CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 4 - MCQExams.com

For a binomial distribution if $$P=\dfrac{2}{3}$$ and $$n=10$$ the probability of mode is 
  • $$^{10}C_{4}\left ( \dfrac{1}{3} \right )^{7}\left ( \dfrac{2}{3} \right )^{3}$$
  • $$^{10}C_{7}\left ( \dfrac{2}{3} \right )^{7}\left ( \dfrac{1}{3} \right )^{3}$$
  • $$\left ( \dfrac{2}{3} \right )^{7}\left ( \dfrac{1}{3} \right )^{3}$$
  • $$\left ( \dfrac{1}{3} \right )^{7}\left ( \dfrac{2}{3} \right )^{3}$$
$$8$$ unbiased coins are tossed at a time $$512$$ times.The expected frequency of getting one head is
  • $$4$$
  • $$8$$
  • $$16$$
  • $$20$$
 Out of $$2000$$ families with $$4$$ children, the number of families you expect to have at least one boy is
  • $$1875$$
  • $$750$$
  • $$1250$$
  • $$625$$
Out of $$2000$$ families with $$4$$ children, the number of families you expect to have two male children
  • $$1875$$
  • $$750$$
  • $$1250$$
  • $$625$$
If for a $$B.D.$$, $$ P=\dfrac{2}{3}$$ , $$n = 8$$, then the probability of mode is
  • $$^{8}C_{6}\left ( \dfrac{2}{3} \right )^{6}\left ( \dfrac{1}{3} \right )^{2}$$
  • $$^{8}C_{4}\left ( \dfrac{2}{3} \right )^{4}\left ( \dfrac{1}{3} \right )^{4}$$
  • $$^{10}C_{6}\left ( \dfrac{2}{3} \right )^{6}\left ( \dfrac{1}{3} \right )^{4}$$
  • $$^{10}C_{4}\left ( \dfrac{2}{3} \right )^{4}\left ( \dfrac{1}{3} \right )^{4}$$
$$201$$ coins each with probability $$P(0<P<1)$$ of showing head are tossed together. If the probability of getting $$100$$ heads is equal to the probability of getting $$101$$ heads, then the value of $$P$$ is
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{6}$$
If $$x$$ is a binomial variable with $$P=\dfrac{1}{4}$$, then the smallest value of $$n$$ so that $$P(x\geq 1)> 0.70$$ is
  • $$3$$
  • $$4$$
  • $$5$$
  • $$6$$
Twenty coins each with probability $$P(0<P<1)$$ of showing head are tossed together. If the probability of getting $$10$$ heads is equal to the probability of getting $$11$$ heads, the value of $$P$$ is
  • $$\dfrac{9}{20}$$
  • $$\dfrac{11}{20}$$
  • $$\dfrac{11}{21}$$
  • $$\dfrac{9}{21}$$
In a market region half of the households is known to use a particular brand of soap. In a household survey, a sample of $$10$$ house holds are alloted to each investigator and $$2048$$ investigators are appointed for the survey. The number of investigators likely to report that there are at least $$4$$ users is
  • $$240$$
  • $$352$$
  • $$1696$$
  • $$120$$
If we take $$1280$$ sets each of $$10$$ tosses of a fair coin, the number of sets we expect to get $$7$$ heads and $$3$$ tails is
  • $$450$$
  • $$300$$
  • $$150$$
  • $$75$$
In a market region half of the households is known to use a particular brand of soap. In a house hold survey, a sample of $$10$$ house holds are alloted to each investigator and $$2048$$ investigators are appointed for the survey. The number of investigators likely to report that there are three households is
  • $$240$$
  • $$352$$
  • $$1696$$
  • $$120$$
An irregular six faced die is thrown and the expectation that in $$10$$ throws it will give $$5$$ even numbers is twice the expectation that it will give $$4$$ even numbers. The number of times in $$10000$$ sets of $$10$$ throws you expect to give no even numbers is
  • $$6$$
  • $$1$$
  • $$2$$
  • $$9$$
The largest possible variance of a binomial variate is
  • $$n$$
  • $$\dfrac{n}{2}$$
  • $$\dfrac{n}{4}$$
  • $$\dfrac{n}{6}$$
Suppose $$A$$ and $$B$$ are two equally strong table tennis players. Which of the following two events is more probable
$$a)$$ $$A$$ beats $$B$$ in exactly $$3$$ games out of $$4$$
$$b)$$ $$A$$ beats $$B$$ in exactly $$5$$ games out of $$8$$
  • $$a$$
  • $$b$$
  • $$a$$ & $$b$$
  • neither$$a$$ nor $$b$$
Assuming that half of the population are consumers of rice so that the chance of an individual being a consumer is $$1/2$$ and assuming that $$1024$$ investigators each take $$10$$ individuals to see whether they are consumers, the number of investigators you expect to report that three or less are consumers is
  • $$360$$
  • $$240$$
  • $$176$$
  • $$60$$
The least number of times a fair coin is to be tossed in order that the probability of getting atleast one head is at least $$0.99$$ is 
  • $$5$$
  • $$6$$
  • $$7$$
  • $$8$$
If the difference between the mean and variance of binomial distribution for $$5$$ trials is $$5/9$$, the distribution is of the form
  • $$\left( 5,\ \cfrac { 1 }{ 2 } ,\ \cfrac { 1 }{ 2 } \right) $$
  • $$\left( 5,\ \cfrac { 1 }{ 3 } ,\ \cfrac { 2 }{ 3 } \right) $$
  • $$\left( 5,\ \cfrac { 3 }{ 4 } ,\ \cfrac { 1 }{ 4 } \right) $$
  • $$\left( 5,\ \cfrac { 3 }{ 5 } ,\ \cfrac { 2 }{ 5 } \right) $$
A coin is tossed $$n$$ times. If the probability of getting head $$6$$ times is equal to the probability of getting head 8 times then $$n \: =$$
  • $$6$$
  • $$8$$
  • $$14$$
  • $$10$$
In a market region half of the households is known to use a particular brand of soap. In a household survey, a sample of $$10$$ house holds are alloted to each investigator and $$2048$$ investigators are appointed for the survey. The number of investigators likely to report that there are not more than $$3$$ households is
  • $$240$$
  • $$352$$
  • $$1696$$
  • $$120$$
A machine manufacturing screws is known to produce $$5\%$$ defectives. In a random sample of $$15$$ screws the probability that there are exactly $$3$$ defectives is
  • $$\dfrac{16}{20}$$
  • $$\dfrac{17}{20}$$
  • $$^{15}C_{3}\left ( \dfrac{1}{20} \right )^{3}\left ( \dfrac{19}{20} \right )^{12}$$
  • $$\dfrac{18}{20}$$
The sum and product of mean and variance of a binomial distribution are $$24$$ and $$128$$ respectively.The binomial distribution is
  • $$\left ( \dfrac{1}{2} +\dfrac{1}{2}\right )^{32}$$
  • $$\left ( \dfrac{3}{10} +\dfrac{7}{10}\right )^{32}$$
  • $$\left ( \dfrac{1}{50} +\dfrac{49}{50}\right )^{32}$$
  • $$\left ( \dfrac{1}{3} +\dfrac{2}{3}\right )^{32}$$
A coin tossed $$n$$ times. If the probability that $$4, 5, 6$$ heads occur are in $$A.P.$$, then $$n =$$
  • $$14$$
  • $$8$$
  • $$15$$
  • $$11$$
A symmetric die is thrown $$(2n+1)$$ times. The probability of getting a prime score on the upturned face at most $$n$$ times is
  • $$\displaystyle\frac{1}{2}$$
  • $$\displaystyle\frac{1}{3}$$
  • $$\displaystyle\frac{1}{4}$$
  • $$\displaystyle\frac{2}{3}$$
If for a binomial distribution with $$n = 5, 4P(X=1)=P(X=2),$$ the probability of success is
  • $$\displaystyle\frac{1}{3}$$
  • $$\displaystyle\frac{2}{3}$$
  • $$\displaystyle\frac{1}{4}$$
  • $$\displaystyle\frac{1}{8}$$
If the difference between the mean and variance of a binomial distribution for $$5$$ trials is $$\dfrac{5}{9}$$, then the distribution is
  • $$\left ( \dfrac{1}{9}+\dfrac{2}{9} \right )^{5}$$
  • $$\left ( \dfrac{1}{4}+\dfrac{3}{4} \right )^{5}$$
  • $$\left ( \dfrac{2}{3}+\dfrac{1}{3} \right )^{5}$$
  • $$\left ( \dfrac{3}{4}+\dfrac{1}{4} \right )^{5}$$
For a binomial distribution : $$n = 6$$ and $$9$$. $$P(X = 4) = P (X = 2)$$. Then we have
  • $$P(X=1)\leq P(X=5)< P(X=3)$$
  • $$P(X=5)> P(X=3)> P(X=1)$$
  • $$P(X=5)< P(X=3)< P(X=1)$$
  • $$P(X=5)+ P(X=3)> P(X=1)$$
If $$X$$ is a binomial variate with $$n=6$$ and $$9P(X=4)=P(X=2)$$, the parameter $$p$$ is
  • $$\dfrac{3}{4}$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{2}$$
For a binomial distribution mean is $$6$$ and S.D. is $$2$$. The distribution is
  • $$\left( 18,\ \cfrac { 1 }{ 3 } ,\  \cfrac { 2 }{ 3 } \right) $$
  • $$\left( 9,\ \cfrac { 2 }{ 5 } ,\ \cfrac { 3 }{ 5 } \right) $$
  • $$\left( 9,\ \cfrac { 1 }{ 4 } ,\ \cfrac { 3 }{ 4 } \right) $$
  • $$\left( 9,\ \cfrac { 5 }{ 6 } ,\ \cfrac { 1 }{ 6 } \right) $$
If for a B.D. with $$n=12$$, the ratio of variance to mean is $$\displaystyle \frac{1}{3}$$, then the probability of $$10$$ successes is
  • $${ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 10 }$$
  • $$_{ 10 }^{ 12 }{ C{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 10 } }$$
  • $$_{ 10 }^{ 12 }{ C{ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 10 } }$$
  • $$_{ 10 }^{ 12 }{ C{ \left( \cfrac { 1 }{ 4 } \right) }^{ 2 }{ \left( \cfrac { 3 }{ 4 } \right) }^{ 10 } }$$
The mean or average number of heads when we toss $$10$$ unbiased coins is
  • $$20$$
  • $$10$$
  • $$5$$
  • $$15$$
If the mean and variance of a binomial variate $$X$$ are respectively  $$\displaystyle \frac{35}{6}$$and $$\displaystyle \frac{35}{36}$$, then the probability of $$X>6$$ is
  • $$1-{ \left( \cfrac { 5 }{ 6 } \right) }^{ 7 }$$
  • $${ \left( \cfrac { 5 }{ 6 } \right) }^{ 7 }$$
  • $$1-{ \left( \cfrac { 1 }{ 6 } \right) }^{ 7 }$$
  • $${ \left( \cfrac { 1 }{ 6 } \right) }^{ 7 }$$
Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. The variance of the number of aces is
  • $$ 24/169 $$
  • $$ \sqrt{24/169} $$
  • $$ \sqrt{24/173} $$
  • $$ 24/173 $$
The probability that a student is not a swimmer is $$\displaystyle\frac{1}{5}$$. Out of $$5$$ students the probability that exactly four are swimmers is
  • $$\left ( \displaystyle\frac{4}{5} \right )^{4}$$
  • $$\left ( \displaystyle\frac{1}{5} \right )^{4}$$
  • $$^{5}C_{4}\left ( \displaystyle\frac{4}{5} \right )^{4}\left ( \displaystyle\frac{1}{5} \right )^{1}$$
  • $$^{5}C_{4}\left ( \displaystyle\frac{1}{5} \right )^{4}\left ( \displaystyle\frac{4}{5} \right )^{1}$$
$$\mathrm{A}$$ and $$\mathrm{B}$$ play a game in which $$\mathrm{A}^{'}\mathrm{s}$$ chance of winning is $$\displaystyle \frac{1}{5} $$ in a series of $$6$$ games, the probability that $$\mathrm{A}$$ will win all the $$6$$ games is
  • $$_{ 2 }^{ 6 }{ C }\left (\displaystyle\frac { 1 }{ 5 } \right )^{ 6 }$$
  • $$_{ 6 }^{ 6 }{ C }\left (\displaystyle\frac { 1 }{ 5 } \right )^{ 6 }\left (\displaystyle\frac { 4 }{ 5 }\right )^{ 0 }$$
  • $${ \left( \displaystyle\frac { 4 }{ 5 } \right) }^{ 6 }$$
  • $$_{ 5 }^{ 6 }{ C }{ \left( \displaystyle\cfrac { 1 }{ 5 } \right) }^{ 5 }\left( \displaystyle\cfrac { 4 }{ 5 } \right) $$
The binomial distribution whose mean is $$9$$ and the variance is $$2.25$$ is 
  • $$\left( 12,\ \cfrac { 1 }{ 2 } ,\ \cfrac { 1 }{ 2 } \right) $$
  • $$\left( 12,\ \cfrac { 2 }{ 3 } ,\ \cfrac { 1 }{ 3 } \right) $$
  • $$\left( 12,\ \cfrac { 5 }{ 6 } ,\ \cfrac { 1 }{ 6 } \right) $$
  • $$\left( 12,\ \cfrac { 3 }{ 4 } ,\  \cfrac { 1 }{ 4 } \right) $$
The probability of answering $$6$$ out of $$10$$ questions correctly in a true or false examination is
  • $$ \displaystyle ^{10}C_{4}\left ( \frac{1}{2} \right )^{4}$$
  • $$ \displaystyle ^{10}C_{6}\left ( \frac{1}{2} \right )^{6}$$
  • $$ \displaystyle ^{10}C_{6}\left ( \frac{1}{2} \right )^{10}$$
  • $$ \displaystyle ^{10}C_{6}\left ( \frac{1}{2} \right )^{8}$$
$$X$$ follows a binomial distribution with parameters $$n$$ and $$p$$, and $$Y$$ follows a binomial distribution with parameters $$m$$ and $$p$$. Then, if $$X$$ and $$Y$$ are independent $$\displaystyle P\left( \frac { X=r }{ X+Y=r+s }  \right) =$$
  • $$\displaystyle \frac { \left( _{ }^{ m }{ { C }_{ r } } \right) \left( _{ }^{ n }{ { C }_{ s } } \right) }{ _{ }^{ m+n }{ { C }_{ r+s } } } $$
  • $$\displaystyle \frac { \left( _{ }^{ m }{ { C }_{ s } } \right) \left( _{ }^{ n }{ { C }_{ r } } \right) }{ _{ }^{ m+n }{ { C }_{ r+s } } } $$
  • $$\displaystyle \frac { \left( _{ }^{ m }{ { C }_{ s } } \right) }{ _{ }^{ m+n }{ { C }_{ r } } } $$
  • none of these
lf $$\mathrm{m}$$ is the variance of a Poisson Distribution, then the sum of the terms in odd places is:
  • $$e^{-m}$$
  • $$e^{-m}\cosh m$$
  • $$e^{-m}\sinh m$$
  • $$e^{-m}\coth m$$
The variance of P.D. with parameter $$\lambda $$ is
  • $$\lambda $$
  • $$\sqrt{\lambda }$$
  • $$\dfrac{1}{\lambda}$$
  • $$\dfrac{1}{\sqrt {\lambda}}$$
A : the sum of the times in odd places in a P.D is $$e^{-\lambda }$$ cosh $$\lambda$$ 
R : cosh $$\lambda =\frac{\lambda ^{1}}{1!}+\frac{\lambda ^{3}}{3!}+\frac{\lambda ^{5}}{5!}+......$$
  • Both A and R are true and R is the correct

    explanation of A
  • Both A and R are true but R is not correct

    explanation of A
  • A is true but R is false
  • A is false but R is true
The probability of r successes in case of poissons distrbution is
  • $$\dfrac{e^{\gamma }m}{\angle \gamma }$$
  • $$\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$$
  • $$\dfrac{e^{m}\gamma }{\angle \gamma }$$
  • $$\dfrac{e^{-m}m^{r}}{\angle \gamma }$$
A box contains 6 tickets. Two of the tickets carry a prize of Rs. 5/- each, the other four a prize of Rs.1/-. If one ticket is drawn. The mean value of prize is
  • Rs. $$2.5$$
  • Rs. $$7/3$$
  • Rs. $$5/3$$
  • Rs.$$4/3$$
A random variable $$X$$ has the following probability distribution
$$X=x:\quad \quad \quad1   \quad \quad 2 \quad\quad    3  \quad\quad  4$$
$$p(X=x): \  \ \ \ \ \ 0.4 \quad 0.3 \quad \  0.2 \quad \  \   0.1$$, then its mean is
  • $$4$$
  • $$3$$
  • $$2$$
  • $$1$$
If m is the variance of P.D., then the ratio of sum of the terms in odd places to the sum of the terms in even places is
  • $$e^{-m}\cosh m$$
  • $$e^{-m}\sinh m$$
  • $$\coth m$$
  • $$\tanh m$$

If $${m}$$ is the variance of Poisson distribution, then sum of the terms in even places is
  • $$e^{-m}$$
  • $$e^{-m}\cosh m$$
  • $$e^{-m}\sinh m$$
  • $$e^{-m}\coth m$$
If m is the variance of P.D., then the ratio of sum of the terms in even places to the sum of the terms in odd places is
  • $$e^{-m}\cosh m$$
  • $$e^{-m}\sinh m$$
  • $$\coth m$$
  • $$\tanh m$$
A random variable $$X$$ follows poisson distribution such that $$P(X=k)=P(X=k+1)$$ then the parameter of the distribution $$\lambda =$$
  • $$K$$
  • $$K+1$$
  • $$\dfrac{K}{2}$$
  • $$\dfrac{K+1}{2}$$
If X is a poisson variate such that $$P(X=2)=9p(X=4)+90p(X=6)$$ , then the mean of x is
  • $$3$$
  • $$2$$
  • $$1$$
  • $$0$$
If a random variable $$X$$ has a poisson distributionsuch that $$P(X=1)=P(X=2)$$, its mean and varianceare
  • $$1,1$$
  • $$2, 2$$
  • $$2, 3$$
  • $$2,4$$
If in a poisson frequency distribution, the frequency of $$3$$ successes is $$\displaystyle \frac{2}{3}$$ times the frequency of $$4$$ successes, the mean of the distribution is
  • $$\displaystyle \frac{2}{3}$$
  • $$\displaystyle \frac{1}{3}$$
  • $$6$$
  • $$\sqrt{6}$$
0:0:1


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