CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 6 - MCQExams.com

A distributor of bean seeds determines from extensive tests that $$5$$% of seeds will not germinate. He sells the seeds in packets of $$200$$ and guarantees $$90$$% germination. The probability that a particular packet will violate the guarantee is
  • $$\displaystyle \dfrac{e^{-10}(10)^{10}}{ 10!}$$
  • $$\displaystyle \frac{e^{-10}(10)^{20}}{10!}$$
  • $$1-\displaystyle \sum_{x=0}^{20}\frac{e^{-10}(10)^{x}}{ x!}$$
  • $$1-\displaystyle \sum_{x=0}^{\infty }\frac{e^{-10}(10)^{x}}{ x!}$$
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 $$6$$ ships in the next half an hour is
  • $$\displaystyle \frac{e^{-6}6^{6}}{ 6!}$$
  • $$\displaystyle \frac{e^{-3}3^{3}}{ 3!}$$
  • $$\displaystyle \frac{e^{-3}3^{6}}{ 6!}$$
  • $$\displaystyle \frac{e^{-6}3^{3}}{ 6!}$$
A coin is tossed $$n$$ times. The probability that head will turn up an odd number of times, is
  • $$\displaystyle \frac { 1 }{ 2 } $$
  • $$\displaystyle \frac { n+1 }{ 2n } $$
  • $$\displaystyle \frac { n-1 }{ 2n }$$
  • $$\displaystyle \frac { { 2 }^{ n-1 }-1 }{ { 2 }^{ n } } $$
A company knows on the basis of past experience that $$2$$% of its blades are defective. The probability of having $$3$$ defective blades in a sample of $$100$$ blades if $$e^{-2}=0.1353$$ is
  • $$0.1353$$
  • $$0.1804$$
  • $$0.2706$$
  • $$0.3606$$
Patients arrive randomly and independently at a Doctor's room from 8 AM at an average rate of one in 5 minutes. The waiting room can accommodate 12 persons. The probability that the room will be full when the doctor arrives at 9AM is
  • $$\displaystyle \frac{{e}^{-12}(12)^{12}}{ 12!}$$
  • $$\displaystyle \sum_{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$$
  • $$1-\displaystyle \sum_{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$$
  • $$1-\displaystyle \sum_{{x}=0}^{\infty }\frac{{e}^{-12}(12)^{{x}}}{ x!}$$
A box contains $$100$$ bulbs out of which $$10$$ are defective. A sample of $$5$$ bulbs is drawn. The probability that none is defective, is
  • $$\displaystyle { \left( \frac { 1 }{ 10 } \right) }^{ 5 }$$
  • $$\displaystyle { \left( \frac { 1 }{ 2 } \right) }^{ 5 }$$
  • $$\displaystyle { \left( \frac { 9 }{ 10 } \right) }^{ 5 }$$
  • $$\displaystyle \frac { 9 }{ 10 } $$
What is the probability of guessing correctly at least 8 out of 10 answers on a true -false examination?
  • $$\displaystyle\frac{11}{128}$$
  • $$\displaystyle\frac{5}{128}$$
  • $$\displaystyle\frac{13}{128}$$
  • $$\displaystyle\frac{7}{128}$$
The constant term of the expansion of $$\left (3x-\dfrac{5}{x^{2}}\right)^{9}$$ is.
  • $$^{9}\textrm{C}_{2}3^{6}5^{3}$$
  • $$^{9}\textrm{C}_{4}3^{4}5^{5}$$
  • $$^{9}\textrm{C}_{3}3^{6}5^{3}$$
  • $$^{9}\textrm{C}_{3}$$
A box contains $$24$$ identical balls of which $$12$$ are white and $$12$$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is
  • $$\displaystyle \frac { 5 }{ 64 } $$
  • $$\displaystyle \frac { 27 }{ 32 } $$
  • $$\displaystyle \frac { 5 }{ 32 } $$
  • $$\displaystyle \frac { 1 }{ 2 } $$
If $$X$$ follows a binomial distribution with parameters $$n=8$$ and $$\displaystyle p=\frac { 1 }{ 2 } $$, then $$P\left( \left| X-4 \right| \le 2 \right) =$$
  • $$\displaystyle \frac { 119 }{ 128 } $$
  • $$\displaystyle \frac { 116 }{ 128 } $$
  • $$\displaystyle \frac { 29 }{ 128 } $$
  • None of these
Numbers are selected at random, one at a time, from the two-digit numbers $$00,01,02,...99$$ with replacement. An even $$E$$ occurs if and only if the product of the two digits of selected number is $$18$$. If four numbers are selected, the probability that the event $$E$$ occurs at least $$3$$ times, is 
  • $$\displaystyle \frac { 99 }{ { \left( 25 \right) }^{ 4 } } $$
  • $$\displaystyle \frac { 86 }{ { \left( 25 \right) }^{ 4 } } $$
  • $$\displaystyle \frac { 74 }{ { \left( 25 \right) }^{ 4 } } $$
  • $$\displaystyle \frac { 97 }{ { \left( 25 \right) }^{ 4 } } $$
Suppose $$X$$ is a binomial variate $$B(5,p)$$ and $$P(X=2)=P(X=3)$$, then $$p$$ is equal to
  • $$1/2$$
  • $$1/3$$
  • $$1/4$$
  • $$1/5$$
A bag contains three white, two black and four red balls. If four balls are drawn at random with replacement, the probability that the sample contains just one white ball is
  • $$\displaystyle \frac { 32 }{ 81 } $$
  • $$\displaystyle \frac { 11 }{ 2 } $$
  • $$\displaystyle \frac { 32 }{ 93 } $$
  • None of these
India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points $$0,1,2$$ are $$0.45,0.05,0.50$$ respectively. Assuming that the outcomes are independent, the probability of India getting at least $$7$$ points is
  • $$0.8750$$
  • $$0.0875$$
  • $$0.0625$$
  • $$0.0250$$
One hundred identical coins, each with probability $$p$$ of showing up heads, are tossed. If $$0<p<1$$ and the probability of heads showing on $$50$$ coins is equal to that of heads showing on $$51$$ coins, the value of $$p$$ is
  • $$\displaystyle \frac { 1 }{ 2 } $$
  • $$\displaystyle \frac { 49 }{ 101 } $$
  • $$\displaystyle \frac { 50 }{ 101 } $$
  • $$\displaystyle \frac { 51 }{ 101 } $$
Six person try to swim across a wide river. It is known that on an average, only three persons out of the ten are successful in crossing the river. what is the probability that at most four of the six persons will cross river safely?
  • $$0.990523$$
  • $$0.890523$$
  • $$0.980523$$
  • None of these
A die is thrown $$7$$ times. What is the chance that an odd number turns up ?
(i) exactly $$4$$ times
(ii) at least $$4$$ times.
  • (i)$$\displaystyle \frac{35}{128}$$, (ii)$$\displaystyle \frac{1}{2}$$
  • (i)$$\displaystyle \frac{45}{128}$$, (ii)$$\displaystyle \frac{1}{2}$$
  • (i)$$\displaystyle \frac{35}{128}$$, (ii)$$\displaystyle \frac{1}{4}$$
  • None of these
A die is thrown $$7$$ times. What is the chance that an odd number turns up at least $$4$$ times?
  • $$\cfrac{1}{3}$$
  • $$\cfrac{1}{4}$$
  • $$\cfrac{1}{2}$$
  • $$\cfrac{35}{128}$$
Suppose X is a binomial variate $$B ( 5, P )$$ and $$P ( X=2 )$$ and $$P ( X=3 )$$, then $$P$$ is equal to
  • $$\displaystyle\frac{1}{2}$$
  • $$\displaystyle\frac{1}{3}$$
  • $$\displaystyle\frac{1}{4}$$
  • $$\displaystyle\frac{1}{5}$$
If $$X$$ and $$Y$$ are independent binomial variate of binomial distribution $$\displaystyle A\left( 5,\frac { 1 }{ 2 }  \right) $$ and $$\displaystyle B\left( 7,\frac { 1 }{ 2 }  \right) $$ respectively, then $$P(X+Y=3)$$ is
  • $$\displaystyle \frac { 55 }{ 1024 } $$
  • $$\displaystyle \frac { 55 }{ 4098 } $$
  • $$\displaystyle \frac { 55 }{ 2048 } $$
  • none of these
The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, is
  • 7
  • 6
  • 5
  • 3
If X follows a binomial distribution with parameters n = 100 and $$P = \displaystyle\frac{1}{3}$$, then $$P (X = r)$$ is maximum when $$r = $$
  • $$33$$
  • $$50$$
  • $$25$$
  • none of these
If the mean and the variance of a binomial variate $$X$$ are $$2$$ and $$1$$ respectively, then the probability that $$X$$ takes a value greater than $$1$$ is equal to.....
  • $$11/16$$
  • $$5/16$$
  • $$9/16$$
  • $$7/16$$
One hundred identical coins , each with probability, $$p$$, of showing up heads are tossed once. If $$0<p<1$$ and the probability of heads showing on $$50$$ coins is equal to that of heads showing on $$51$$ coins, then the value of $$p$$ is:
  • $$\displaystyle \frac{1}{2}$$
  • $$\displaystyle \frac{49}{101}$$
  • $$\displaystyle \frac{50}{101}$$
  • $$\displaystyle \frac{51}{101}$$
A biased coin with probability p, 0 < p < 1, of head is tossed until a head appears for the first time. If the probability that the number of tossess required is even is 2/5, then p = .......
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{6}$$
  • $$\dfrac{1}{3}$$
$$6$$ ordinary dice are rolled. The probability that at least half of them will show at least $$3$$ is
  • $$\displaystyle 41 \times \frac {2^4}{3^6}$$
  • $$\displaystyle \frac {2^4}{3^6}$$
  • $$\displaystyle 20 \times \frac {2^4}{3^6}$$
  • none of these
A coin is tossed $$n$$ times, the probability that head will turn up an even number of times is
  • $$ \displaystyle \frac{n+1}{2n} $$
  • $$ \displaystyle \frac{n}{n+1} $$
  • $$ \displaystyle \frac{1}{2} $$
  • $$\displaystyle 2^{n-1}$$
Let $$X$$ denote the number of time tail appear in $$n$$ tosses of a fair coin. If $$ P\left ( X\:=\:1 \right ), \: P\left ( X\:= \:2 \right ) $$ and $$ P\left ( X\:=\:3 \right ) $$ are in A.P., then value of $$n$$ is
  • $$9$$
  • $$2$$
  • $$7$$
  • None of these
The sum mean and variance of a B.D. for $$5$$ trials is $$1.8$$, then the Binomial distribution (B.D.) is given by
  • $$ \displaystyle\left (0.2+0.8 \right )^{5} $$
  • $$ \displaystyle\left (0.3+0.7 \right )^{5} $$
  • $$ \displaystyle\left (0.4+0.6 \right )^{5} $$
  • None of these
For a Poission distribution which pair has same value.
  • (Mean, Std. Deviation)
  • (Variance, Standard Deviation)
  • (Mean, Variance)
  • None of these
Numbers are selected at random one at a time, from the number $$00,01,02,...,99$$ with replacement. An event $$E$$ occurs if and only if the product of the two digits of a selected number is $$18.$$ If four numbers are selected, then the probability that $$E$$ occurs at least $$3$$ times, is
  • $$ \displaystyle \frac{97}{390625} $$
  • $$ \displaystyle \frac{68}{390625} $$
  • $$ \displaystyle \frac{72}{390625} $$
  • None of these
If X follows a B.D. with parameter $$ n= 6 $$. If $$ \:4P\left ( X= 4 \right )= P\left ( X= 2 \right ) $$ then P equals
  • $$ \displaystyle \frac{1}{6} $$
  • $$ \displaystyle \frac{1}{4} $$
  • $$ \displaystyle \frac{1}{8} $$
  • $$ \displaystyle \frac{1}{3} $$
If $$X$$ and $$Y$$ are the independent random variables $$\displaystyle B\left( 5,\frac { 1 }{ 2 }  \right) $$ and $$\displaystyle B\left( 7,\frac { 1 }{ 2 }  \right) $$, then $$P\left( X+Y\ge 1 \right) =$$
  • $$ \displaystyle \frac { 4095 }{ 4096 } $$
  • $$ \displaystyle \frac { 309 }{ 4096 } $$
  • $$ \displaystyle \frac { 4032 }{ 4096 } $$
  • None of these
For a Binomial distribution whose mean is 9 and S.D. is $$\displaystyle \frac{3}{2}$$, then the value of n equals
  • $$12$$
  • $$36$$
  • $$9$$
  • none of these
A die is thrown $$ 2n $$ times. The probability that the number greater than 4 appears at least once in $$ 2n $$ throws is
  • $$ \displaystyle \left ( \frac{1}{3} \right )^{2n} $$
  • $$ 1-\displaystyle \left ( \frac{1}{3} \right )^{2n} $$
  • $$ \displaystyle \frac{3^{2n}-\:2^{2n}}{3^{2n}} $$
  • None of these
For a poission distribution variable $$X$$ is such that $$P(X = 2) = 9 P(X= 4) + 90 P(X= 6)$$ the mean is
  • $$2$$
  • $$3$$
  • $$1$$
  • None of these
For a B,D, variance is given by the formula
  • $$ \displaystyle \sigma = npq $$
  • $$ \displaystyle \sigma = \frac{\sqrt{npq}}{2} $$
  • $$ \displaystyle \sigma ^{2} = npq $$
  • $$ \displaystyle \sigma = n^{2}p^{2}q^{2} $$
For a Poission distribution, which of the following is true
  • $$Mean = Mode$$
  • $$Median = S.D.$$
  • $$Mean = Variance$$
  • $$Median = Variance$$
A pair of dice is thrown 4 times, then the probability of getting doublets at least twice is
  • $$\displaystyle \frac{19}{44}$$
  • $$\displaystyle \frac{21}{44}$$
  • $$\displaystyle \frac{31}{44}$$
  • $$\displaystyle \frac{39}{44}$$
A die is thrown 5 times. Getting an odd number considered as success, then variance of distribution is
  • $$\displaystyle \frac{8}{3}$$
  • $$\displaystyle \frac{3}{8}$$
  • $$\displaystyle \frac{4}{5}$$
  • $$\displaystyle \frac{5}{4}$$
The mean and variance of Binomial Distribution are $$4$$ and $$2$$ respectively, then the probability of success is
  • $$ \displaystyle \frac{128}{256} $$
  • $$ \displaystyle \frac{219}{256} $$
  • $$ \displaystyle \frac{37}{256} $$
  • $$ \displaystyle \frac{28}{256} $$
The mean and the variance of a random variable X having Binomial Distribution are  4 and 2 respectively then 
  • $$ \displaystyle \frac{1}{16} $$
  • $$ \displaystyle \frac{1}{8} $$
  • $$ \displaystyle \frac{1}{4} $$
  • $$ \displaystyle \frac{1}{32} $$
In $$n$$ independent trials (finite) of a random experiment, let $$X$$ be the number of times an event $$A$$ occurs. If the probability of success of one trial say $$p$$ and we get the probability of failure of event $$A$$ i.e. $$P\left ( \bar{A} \right )=1-p=q$$. The probability of $$r$$ success in $$n$$ trials is denoted by$$ P(X = r)$$ such that $$P(X = r) = ^{n}C_{r}p^{r}q^{n-r}$$, known as binomial distribution of random variable $$X$$. We also have the following result-
(i) Probability of getting at least $$k$$ successes is
$$P\left ( X\geq K \right )=\sum_{r=k}^{n}\ ^{n}C_{r}p^{r}q^{n-r}$$
(ii) Probability of getting at the most $$k$$ successes is
$$P\left ( X\leq  K \right )=\sum_{r=0}^{n}\ ^{n}C_{r}p^{r}q^{n-r}$$
(iii) $$\sum_{r=0}^{n}=^nC_{r}p^{r}q^{n-r}=\left ( p+q \right )^{n}=1$$

On the basis of the above information answer the following question.

A product is supposed to contain 5% defective items. What is the probability that a sample of 8 items will contain less than 2 defeclive items?
  • $$\displaystyle \frac{25\times \left ( 19 \right )^{7}}{\left ( 20 \right )^{8}}$$
  • $$\displaystyle \frac{27\times \left ( 19 \right )^{7}}{\left ( 20 \right )^{8}}$$
  • $$\displaystyle \frac{ \left ( 19 \right )^{7}}{\left ( 20 \right )^{8}}$$
  • None of these
The mean variance of a binomial variable $$X$$ are $$2$$ and $$1$$ respectively. The probability that $$X$$ takes values greater than $$1$$, is
  • $$ \displaystyle \dfrac { 5 }{ 16 } $$
  • $$ \displaystyle \dfrac { 9 }{ 16 } $$
  • $$ \displaystyle \dfrac { 11 }{ 16 } $$
  • None of these
A random variable $$X$$ has Poisson distribution with mean $$2$$ The $$\displaystyle P(X>1.5)$$ equals
  • $$0$$
  • $$\displaystyle 2/e^{2}$$
  • $$\displaystyle 3/e^{2}$$
  • $$\displaystyle 1-\frac{3}{e^{2}}$$
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
  • $$\displaystyle \frac{6}{5^{e}}$$
  • $$\displaystyle \frac{5}{6}$$
  • $$\displaystyle \frac{6}{55}$$
  • $$\displaystyle \frac{6}{e^{5}}$$
The probability that a student is not a swimmer is $$1/5$$. The probability that out of $$5$$ students exactly $$4$$ are swimmer is
  • $$\left (\dfrac {4}{5}\right )^3$$
  • $$\left (\dfrac {4}{5}\right )^4$$
  • $$5\left (\dfrac {4}{5}\right )^4$$
  • $$3\left (\dfrac {4}{5}\right )^3$$
If the mean and variance of a binomial variate $$X$$ and $$2$$ and $$1 $$ respectively, then the probability that $$X$$ takes a value at most $$3$$ is
  • $$2/3$$
  • $$4/5$$
  • $$7/8$$
  • $$15/16$$
An unbaised die with faces marked $$1, 2, 3, 4, 5$$ and $$6$$ is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than $$2$$ and the maximum face value is not greater than $$5$$ is then
  • $$16/81$$
  • $$1/81$$
  • $$80/81$$
  • $$65/81$$
Suppose $$X$$ follows a binomial distribution with parameters $$n$$ and $$p$$, where $$0 < p < 1$$. If $$P(X-r)/P(X=n-r)$$ is independent of $$n$$ for every value of $$r$$, then
  • $$p=1/2$$
  • $$p=1/3$$
  • $$p=1/4$$
  • $$p=1/5$$
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