CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 2 - MCQExams.com

If X following the Binomial distribution with parameters n=6  and p and 9P (X = 4)  =  P (X = 2), then p is:
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{2}{3}$$
The probability of A's a Tenis game against $$B$$ is $$\dfrac{1}{3}$$. Find the probability that A wins at least $$1$$ of the total $$3$$ set of games. 
  • $$\dfrac{1}{19}$$
  • $$\dfrac{19}{27}$$
  • $$\dfrac{1q}{19}$$
  • $$\dfrac{7}{19}$$
If the mean and variance of a binomial distribution are  $$4$$ and $$2$$ respectively, then the probability of $$2$$ successes of that binomial variate $$X$$, is
  • $$\dfrac {1}{2}$$
  • $$\dfrac {219}{256}$$
  • $$\dfrac {37}{256}$$
  • $$\dfrac {7}{64}$$
$$\sum_{r=1}^{11} r.5^{r} =\dfrac{(43\times 5^{a}+5)}{b}$$, then $$(a+b)$$ is
  • $$18$$
  • $$28$$
  • $$15$$
  • $$38$$
$$X$$ follows a binomial distribution with parameters $$n=6$$ and $$p$$. If $$4P(X=4)=P(X=2)$$, then $$p$$ is:
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{6}$$
  • $$\dfrac{1}{3}$$
Find the Binomial probability distribution whose mean is $$3$$ and variance is $$2$$
  • $${\left( \dfrac { 2 } { 3 } + \dfrac { 1 } { 3 } \right) }^ { 9 }$$
  • $${\left( \dfrac { 5 } { 3 } + \dfrac { 2 } { 3 } \right)} ^ { 9 }$$
  • $${\left( \dfrac { 3 } { 3 } + \dfrac { 1 } { 2 } \right) }^ { 9 }$$
  • None of these
The probability that a certain kind of component will survive a given shock test is $$3/4$$. Find the probability that among $$5$$ components tested exactly $$2$$ will survive.
  • $$ \cfrac{90}{124}$$
  • $$ \cfrac{9}{14}$$
  • $$ \cfrac{90}{1024}$$
  • None of these
In a poisson distribution, the variance is $$m$$ . The sum of terms in odd places in the distribution is 
  • $$e^{-m}$$
  • $$e^{-m} \cos \, h \, (m)$$
  • $$e^{-m} \sin \, h \, (m)$$
  • $$e^{-m} \cot \, h \, (m)$$
The probability that a certain kind of component will survive a given shock test is $$3/4$$. Find the probability that among $$5$$ components tested at most $$3$$ will survive.
  • $$\cfrac{75}{1024}$$
  • $$\cfrac{375}{1024}$$
  • $$\cfrac{275}{1024}$$
  • None of these
Eight coins are tossed at a time, the probability of getting at least 6 heads up, is
  • $$\frac{7}{{64}}$$
  • $$\frac{{57}}{{64}}$$
  • $$\frac{{37}}{{256}}$$
  • $$\frac{{229}}{{256}}$$
If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, then p=......
  • 0.18
  • 0.2
  • 0.8
  • 0.4
In a binomial distribution, the probability of getting a success is  $$\dfrac{1}{4}$$  and the standard deviation is $$3$$. Then its mean is.  (Eamcet 2002)
  • $$12$$
  • $$10$$
  • $$8$$
  • $$6$$
For a binomial variate with n = 6, if
$$P(X=2)=9P(X=4),$$then the variance is
  • $$\dfrac{8}{9}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{9}{8}$$
  • $$4$$
If $$X$$ follows a binomial distribution with parameters $$n=8$$ and $$\displaystyle p=\frac { 1 }{ 2 } $$, then $$P\left( \left| X-4 \right| \le 2 \right) $$ is
  • $$\dfrac{119}{128}$$
  • $$\dfrac{119}{228}$$
  • $$\dfrac{19}{128}$$
  • $$\dfrac{18}{28}$$
The probability that a bomb dropped from a plane strikes the target is $$\displaystyle \frac{1}{5}$$. The probability that out of $$6$$ bombs dropped atleast $$2$$ bombs strike the target
  • $$0.345$$
  • $$0.246$$
  • $$0.543$$
  • $$0.426$$
A binomial distribution has a mean of $$5$$ and variance $$4$$. The number of trials is
  • $$20$$
  • $$16$$
  • $$25$$
  • $$10$$
If $$X$$ is binomial variate with $$E(X) = 5$$ and variance $$4$$. The parameters of the distribution are
  • $$\dfrac{1}{4},20$$
  • $$\dfrac{1}{5},20$$
  • $$25,\dfrac{1}{5}$$
  • $$\dfrac{1}{25},\dfrac{1}{5}$$
If the mean of the binomial distribution is $$25$$. Then standard deviation lies in the interval 
  • $$[0, 5)$$
  • $$(0, 5]$$
  • $$[0, 25)$$
  • $$(0, 25]$$
The mean of binomial distribution is 6 and its S.D. is $$\sqrt{2}$$ , then the number of trials $$n$$ is
  • $$7$$
  • $$8$$
  • $$9$$
  • $$10$$
A man take a step forward with probability $$0.4$$ and backward with probability $$0.6$$. The probability that at the end of eleven steps he is one step away from the starting point, is
  • $$0.37$$
  • $$0.57$$
  • $$0.3$$
  • None of these
A symmetrical die is rolled $$720$$ times. Getting a face with four points is considered to be a success. The mean and variance of the number of successes is
  • $$20, 120$$
  • $$120, 100$$
  • $$100, 100$$
  • $$50, 50$$
The mean and the variance of a random variable $$X$$ having a binomial distribution are $$4$$ and $$2$$ respectively, then $$P(X = 1)$$ is  
  • $$1/32$$
  • $$1/16$$
  • $$1/8$$
  • $$1/4$$
The probability of obtaining $$2$$ heads when an unbiased coin is tossed $$5$$ times is
  • $$\displaystyle \frac{5}{8}$$
  • $$\displaystyle \frac{4}{9}$$
  • $$\displaystyle \frac{5}{16}$$
  • $$\displaystyle \frac{4}{16}$$
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
  • $$\dfrac{128}{256}$$
  • $$\dfrac{219}{256}$$
  • $$\dfrac{37}{256}$$
  • $$\dfrac{28}{256}$$
For a binomial distribution $$n = 10$$, $$q = 0.4$$, then its mean is
  • $$1$$
  • $$4$$
  • $$6 $$
  • $$10$$
If for a binomial distribution mean is $$ \dfrac{10}{3}$$ and sum of mean and variance is $$\dfrac{40}{9}$$. Then the parameters are
  • $$\dfrac{2}{3},10$$
  • $$\dfrac{2}{3},20$$
  • $$5,\dfrac{2}{3}$$
  • $$4,\dfrac{2}{3}$$
If for a binomial distribution with n = 16, the ratio of mean to variance is $$\dfrac{5}{4}$$, then the probability of success is
  • $$\dfrac{4}{5}$$
  • $$\dfrac{2}{3}$$
  • $$\dfrac{1}{5}$$
  • $$\dfrac{1}{4}$$
If the mean of binomial distribution with $$9$$ trials is $$6$$, then its variance is
  • $$ 2 $$
  • $$3 $$
  • $$4 $$
  • $$\sqrt{2}$$
The mean and variance of a random variable $$X$$ having binomial distribution are $$4$$ and $$2$$ respectively. Then $$P(X>6)$$ = 
  • $$\dfrac{9}{128}$$
  • $$\dfrac{81}{128}$$
  • $$\dfrac{9}{256}$$
  • $$\dfrac{3}{11}$$
The probability of happening of an event in an experiment is 0.The probability of happening of the event atleast once, if the experiment is repeated 3 times under similar conditions is
  • $$0.216$$
  • $$0.784$$
  • $$0.64$$
  • $$0.32$$
If in a binomial distribution the mean is $$20$$, standard deviation is $$\sqrt{15}$$, then $$p=$$
  • $$\dfrac{3}{4}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{3}$$
If the mean and variance of a binomial distribution are $$\dfrac{15}{4}$$ and $$\dfrac{15}{16}$$. The number of trials is
  • $$5$$
  • $$2$$
  • $$4 $$
  • $$ 6$$
A symmetrical die is rolled $$6$$ times. If getting an odd number is a success, the probability of atmost $$5$$ successes is
  • $$\dfrac{60}{64}$$
  • $$\dfrac{63}{64}$$
  • $$\dfrac{120}{164}$$
  • $$\dfrac{10}{14}$$
A symmetrical die is thrown four times and getting
a multiple of 2 is considered to be a success.
The mean and variance of success are
  • $$4, 2$$
  • $$2, 1$$
  • $$0, 2$$
  • $$1, 2$$
If $$A$$ and $$B$$ are two equally strong table tennis players, the probability that $$A$$ beats $$B$$ in exactly three games out of $$4$$ games is
  • $$\displaystyle\frac{1}{6}$$
  • $$\displaystyle\frac{1}{5}$$
  • $$\displaystyle\frac{1}{4}$$
  • $$\displaystyle\frac{1}{2}$$
In a binomial distribution mean is $$5$$, and variance $$4$$, then the number of trials is
  • $$100$$
  • $$50$$
  • $$25$$
  • $$24$$
In six throws of a die, getting $$4$$ or $$5$$ is considered a success. The mean number of successes is
  • $$4$$
  • $$3$$
  • $$2$$
  • $$1$$
If for a binomial distribution $$\bar{x}=\dfrac{6}{5}$$and the difference between mean and variance is $$\dfrac{6}{25}$$. The number of trials is
  • $$8$$
  • $$7$$
  • $$6$$
  • $$5$$
If the standard deviation of the binomial distribution $$(q+p)^{16}$$ is 2, then mean is
  • $$2$$
  • $$4$$
  • $$8$$
  • $$7$$
In a binomial distribution mean is $$4.8$$ and variance is $$2.88$$, then the parameter $$n$$ is
  • $$8$$
  • $$12$$
  • $$16$$
  • $$20$$
Six unbiased coins are tossed once. The probability of obtaining atleast two heads is :
  • $$63/64$$
  • $$57/64$$
  • $$1/64$$
  • $$1/32$$
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
  • $$\left (\displaystyle \frac{1}{2}\right )^3$$
  • $$\left (\displaystyle \frac{19}{20}\right )^3$$
  • $$ ^{15}C_{3} \left (\displaystyle \frac{19}{20}\right )^{12} \left (\displaystyle \frac{1}{20}\right )^3$$
  • $$\left (\displaystyle \frac{1}{20}\right )^3$$
The mean and variance of Binomial Distribution are 6,Then the parameters of the distribution are 
  • $$ \displaystyle 12, \frac{1}{2}$$
  • $$ \displaystyle 9, \frac{2}{3}$$
  • $$ \displaystyle 10, \frac{3}{5}$$
  • $$ \displaystyle 18, \frac{1}{3}$$
$$6$$ symmetrical dice are thrown $$729$$ times. The number of times you expect exactly three dice showing a four or five is
  • $$160$$
  • $$240$$
  • $$12$$
  • $$6$$
A fair coin is tossed 6 times. The variance of the number of heads is
  • $$3$$
  • $$3/2$$
  • $$3/4$$
  • $$2$$
In a binomial distribution mean $$=\dfrac{11}{4}$$ and variance $$=\dfrac{15}{16}$$ , then the probability of success is
  • $$\dfrac{1}{2}$$
  • $$\dfrac{29}{44}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{3}{4}$$
In a binomial distribution with $$n=10, P=\displaystyle \frac{2}{5}$$, the mode of the B.D. is
  • $$6$$
  • $$5$$
  • $$4$$
  • $$3$$
$$A$$ student is given $$6$$ questions in a true or false examination. If he gets $$4$$ or more correct answers he passes the examination. The probability that he passes the examination is
  • $$5/32$$
  • $$7/32$$
  • $$11/32$$
  • $$3/32$$
Team $$A$$ has probability $$\displaystyle \frac{2}{3}$$ of winning whenever  it plays. If $$A$$ plays $$4$$ games the probability that $$A$$ loses all the games is
  • $$\left (\displaystyle \frac{2}{3}\right )^4$$
  • $$\left (\displaystyle \frac{1}{3}\right )^4$$
  • $$\left (\displaystyle \frac{3}{4}\right )^4$$
  • $$\left (\displaystyle \frac{1}{4}\right )^4$$
The mean of B.D. is 6 and its S.D. is Then the probability of 'x' successes is
  • $$ \displaystyle ^{ 18 }C_{ x }{ \left( \frac { 1 }{ 3 } \right) }^{ 18-x }{ \left( \frac { 2 }{ 3 } \right) }^{ x }$$
  • $$ \displaystyle ^{ 12 }C_{ x }{ \left( \frac { 2 }{ 3 } \right) }^{ 12-x }{ \left( \frac { 1 }{ 3 } \right) }^{ x }$$
  • $$ \displaystyle ^{ 18 }C_{ x }{ \left( \frac { 2 }{ 3 } \right) }^{ 18-x }{ \left( \frac { 1 }{ 3 } \right) }^{ x }$$
  • $$ \displaystyle ^{ 18 }C_{ x }{ \left( \frac { 1 }{ 4 } \right) }^{ 18-x }{ \left( \frac { 3 }{ 4 } \right) }^{ x }$$
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