$

$= 0$

either

$\mu = 0 or \mu = 2$

$P(X=0) =$

$\dfrac { { e }^{-\mu}{\mu}^{0}}{0!} = { e }^{-\mu}$

So, either

${ e }^{-0}$ or

${ e }^{-2}$
that is 1 or

$\dfrac {1}{{ e }^{2}}$

For a Poisson variate

$X$ if

$P(X=2)=3P(X=3)$ , then the mean of

$X$ is

0%
$1$
0%
$1/2$
0%
$1/3$
0%
$1/4$

Explanation

Poisson's distribution implies

$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=

$\lambda$
Hence

$P(x=2)=3P(x=3)$

$3\dfrac{\lambda^{3}e^{-\lambda}}{3!}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$

$3\lambda=3$

$\lambda=1$
Hence mean=variance=

$1$

If for a poisson variable

$X$ ,

$P(X=1)=2.\ P(X=2)$ , then the parameter

$\lambda$ is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

For Poisson's distribution,

$P(X)$ is given by

$\displaystyle P(X)=\frac { { e }^{ -\lambda }.{ \lambda }^{ x } }{ x! }$
Hence

$\displaystyle P(X=1) =\displaystyle \frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! }$ .
Similarly

$\displaystyle P(X=2)=\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! }$ .
Given that,

$P(X=1) = 2 P(X=2)$

$\Rightarrow \displaystyle \frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } =2.\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! }$
Simplifying we get

$\lambda = 1$

In a poisson distribution

$P(X=0)=P(X=1)=k$ , then the value of

$k$ is

0%
$1$
0%
$\displaystyle\frac{1}{e}$
0%
$e$
0%
$\sqrt{2}$

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

${ e }^{ -\mu } \neq 0$ , =>

$\mu = 1$

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

If in a poisson distribution

$P(X=1)=P(X=2)$ ; the mean of the distribution

$f(x)=e^{-x}\dfrac{\lambda ^{x}}{\angle x}$ is

0%
$1$
0%
$2$
0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{2}$

Explanation

Poisson's distribution implies

$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=

$\lambda$
Hence

$P(x=1)=P(x=2)$

$\frac{\lambda^{1}e^{-\lambda}}{1!}=\frac{\lambda^{2}e^{-\lambda}}{2!}$

$\lambda=2$
Hence mean=variance=

$2$

$X$ is a Poisson variate with parameter

$\displaystyle \frac{3}{2}$ , find

$P(X\geq 2)$

0%
$\displaystyle \frac{5}{2}e^{\frac{-3}{2}}$
0%
$\displaystyle 1-\frac{5}{2}e^{\frac{-3}{2}}$
0%
$\displaystyle 1-e^{\frac{-3}{2}}$
0%
$\displaystyle e^{\frac{-3}{2}}$

Explanation

Here

$\lambda = \cfrac{3}{2}$

$\therefore P(X\geq 2) = 1-P(X=0)-P(X=1)$

$\displaystyle =1-\cfrac{e^{-3/2}(3/2)^0}{0!}-\cfrac{e^{-3/2}(3/2)^1}{1!}=1-\cfrac{5}{2}e^{-3/2}$

$X$ is a Poisson variate with parameter

$1.5$ , then

$P(X>1)$ is

0%
$1-e^{-1.5}$
0%
$e^{-1.5}(2.5)$
0%
$1-e^{-1.5}(2.5)$
0%
$1-e^{-1.5}(3.5)$

Explanation

$P(X>1)$

$=1-[P(X=0)+P(X=1)]$

$=1-[e^{-1.5}+1.5e^{-1.5}]$

$=1-e^{1.5}(1+1.5)$

$=1-e^{1.5}(2.5)$

Suppose

$X$ is a poisson variable such that

$P(X=2)=\frac{2}{3}P(X=1)$ , then

$P(x=0)$ is

0%
$\dfrac{3}{4}$
0%
$e^{\dfrac{4}{3}}$
0%
$e^{\dfrac{-4}{3}}$
0%
$\dfrac{1}{2}$

Explanation

Poisson's distribution implies

$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$

Where mean=variance=

$\lambda$

Hence,

$P(X=2)=\dfrac{2}{3}P(x=1)$

$\dfrac{\lambda^{2}e^{-\lambda}}{2!}=\dfrac{2\lambda^{1}e^{-\lambda}}{3.1!}$

$\dfrac{\lambda}{2}=\dfrac{2}{3}$

$\lambda=\dfrac{4}{3}$
Hence,

$P(X=0)=e^{\dfrac{-4}{3}}$

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

0%
$4$
0%
$3$
0%
$1.5$
0%
$0$

Explanation

Given

$P(X\geq 1) = 1-e^{-1.5}$

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

$X$ is a poisson variable such that

$P(X=2)=\frac{2}{3}P(X=1)$ , then

$P(x=3)$ is

0%
$e^{\frac{-4}{3}}$
0%
$\frac{64}{162}e^{\frac{-4}{3}}$
0%
$e^{\frac{-3}{4}}$
0%
$e^{\frac{3}{4}}$

Explanation

$P(X=2)= \cfrac {2}{3}P(X=1)$

$P$ $(2; \mu) = \cfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \cfrac {2}{3} P(1; \mu)$ = $\cfrac {2}{3} \cfrac { { e }^{-\mu}{\mu}^{1}}{1!}$

${ e }^{-\mu} \mu (\cfrac {\mu}{2} - \cfrac {2}{3})=0$

either $\mu = 0\ or \mu = \cfrac{4}{3}$

but here $\mu = \cfrac{4}{3}$ $P(X=3) =$ $\cfrac { { e }^{-\mu}{\mu}^{3}}{3!} = \cfrac { { e }^{ \frac { -4 }{ 3 } }{( \frac { 4 }{ 3 } })^{ 3 } }{ 6 } = \cfrac { 64 }{ 162 } { e }^{ \frac{-4}{3} }$

$X$ is a random Poisson variate such that

$P(X=0)=\displaystyle\frac{1}{e}$ , then the variance of the same distribution is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

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=

$1$

In a Poisson distribution, the probability

$P(X=0)$ is twice the probability

$P(X=1)$ . The mean of the distribution is

0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{3}{4}$

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=

$0.5$

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,

$\displaystyle P(X=4)=\frac{512}{3e^{8}}$

0%
both A and R are true and R is the correct explanation of A
0%
both A and R are true and R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

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.

$X$ is a random poission variate such that

$2P(X=0)+P(X=2)=2P(X=1)$ then

$E(X)=$

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

Poisson's distribution implies

$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=

$\lambda$
Hence

$2.P(x=0)+P(x=2)=2P(x=1)$

$2\dfrac{\lambda^{0}e^{-\lambda}}{0!}+\dfrac{\lambda^{2}e^{-\lambda}}{2!}=2\dfrac{\lambda^{1}e^{-\lambda}}{1!}$

$2e^{-\lambda}+\dfrac{\lambda^{2}e^{-\lambda}}{2!}=2\lambda\: e^{-\lambda}$

$2\lambda\: e^{-\lambda}-2e^{-\lambda}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$

$4\lambda-4=\lambda^{2}$

$\lambda^{2}-4\lambda+4=0$

$(\lambda-2)^{2}=0$

$\lambda=2$
Hence mean=variance=

$2$

$3$ % of electric bulbs manufactured by a company are defective, the probability that a sample of

$100$ bulbs has no defective bulbs is

0%
0
0%
$e^{-3}$
0%
$1-e^{-3}$
0%
$3e^{-3}$

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)

$P(0; 3)=\dfrac { { e }^{ -3 }{ 3 }^{ 0 } }{ 0! }$

$P(0;3)={ e }^{ -3 }$

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

0%
$\displaystyle \frac{e^{-0.02}.2^{1}}{1!}$
0%
$1-e^{-0.2}$
0%
$e^{-0.2}$
0%
$1-e^{1.2}$

Explanation

Here Poisson parameter

$\lambda = \cfrac{10}{50}=0.2$

$\therefore P(x \geq 1)=1-P(X=0)=1-\cfrac{e^{-0.2}.(.2)^0}{0!}=1-e^{-0.2}$

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

0%
$e^{-1.5}$
0%
$1.5\times e^{-1.5}$
0%
$1-2.5\times e^{-1.5}$
0%
$1-1.5\times e^{-1.5}$

Explanation

Here

$\lambda = 1.5$
Hence probability that only one car is used is

$=P(X=1) = \cfrac{e^{-1.5}(1.5)}{1!}=1.5e^{-1.5}$

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

0%
$1.12 \times e^{-1.5}$
0%
$1-2.5 \times e^{-1.5}$
0%
$1-3.625 \times e^{-1.5}$
0%
$3.625 \times e^{-1.5}$

Explanation

Here

$\lambda = 1$
Hence probability that both the cars are used

$=1-P(X=0)-P(X=1)=1-\cfrac{e^{-1.5}(1.5)^0}{0!}-\cfrac{e^{-1.5}(1.5)^1}{1!}=1-2.5e^{-1.5}$

Cycle tyres are supplied in lots of

$10$ and there is a chance of

$1$ in

$500$ to be defective. Using poisson distribution, the approximate number of lots containing no defectives in a consignment of

$10,000$ lots if

$e^{-0.02}=0.9802$ is

0%
$9980$
0%
$9998$
0%
$9802$
0%
$9982$

Explanation

Here

$\lambda = \cfrac{1}{500}\times 10 = 0.02$
Thus probability that lot is not defective is

$=P(X=0)=\cfrac{e^{-0.002}(.0020^0}{0!}=e^{-.002}=0.9802$
Hence number of no defective lots out of

$10,000$ is

$=.9802\times 10,000=9802$

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

0%
$\displaystyle e^{-10}\left [ 1+\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$
0%
$\displaystyle e^{-10}\left [ 1+\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$
0%
$\displaystyle e^{-10}\left [ 1-\frac{10}{1!}+\frac{10^{2}}{2!}+\frac{10^{3}}{3!}+\frac{10^{4}}{4!} \right ]$
0%
$\displaystyle e^{-10}\left [ 1-\frac{100}{1!}+\frac{100^{2}}{2!}+\frac{100^{3}}{3!}+\frac{100^{4}}{4!} \right ]$

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

$=\sum_{k=0} ^{k=4} \dfrac{e^{-10}.10^{k}}{k!}$

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

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

0%
$e^{-2}$
0%
$1-e^{-2}$
0%
$\displaystyle \frac{e^{-2}2^{1}}{\angle 1}$
0%
$e^{-002}$

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)

$P(0; 2)=\dfrac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } = {e}^{-2}$

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

0%
$e^{-2}$
0%
$1-e^{-2}$
0%
$\displaystyle \frac{e^{-2}2^{1}}{1!}$
0%
$e^{-0.02}$

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)

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

Suppose

$300$ misprints are distributed randomly throughout a book of

$500$ pages. The probability that a given page contains, at least one misprint is

0%
$1.e^{-0.6}$
0%
$1-e^{-0.6}$
0%
$(0.6)e^{-0.6}$
0%
$(0.06)e^{-0.6}$

Explanation

Here

$\lambda = \cfrac{300}{500}=0.6$
Hence

$P(X\geq 1) = 1-P(X=0)=1-\cfrac{e^{-0.6}(0.6)^0}{0!}=1-e^{-0.6}$

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

0%
$e^{-15}$
0%
$1-e^{-6}$
0%
$1-e^{-15}$
0%
$e^{-6}$

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

$=1-e^{-15}$ .

Suppose there is an average of

$2$ suicides per year per

$50,000$ population. In a city of population

$1,00,000$ , the probability that in a given year there are, zero suicides is

0%
$1.e^{-2}$
0%
$1-e^{-2}$
0%
$e^{-4}$
0%
$1-e^{-4}$

Explanation

P.D parameter

$\lambda = \cfrac{2}{50000}\times 100000 = 4$
Thus probability that in a year there is no suicide is

$=P(X=0)=e^{-4}$

Suppose

$2$ % of the people on an average are left handed. The probability of 3 or more left handed among 100 people is

0%
$3e^{-2}$
0%
$4e^{-2}$
0%
$1-5e^{-2}$
0%
$5 e^{-2}$

Explanation

Here P.D parameter

$\lambda = \cfrac{2}{100}\times 100=2$
Hence probability that 3 or more people are left handed is

$=1-P(X=0)-P(X=1)-P(X=2)$

$=1-e^{-2}-\cfrac{e^{-2}2}{1!}-\cfrac{e^{-2}2^2}{2!}=1-5e^{-2}$

A manufacturer who produces medicine bottles finds that

$0.1$

$\%$ of the bottles are defective. The bottles are packed in boxes containing

$500$ bottles. A drug manufacturer buys

$100$ boxes from the producer of bottles. Using Poisson distribution, the number of boxes with no defective bottle is

0%
$100\times e^{-0.1}$
0%
$100\times e^{-0.5}$
0%
$100\times e^{-0.05}$
0%
$100\times e^{-0.01}$

Explanation

$\displaystyle p=\frac { 0.1 }{ 100 } =0.001,n=500\\ \lambda =np=500\times 0.001=0.5\\ N=100$
We know that

$\displaystyle p\left( r \right) =e^{ -1 }\frac { { \lambda }^{ r } }{ r! }$
Number of boxes containing no defective bottle

$\displaystyle =N.P.\left( r=0 \right) =1000\times { e }^{ 0.5 }\frac { \left( 0.5 \right) ^{ 0 } }{ 0! } =1000\times { e }^{ -0.5 }$

It is known that the probability that an item produced by a certain machine will be defective is

$0.01$ . Use Poisson distribution, to find the probability in a sample of

$100$ items selected at random from the total output, that there are not more than one defective item.

0%
$\displaystyle \frac{3}{e}$
0%
$\displaystyle \frac{2}{e}$
0%
$\displaystyle \frac{1}{e}$
0%
$\displaystyle \frac{4}{e}$

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,

$\mu$ =

$100 \times 0.01 = 1$

$P(x ; \mu ) = P (x =0 ; 1) + P (x=1 ;1)$

$P(x ; 1)=\dfrac { { e }^{ -1}{ 1 }^{0 } }{ 0! } + \dfrac { { e }^{ -1}{ 1 }^{1 } }{ 1! }$

$= { e }^{ -1} +{ e }^{ -1} = 2 { e }^{ -1} = \dfrac{2}{e}$

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

0%
$e^{-2}$
0%
$1-e^{-2}$
0%
$2.e^{-2}$
0%
$3.e^{-2}$

Explanation

Here P.D parameter

$\lambda = 2$
Hence probability of obtaining zero calls during 10 AM to 11 AM is

$=(P(X=0)=e^{-2}$

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

0%
$e^{-2}$
0%
$(1-e^{-2})$
0%
$2e^{-2}$
0%
$3e^{-2}$

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)

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

A manufacturing concern employing a large number of workers finds that, over a period of time, the average absentee rate is

$2$ workers per shift. The probability that exactly

$2$ workers will be absent in a chosen shift at random is

0%
$\displaystyle \frac{e^{-2}2^{2}}{ 2!}$
0%
$\displaystyle \frac{e^{-2}2^{3}}{3!}$
0%
$e^{-2}$
0%
$e^{-3}$

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,

$\mu$ = 2
x = 2 (exact 2 workers)

$P(2;2)=\dfrac { { e }^{ -2 }{ 2 }^{2 } }{ 2! }$

Suppose

$220$ misprints are distributed randomly throughout a book of

$200$ pages. The probability that a given page contains, no misprint is

0%
$1.e^{-02.2}$
0%
$1-e^{-02.2}$
0%
$e^{-1.1}$
0%
$e^{-2.1}$

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,

$\mu = \dfrac{220}{200} = 1.1$

$x = 0$ (no misprint)

$P(0;1.1)=\dfrac { { e }^{ -1.1 }{ 1.1 }^{0 } }{ 0! } = {e}^{-1.1}$

A manufacturer who produces medicine bottles finds that

$0.1$ % of the bottles are defective. The bottles are packed in boxes containing

$500$ bottles. A drug manufacturer buys

$100$ boxes from the producer of bottles. Using poisson distribution,the number of boxes with at least one defective bottle is

0%
$100(1-e^{-0.1})$
0%
$100(1-e^{-0.5})$
0%
$100(1-e^{-0.05})$
0%
$100(1-e^{-0.01})$

Explanation

Here

$\lambda = \cfrac{0.1}{100}\times 500=0.5$
Hence number of boxes out of 100 which contain at least one defective bottle is,

$=100\left(1-P(X=0)\right)=100\left(1-e^{-0.5}\right)$

The chance of a traffic accident in a day attributed to a taxi driver is

$0.001$ . Out of a total of

$1000$ days the number of days with no accident is

0%
$1000\times e^{-1}$
0%
$1000\times e^{-0.1}$
0%
$1000\times e^{-0.001}$
0%
$1000\times e^{-0.0001}$

Explanation

Here

$\lambda = 0.001$
Hence number of day out of 1000 days without accident is

$1000\times P(X=0)=1000\times e^{-0.001}$

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

0%
$\displaystyle \frac{e^{-12}12^{4}}{ 4!}$
0%
$\displaystyle \frac{e^{-4}12^{12}}{ 3!}$
0%
$\displaystyle \frac{e^{-6}12^{4}}{ 4!}$
0%
$\displaystyle \frac{e^{-3}12^{2}}{ 4!}$

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)

$P(4;12)=\dfrac { { e }^{ -12 }{12 }^{4 } }{ 4! }$

The number of accidents in a year attributed to a taxi driver in a city follows Poisson distribution with mean

$3$ . Out of

$1000$ taxi drivers, the approximate number of drivers with no accident in a year given that

$e^{-3}=0.0498$ is

0%
$4.98$
0%
$49.8$
0%
$498$
0%
$4.8$

Explanation

Here

$\lambda = 3$
Hence Number of drivers with no accident out of 1000 is

$=1000\times P(X=0)=1000\times e^{-3}=1000\times 0.0498=49.8$

Four coins are tossed together. What is the probability that head will appear on at least one of the four?

0%
$\displaystyle\frac{15}{16}$
0%
$\displaystyle\frac{1}{16}$
0%
$\displaystyle\frac{1}{4}$
0%
None of these

Explanation

Probability of getting head if one coin is tossed is

$p = \cfrac{1}{2}$
Here 4 coin are tossed together i.e

$n =4$
Hence using binomial distribution,
Probability that head will appear at least of the four is

$=1-P(X = 0) =1-^4C_0\left(\cfrac{1}{2}\right)^4=1-\cfrac{1}{16}=\cfrac{15}{16}$

A manufacturer of cotter pins knows that

$5$ % of his product is defective. If he sells cotter pins in boxes of

$100$ and guarantees that not more than

$10$ pins will be defective, the approximate probability that a box will fail to meet the guaranteed quality is

0%
$\displaystyle \frac{e^{-5}5^{10}}{ 10!}$
0%
$1-\displaystyle \sum_{x=0}^{10}\frac{e^{-5}5^{x}}{ x!}$
0%
$1-\displaystyle \sum_{x=0}^{\infty }\frac{e^{-5}5^{x}}{ x!}$
0%
$\displaystyle \sum_{x=0}^{\infty }\frac{e^{-5}5^{x}}{ x!}$

Explanation

we are given

$n=100$
let

$p=$ probability of a defective bulb

$=5$ %

$=0.05$

$\therefore m=$ mean number of defective bulbs in a box of

$100=np=100\times 0.05=5$
Since p is small , we can use poison's distribution.
Probability of

$x$ defective bulbs in a box of

$100$ is

$\displaystyle P\left( X=x \right) =\frac { { e }^{ -m }{ m }^{ x } }{ x! } =\frac { { e }^{ -5 }{ 5 }^{ x } }{ x! } ,x=0,1,2...$
Probability that is box will fail to meet the guarented quality is

$\displaystyle P\left( X>10 \right) =1-P\left( X\le 10 \right) =1-\sum _{ x=0 }^{ 10 }{ \frac { { e }^{ -5 }{ 5 }^{ x } }{ x! } } =1-{ e }^{ -5 }\sum _{ x=0 }^{ 10 }{ \frac { { 5 }^{ x } }{ x! } }$

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

0%
$1-e^{-2}$
0%
$1-2e^{-2}$
0%
$1-3e^{-2}$
0%
$1-4e^{-2}$

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

$P=1-e^{-\lambda}-\dfrac{\lambda^{1}e^{-\lambda}}{1!}$

$=1-e^{-2}-2e^{-2}$

$=1-3e^{-2}$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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