CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 3 - MCQExams.com

$$A$$ and $$B$$ play a game in which $$A'$$s chance of winning is $$1/5$$. In a series of $$6$$ games, the probability that $$A$$ will win at least three games is

  • $$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }$$
  • $$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }$$
  • $$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }+{ \left( \cfrac { 1 }{ 5 } \right) }^{ 6 }$$
  • $$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }$$
If the sum of mean and variance of $$B.D.$$ for $$5$$ trials is $$1.8$$, the binomial distribution is
  • $$(5, 0.8, 0.2)$$
  • $$(5, 0.2, 0.8)$$
  • $$(10, 0.8, 0.2)$$
  • $$(10, 0.2, 0.8)$$
A family has six children. The probability that there are fewer boys than girls, if the probability of any particular child being a boy is $$\displaystyle \frac{1}{2}$$ is
  • $$\displaystyle \frac{5}{32}$$
  • $$\displaystyle \frac{7}{32}$$
  • $$\displaystyle \frac{11}{32}$$
  • $$\displaystyle \frac{9}{32}$$
On an average if it rains on $$10$$ days in every $$30$$, the probability that there will be rain on at least three days of a given week is
  • $$_{ 3 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 3 }$$
  • $$1-_{ 0 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 7 }-_{ 1 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }-_{ 2 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }$$
  • $$_{ 3 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 3 }+_{ 4 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 4 }+_{ 5 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 5 }$$
  • $$_{ 0 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 7 }+_{ 1 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }+_{ 2 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }$$
If for a binomial distribution $$n = 4$$ and $$6P(X=4)=P(X=2)$$, the probability of success is
  • $$3/4$$
  • $$1/2$$
  • $$1/3$$
  • $$1/4$$
A machine manufacturing screws is known to produce $$5$$% defectives. In a random sample of $$15$$ screws the probability that there are not more
than $$3$$ defectives is

  • $$_{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 12 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 3 }$$
  • $$\\ _{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 12 }+_{ 2 }^{ 15 }{ C{ \left( \cfrac { 19 }{ 20 } \right) }^{ 2 } }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 13 }+_{ 1 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 14 }+_{ 0 }^{ 15 }{ C }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 15 }\\ \\ $$
  • $$_{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 12 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 3 }+_{ 2 }^{ 15 }{ C{ \left( \cfrac { 19 }{ 20 } \right) }^{ 13 } }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }+_{ 1 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 14 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }+_{ 0 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 15 }$$
  • $$_{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 12 }$$
The incidence of occupational disease in an industry is such that the workers have a $$20$$ percent chance of suffering from it. The probability that out of $$6$$ workers chosen at random, exactly four will suffer is
  • $$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 2 }{ 5 } \right) }^{ 2 } }{ \left( \cfrac { 3 }{ 5 } \right) }^{ 4 }$$
  • $$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 2 }{ 5 } \right) }^{ 4 } }{ \left( \cfrac { 3 }{ 5 } \right) }^{ 2 }$$
  • $$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 } }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }$$
  • $$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 1 }{ 5 } \right) }^{ 2 } }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 4 }\\ \\ $$
The incidence of occupational disease in an industry is such that the workers have a $$20$$% chance of suffering from it. The probability that out of 6 workers chosen at random, not even one will suffer from that disease is

  • $${ \left( \cfrac { 1 }{ 5 } \right) }^{ 6 }\\ \\ $$
  • $${ \left( \cfrac { 4 }{ 5 } \right) }^{ 6 }\\ \\ $$
  • $$_{ 1 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }\\ \\ $$
  • $$_{ 1 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 5 }\\ \\ $$
A fair coin is tossed four times. The probability that the tails exceed the heads in number is
  • $$_{ 3 }^{ 4 }{ C }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }$$
  • $$_{ 3 }^{ 4 }{ C }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }+{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }$$
  • $${ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }$$
  • $$_{ 3 }^{ 4 }{ C }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 1 }$$
The chance that a person with two dices, the faces of each being numbered $$1$$ to $$6$$, will throw aces exactly $$4$$ times in $$6$$ trials is
  • $${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }$$
  • $${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$$
  • $$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$$
  • $$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 2 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 4 }$$
If on an average $$1$$ vessel in every $$10$$ is wrecked, the probability that out of $$5$$ vessels expected to arrive, $$4$$ at least will arrive safely is
  • $$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 9 }{ 10 } \right) }^{ 4 }\left( \frac { 1 }{ 10 } \right) \ + _{ 0 }^{ 5 }C{ { \left( \frac { 9 }{ 10 } \right) } }^{ 5 }$$
  • $$\displaystyle 1 - _{ 1 }^{ 5 }{ C }{ \left( \frac { 9 }{ 10 } \right) }^{ 4 }\left( \frac { 1 }{ 10 } \right) - _{ 0 }^{ 5 }C{ { \left( \frac { 9 }{ 10 } \right) } }^{ 5 }$$
  • $$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 9 }{ 10 } \right) }^{ 4 }\left( \frac { 1 }{ 10 } \right) $$
  • $$\displaystyle \quad _{ 0 }^{ 5 }C{ { \left( \frac { 9 }{ 10 } \right) } }^{ 5 }$$
Suppose on an average $$5$$ out of $$2000$$ houses get damaged due to fire accident during summer. Out of $$10,000$$ houses in a locality, the probability that exactly $$10$$ houses will get damaged during summer is
  • $$\displaystyle \frac{e^{-5}5^{10}}{ 10!}$$
  • $$\displaystyle \frac{e^{-10}10^{10}}{ 10!}$$
  • $$\displaystyle \frac{e^{-25}25^{10}}{10!}$$
  • $$\displaystyle \frac{e^{-15}15^{10}}{10!}$$
The probability of getting a total of $$9$$ exactly twice in $$6$$ tosses of a pair of dice is
  • $$_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9} \right) }^{ 2}+_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 3 }$$
  • $$_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$$
  • $$1-_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$$
  • $${ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$$
If $$X$$ is a binomial variable with $$E(X)=2$$ and $$V(X)=\cfrac { 4 }{ 3 } $$,
the probability of $$x$$ successes is

  • $${ ^{ 6 } }C_{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6-x }$$
  • $${ ^{ 6 } }C_{ x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 6-x }$$
  • $${ ^{ 9 } }C_{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6-x }$$
  • $${ ^{ 9 } }C_{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 6-x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ x }$$
The probability that a student is not a swimmer is $$\frac { 1 }{ 5 } $$. Out of $$5$$ students the probability that at least four are swimmers is
  • $$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }\left( \frac { 1 }{ 5 } \right) $$
  • $$\displaystyle _{ 0 }^{ 5 }{ C{ \left( \frac { 4 }{ 5 } \right) }^{ 5 }+ }\quad _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }\left( \frac { 1 }{ 5 } \right) $$
  • $$\displaystyle _{ 0 }^{ 5 }{ C{ \left( \frac { 1 }{ 5 } \right) }^{ 5 }+ }\quad _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }$$
  • $$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }$$
Team A has the probability $$ \displaystyle \frac{2}{5}$$ of winning, whenever it plays. If A plays 4 games, the probability that A wins more than half of the games is
  • $$ \displaystyle \frac{22}{125}$$
  • $$ \displaystyle \frac{112}{625}$$
  • $$ \displaystyle \frac{116}{625}$$
  • $$ \displaystyle \frac {110}{625}$$
A production process is supposed to contain $$5$$% defective items. The probability that a sample of $$8$$ items will contain less than $$2$$ defective items is

  • $$_{ }^{ 8 }{ C_2 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }$$
  • $$_{  }^{ 8 }{ C_2}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }$$
  • $${ \left( \cfrac { 19 }{ 20 } \right) }^{ 8 }+\,_{ }^{ 8 }{ C_1 }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 7 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$
  • $${ \left( \cfrac { 19 }{ 20 } \right) }^{ 8 }+\,_{ }^{ 8 }{ C_1 }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 7 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }+\,_{}^{ 8 }{ C_2}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }$$
A fair die is tossed $$1620$$ times. If getting a face with $$6$$ points is considered as a success, then $$E(x) = ........, V(x) = .........$$
  • $$270, 270$$
  • $$270, 225$$
  • $$270, 25$$
  • $$27, 250$$
If on an average $$1$$ vessel in every $$10$$ is wrecked, the probability that out of $$5$$ vessels expected to arrive, at least $$4$$ will arrive safely is
  • $$1-_{ 4 }^{ 5 }{ C }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 1 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 4 }-{ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$$
  • $${ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$$
  • $$_{ 4 }^{ 5 }{ C }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 1 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 4 }+{ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$$
  • $$_{ 4 }^{ 5 }{ C }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 1 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 4 }$$
The probability of a man hitting the target is $$ \displaystyle \frac{1}{4}$$. If he fires 7 times, the probability of hitting the target exactly six times is
  • $$ \displaystyle ^{ 7 }C_{ 5 }{ \left( \frac { 1 }{ 4 } \right) }^{ 6 }{ \left( \frac { 3 }{ 4 } \right) }$$
  • $$ \displaystyle ^{ 7 }C_{ 6 }{ \left( \frac { 3 }{ 4 } \right) }^{ 6 }{ \left( \frac { 1 }{ 4 } \right) }$$
  • $$ \displaystyle ^{ 7 }C_{ 6 }{ \left( \frac { 1 }{ 4 } \right) }^{ 6 }{ \left( \frac { 3 }{ 4 } \right) }$$
  • $$ \displaystyle ^{ 7 }C_{ 6 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }{ \left( \frac { 1 }{ 2 } \right) }$$
An experiment succeeds twice as often as it fails. The chance that in the next six trials, there shall be at least four successes is
  • $$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }$$
  • $$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }$$
  • $$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }+{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }\\ $$
  • $$1-_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }-_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }-{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }\\ $$
If in a random experiment the probability of getting a success is twice that of a failure, then the probability of getting $$6$$ successes in $$10$$ trials is
  • $$^{ 10 }C_{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }$$
  • $$^{ 10 }C_{ 6 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 4 }$$
  • $$^{ 10 }C_{ 6 }{ \left( \cfrac { 1 }{ 4 } \right) }^{ 6 }{ \left( \cfrac { 3 }{ 4 } \right) }^{ 4 }$$
  • $$^{ 10 }C_{ 6 }{ \left( \cfrac { 3 }{ 4 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 4 } \right) }^{ 4 }$$
For a binomial distribution the mean and variance are respectively $$4$$ and $$3$$. The probability of getting a non-zero success is
  • $$1-{ \left( \cfrac { 3 }{ 4 } \right) }^{ 16 }$$
  • $${ \left( \cfrac { 1 }{ 4 } \right) }^{ 16 }$$
  • $${ \left( \cfrac { 3 }{ 4 } \right) }^{ 16 }$$
  • $$1-{ \left( \cfrac { 1 }{ 4 } \right) }^{ 16 }$$
The probability of getting a total of $$9$$ at least twice in $$6$$ tosses of a pair of dice is
  • $$_{ }^{ 6 }{ C_2 }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$$
  • $${ \left( \cfrac { 8 }{ 9 } \right) }^{ 6 }+\ _{ }^{ 6 }{ C_1 }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 1 }$$
  • $$1-{ \left( \cfrac { 8 }{ 9 } \right) }^{ 6 }-\ _{  }^{ 6 }{ C_1 }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 1 }$$
  • $$1-{ \left( \cfrac { 8 }{ 9 } \right) }^{ 6 }$$
The probability that a bulb produced by a factory will fuse after $$100$$ days of use is $$0.05$$. The probability that out of $$5$$ such bulbs none of them fuse after $$100$$ days is
  • $$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$
  • $${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$
  • $$1-{ \left( \cfrac { 1 }{ 20 } \right) }^{ 5 }$$
  • $${ \left( \cfrac { 1 }{ 20 } \right) }^{ 5 }$$
If the probability of a defective bolt is $$0.1$$. The mean and standard deviation for the distribution of defective bolts in a total of $$400$$.... & ....
  • $$40, 36$$
  • $$40, 6$$
  • $$20, 6$$
  • $$10, 6$$
A random variable $$X$$ is binomially distributed with mean $$12$$ and variance $$8$$. The parameters of the distribution are ...... & .......
  • $$ \displaystyle 36, \frac{2}{3}$$
  • $$ \displaystyle 36, \frac{1}{3}$$
  • $$ \displaystyle 24, \frac{1}{3}$$
  • $$ \displaystyle 24, \frac{2}{3}$$
The probability that a bulb produced by a factory will fuse after $$100$$ days of use is $$0.05$$. The probability that out of $$5$$ such bulbs not more than one will fuse after $$100$$ days is
  • $$^{ 5 }{ C }_1{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$
  • $${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }+^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$
  • $$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$
  • $$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }-^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$
The probability that a bulb produced by a factory will fuse after $$100$$ days of use is $$0.05$$. The probability that out of $$5$$ such bulbs at least one will fuse after $$100$$ days of use is
  • $${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$
  • $$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }-_{ 1 }^{ 5 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$
  • $$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$
  • $${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }+_{ 1 }^{ 5 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$
In a binomial distribution $$n=9$$ and $$p=\cfrac { 1 }{ 3 } $$.The mode of the binomial distribution is 
  • $$3$$
  • $$4$$
  • $$3$$ and $$4$$
  • $$3.5$$
Let X be a binomially distributed variate with mean 10 and varianceThen p(x>10) is
  • $$\dfrac{1}{2^{20}}\sum_{k=11}^{20}$$$$^{20}$$$$C_{k}$$
  • $$\dfrac{1}{2^{20}}\sum_{k=1}^{11}$$$$^{20}C_{k}$$
  • $$\dfrac{1}{2^{20}}\sum_{k=11}^{20}$$$$^{10}C_{k}$$
  • $$\sum_{k=11}^{20}$$ $$^{20}C_{k}\left ( \dfrac{2}{3} \right )^{30-k}$$
In an experiment success is twice that of failure. If the experiment is repeated $$6$$ times, the probability that atleast $$4$$ times success is :
  • $$\dfrac{64}{729}$$
  • $$\dfrac{192}{729}$$
  • $$\dfrac{240}{729}$$
  • $$\dfrac{496}{729}$$
For a $$B.D.$$ $$ \bar{x}=4,\sigma =\sqrt{3}$$, then $$P(X=r)$$=
  • $$^{16}C_{r}\left ( \dfrac{1}{4} \right )^{r}\left ( \dfrac{3}{4} \right )^{16-r}$$
  • $$^{12}C_{r}\left ( \dfrac{1}{4} \right )^{r}\left ( \dfrac{3}{4} \right )^{12-r}$$
  • $$^{12}C_{r}\left ( \dfrac{2}{3} \right )^{r}\left ( \dfrac{1}{3} \right )^{12-r}$$
  • $$^{12}C_{r}\left ( \dfrac{3}{4} \right )^{r}\left ( \dfrac{1}{4} \right )^{12-r}$$
A symmetrical die is thrown $$ $$6 times. If getting an odd number is a success, the probability of at the most $$5$$ successes is
  • $$\displaystyle \frac{5}{64}$$
  • $$\displaystyle \frac{15}{64}$$
  • $$\displaystyle \frac{63}{64}$$
  • $$\displaystyle \frac{36}{64}$$
The binomial distribution for which,
mean$$+$$$$2\times$$variance $$=$$4, mean$$+$$variance $$= 3$$ is
  • $$\left( 6,\ \dfrac { 1 }{ 3 } ,\ \dfrac { 2 }{ 3 } \right) $$
  • $$\left( 6,\ \dfrac { 1 }{ 2 } ,\  \dfrac { 1 }{ 2 } \right) $$
  • $$\left( 4,\ \dfrac { 1 }{ 2 } ,\ \dfrac { 1 }{ 2 } \right) $$
  • $$\left( 4,\ \dfrac { 1 }{ 3 } ,\ \dfrac { 2 }{ 3 } \right) $$
If the mean and variance of binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to
  • $$\displaystyle \frac{5}{16}$$
  • $$\displaystyle \frac{7}{16}$$
  • $$\displaystyle \frac{11}{16}$$
  • $$\displaystyle \frac{9}{16}$$
$$X$$ follows a binomial distribution with parameters $$n = 6$$ and $$P$$. If $$4P(x=4)=P(x=2)$$, then $$P=$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{6}$$
  • $$\dfrac{1}{3}$$
In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. The parameter p of the distribution is
  • $$1/3$$
  • $$1/4$$
  • $$1/5$$
  • $$1/6$$
A variable takes the values 0, 1, 2, 3, ....n with frequencies proportional to the binomial coefficients  $$^{n}c_{0},^{n}c_{1},^{n}c_{2},^{n}c_{3},.......^{n}c_{n}$$ then variance $$\sigma ^{2}=$$
  • $$n$$
  • $$\dfrac{n}{2}$$
  • $$\dfrac{n}{4}$$
  • $$\dfrac{n}{6}$$
For a binomial distribution if $$ P=\dfrac{1}{4}, n=20$$, the probability of mode is
  • $$^{20}C_{5}\left ( \dfrac{3}{4} \right )^{5}$$
  • $$^{20}C_{5}\left ( \dfrac{1}{4} \right )^{5}\left ( \dfrac{3}{4} \right )^{15}$$
  • $$^{10}C_{10}\left ( \dfrac{1}{4} \right )^{10}\left ( \dfrac{3}{4} \right )^{10}$$
  • $$^{10}C_{10}\left ( \dfrac{3}{4} \right )^{10}$$
Out of $$800$$ families with $$4$$ children each, the expected number of families having atleast one boy is
  • $$550$$
  • $$50$$
  • $$750$$
  • $$300$$
The probability that a candidate secure a seat in engineering through EAMCET is $$\dfrac{1}{10}$$. Seven candidates are selected at random from a center. The probability that exactly two will get seats is
  • $$15(0.1)^{2}(0.9)^{5}$$
  • $$20(0.1)^{2}(0.9)^{5}$$
  • $$21(0.1)^{2}(0.9)^{5}$$
  • $$23(0.1)^{2}(0.9)^{5}$$
If $$X$$ follows a binomial distribution with parameters $$n = 8$$ and $$P=\frac{1}{2},$$ than $$P(\left | x-4 \right |\leq 2)=$$
  • $$\dfrac{119}{128}$$
  • $$\dfrac{9}{128}$$
  • $$\dfrac{101}{128}$$
  • $$\dfrac{11}{128}$$
4 unbiased coins are tossed $$256$$ times. The expected frequency of $$x$$ heads
  • $$16^{4}C_{x}$$
  • $$12^{4}C_{x}$$
  • $$12^{4}C_{0}$$
  • $$8^{4}C_{0}$$
Suppose $$X$$ follows binomial distribution with parameters $$n$$ and $$p$$, where $$0<p<1$$. If $$\dfrac{P(x=r)}{P(x=n-r)}$$ is independent of $$n$$ and $$r$$, then p$$=$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{4}$$
  • $$1$$
Out of $$800$$ families with $$4$$ children each, the expected number of families having $$2$$ boys and $$2$$ girls
  • $$100$$
  • $$200$$
  • $$300$$
  • $$400$$
The probability of expected number of boys in a family with 8 children assuming that the sex distributions are equally probable is
  • $$^{8}C_{2}\left ( \dfrac{1}{2} \right )^{8}$$
  • $$^{8}C_{3}\left ( \dfrac{1}{2} \right )^{8}$$
  • $$^{8}C_{4}\left ( \dfrac{1}{2} \right )^{8}$$
  • $$^{8}C_{5}\left ( \dfrac{1}{2} \right )^{8}$$
If X be B.V. with  $$E(X)=5$$  and  $$E(X^{2})-{E(X)}^{2}=4$$  then the parameters of distribution are
  • $$\dfrac{1}{4}, 20$$
  • $$\dfrac{1}{5}, 20$$
  • $$\dfrac{1}{5}, 25$$
  • $$\dfrac{4}{5}, 25$$
If a sex ratio of births is 49 girls to 51 boys, the probability that there will be 8 girls amongst 10 babies born on the same day in a maternity hospital

  • $$^{10}C_{8}(0.51)^{8}(0.49)^{2}$$
  • $$^{10}C_{8}(0.49)^{8}(0.51)^{2}$$
  • $$^{10}C_{8}(0.49X0.51)^{8}$$
  • $$^{10}C_{8}(0.49X0.51)^{2}$$
If x is $$B\left ( x,n,\dfrac{1}{3} \right ),P(x\geq 1)> 0.8$$n  the least value of n is
  • $$3$$
  • $$4$$
  • $$5$$
  • $$6$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers