MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 3

$A$ 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

0%
$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }$
0%
$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }$
0%
$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }+{ \left( \cfrac { 1 }{ 5 } \right) }^{ 6 }$
0%
$_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 3 }+_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }$

Explanation

Probability of

$A$ winning the game is

$\displaystyle P(A)=(\frac{1}{5})$ ,
Probability of

$B$ winning the game is

$\displaystyle P(B)=(\frac{4}{5})$
Any three games out of total

$6$ games can be chosen in

$^{ 6 }C_{ 3 }$ ways.
Probability of

$A$ winning any three particular games and losing the remaining is

$\displaystyle { \left( \frac { 1 }{ 5 } \right) }^{ 3 }{ \left( \frac { 4 }{ 5 } \right) }^{ 3 }$
Hence probability that

$A$ will win exactly any

$3$ of the

$6$ games, using multiplication theorem is

$P(A=3) =\displaystyle {}^{ 6 }C_{ 3 }{ \left( \frac { 1 }{ 5 } \right) }^{ 3 }{ \left( \frac { 4 }{ 5 } \right) }^{ 3 }$

Similarly probability that

$A$ will win exactly

$4, 5, 6$ games are

$\displaystyle ^{ 6 }C_{ 4 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 },\; ^{ 6 }C_{ 5 }{ \left( \frac { 1 }{ 5 } \right) }^{ 5 }{ \left( \frac { 4 }{ 5 } \right) }^{ 1 },\; ^{ 6 }C_{ 6 }{ \left( \frac { 1 }{ 5 } \right) }^{ 6 }$ respectively.
Hence the probability of

$A$ winning at least

$3$ games

$\displaystyle = P(A=3)+P(A=4)+P(A=5)+P(A=6) \\ \displaystyle ={}^{6 }C_{ 3 }{ \left( \frac { 1 }{ 5 } \right) }^{ 3 }{ \left( \frac { 4 }{ 5 } \right) }^{ 3 }+{}^{ 6 }C_{ 4 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 }+{}^{ 6 }C_{ 5 }{ \left( \frac { 1 }{ 5 } \right) }^{ 5 }{ \left( \frac { 4 }{ 5 } \right) }^{ 1 }+{}^{ 6 }C_{ 6 }{ \left( \frac { 1 }{ 5 } \right) }^{ 6 }$

If the sum of mean and variance of

$B.D.$ for

$5$ trials is

$1.8$ , the binomial distribution is

0%
$(5, 0.8, 0.2)$
0%
$(5, 0.2, 0.8)$
0%
$(10, 0.8, 0.2)$
0%
$(10, 0.2, 0.8)$

Explanation

It is given that

$n=5$
Therefore
Sum of Mean and variance implies

$np+np(1-p)$

$=np(2-p)$

$=5p(2-p)$

$=1.8$

$p(2-p)=\dfrac{9}{25}$

$p(2-p)=0.36$

$p(2-p)=0.2(2-0.2)$
Hence

$p=0.2$

$1-p=0.8$
And

$n=5$

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

0%
$\displaystyle \frac{5}{32}$
0%
$\displaystyle \frac{7}{32}$
0%
$\displaystyle \frac{11}{32}$
0%
$\displaystyle \frac{9}{32}$

Explanation

$P(B)=P(G)=\dfrac{1}{2}$
Now the probability of having fewer boys in a family of 6 children is

$\:^{6}C_{4}(P(G))^{4}(P(B))^{2}+\:^{6}C_{5}(P(G))^{5}(P(B))^{1}+\:^{6}C_{6}(P(G))^{6}(P(B))^{0}$

$=\dfrac{1}{2^{6}}[15+6+1]$

$=\dfrac{22}{64}$

$=\dfrac{11}{32}$

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

0%
$_{ 3 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 3 }$
0%
$1-_{ 0 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 7 }-_{ 1 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }-_{ 2 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }$
0%
$_{ 3 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 3 }+_{ 4 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 4 }+_{ 5 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 5 }$
0%
$_{ 0 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 7 }+_{ 1 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }+_{ 2 }^{ 7 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }$

Explanation

Probability of raining is

$\dfrac{10}{30}$

$=\dfrac{1}{3}$
Probability of not raining is

$\dfrac{2}{3}$
Now the probability of rainfall on 3 or more days out of 7 days is

$\sum ^{7} _{k=3}\:^{7}C_{k}(\dfrac{1}{3})^{k}(\dfrac{2}{3})^{7-k}$

If for a binomial distribution

$n = 4$ and

$6P(X=4)=P(X=2)$ , the probability of success is

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

Explanation

If for a binomial distribution n=4

$6 P(X=4)=P(X=2) P(X = k) = ^{n}C_{k} {p}^{k} ({1-p})^{n-k}$

$6^{4}C_{4} {p}^{4} ({1-p})^{4-4} = ^{4}C_{2} {p}^{2} ({1-p})^{4-2}$

$6 {p}^{2} = 6 {(1-p)}^{2}$

$1 - 2p = 0$

$p =$

$\dfrac {1}{2}$

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

0%
$_{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 12 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 3 }$
0%
$\\ _{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 12 }+_{ 2 }^{ 15 }{ C{ \left( \cfrac { 19 }{ 20 } \right) }^{ 2 } }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 13 }+_{ 1 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 14 }+_{ 0 }^{ 15 }{ C }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 15 }\\ \\$
0%
$_{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 12 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 3 }+_{ 2 }^{ 15 }{ C{ \left( \cfrac { 19 }{ 20 } \right) }^{ 13 } }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }+_{ 1 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 14 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }+_{ 0 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 15 }$
0%
$_{ 3 }^{ 15 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 12 }$

Explanation

Here

$p=\dfrac{5}{100}$

$=\dfrac{1}{20}$

$1-p=\dfrac{19}{20}$
Now for not more than 3 defective screws, we get the required probability as

$\sum\:^{3}_{k=0}\:^{15}C_{k}(\dfrac{1}{20})^{k}(\dfrac{19}{20})^{15-k}$

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

0%
$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 2 }{ 5 } \right) }^{ 2 } }{ \left( \cfrac { 3 }{ 5 } \right) }^{ 4 }$
0%
$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 2 }{ 5 } \right) }^{ 4 } }{ \left( \cfrac { 3 }{ 5 } \right) }^{ 2 }$
0%
$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 4 }{ 5 } \right) }^{ 2 } }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 4 }$
0%
$_{ 4 }^{ 6 }{ C{ \left( \cfrac { 1 }{ 5 } \right) }^{ 2 } }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 4 }\\ \\$

Explanation

Probability of suffering from occupational disease

$=\dfrac{20}{100}=\dfrac{1}{5}$

Probability of not suffering from occupational disease

$=1-{\dfrac{1}{5}}=\dfrac{4}{5}$

Selecting exactly four workers out of six those are suffering:

$_{ 4 }^6C$

Probability that out of 6 workers, exactly four will suffer:

$_{ 4 }^{ 6 }{ C{ \left( \dfrac { 4 }{ 5 } \right) }^{ 2 } }{ \left( \dfrac { 1 }{ 5 } \right) }^{ 4 }$

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

0%
${ \left( \cfrac { 1 }{ 5 } \right) }^{ 6 }\\ \\$
0%
${ \left( \cfrac { 4 }{ 5 } \right) }^{ 6 }\\ \\$
0%
$_{ 1 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 5 }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 1 }\\ \\$
0%
$_{ 1 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 5 } \right) }^{ 1 }{ \left( \cfrac { 4 }{ 5 } \right) }^{ 5 }\\ \\$

Explanation

Here

$20$ % of the workers have a chance of suffering from the disease.
Therefore

$p=\dfrac{20}{100}$

$=\dfrac{1}{5}$

$(1-p)=\dfrac{4}{5}$ . ...(probability of not suffering from the disease).
Now the required probability, where out of 6, none suffers from the disease is

$\:^{6}C_{0}(1-p)^{6}(p)^{0}$

$=(\dfrac{4}{5})^{6}$

A fair coin is tossed four times. The probability that the tails exceed the heads in number is

0%
$_{ 3 }^{ 4 }{ C }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }$
0%
$_{ 3 }^{ 4 }{ C }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }+{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }$
0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }$
0%
$_{ 3 }^{ 4 }{ C }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 1 }$

Explanation

Probability of getting head

$=$ probability of getting a tail

$=$

$\dfrac{1}{2}$
Hence the probability that the no. of tails exceeds the no. of heads where the coin has been tossed four times is

$\:^{4}C_{3}(\dfrac{1}{2})^{1}(\dfrac{1}{2})^{3}+\:^{4}C_{4}(\dfrac{1}{2})^{0}(\dfrac{1}{2})^{4}$

$=\dfrac{1}{2^{4}}[4+1]$

$=\dfrac{5}{2^{4}}$

$=\dfrac{5}{16}$

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

0%
${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }$
0%
${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$
0%
$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$
0%
$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 2 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 4 }$

Explanation

Total cases for two dice:

$6\times6=36$
Cases when the faces of each being numbered

$1$ to

$6$

$=1$
Probability when the faces of each being numbered

$1$ to

$6$

$P=\frac{1}{36}$
Probability for

$4$ times in

$6$ trials

$=_{ 4 }^{ 6 }{ C }{ \left( P \right) }^{ 4 }{ \left( (1-P) \right) }^{ 2 }$

$=_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$

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

0%
$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 9 }{ 10 } \right) }^{ 4 }\left( \frac { 1 }{ 10 } \right) \ + _{ 0 }^{ 5 }C{ { \left( \frac { 9 }{ 10 } \right) } }^{ 5 }$
0%
$\displaystyle 1 - _{ 1 }^{ 5 }{ C }{ \left( \frac { 9 }{ 10 } \right) }^{ 4 }\left( \frac { 1 }{ 10 } \right) - _{ 0 }^{ 5 }C{ { \left( \frac { 9 }{ 10 } \right) } }^{ 5 }$
0%
$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 9 }{ 10 } \right) }^{ 4 }\left( \frac { 1 }{ 10 } \right)$
0%
$\displaystyle \quad _{ 0 }^{ 5 }C{ { \left( \frac { 9 }{ 10 } \right) } }^{ 5 }$

Explanation

Probability of a vessel reacing safely is

$0.9$ .
The question asks for the probability of at least

$4$ arriving safely. i.e. either 4 arrive or all the 5 arrive

$P(4) =$

$5{C}_{4} \times {(\dfrac{9}{10})}^{4} \times \dfrac{1}{10}$

$P(5)$ =

${(\dfrac{9}{10})}^{5}$
So the probability of at least

$4$ reaching safely is

$P(4) + P(5)$
=

$5{C}_{4} \times {(\dfrac{9}{10})}^{4} \times \dfrac{1}{10}$ +

${(\dfrac{9}{10})}^{5}$ = Probability of a vessel reacing safely is

$0.9$
The question asks for the probability of at least

$4$ arriving safely. i.e. either

$4$ arrive or all the

$5$ arrive

$P(4) =$

$5{C}_{4} \times {(\frac{9}{10})}^{4} \times \dfrac{1}{10}$

$P(5) =$

${(\dfrac{9}{10})}^{5}$
So the probability of at least

$4$ reaching safely is

$P(4) + P(5)$
=

$5{C}_{1} \times {(\dfrac{9}{10})}^{4} \times \dfrac{1}{10}$ +

${(\dfrac{9}{10})}^{5}$

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

0%
$\displaystyle \frac{e^{-5}5^{10}}{ 10!}$
0%
$\displaystyle \frac{e^{-10}10^{10}}{ 10!}$
0%
$\displaystyle \frac{e^{-25}25^{10}}{10!}$
0%
$\displaystyle \frac{e^{-15}15^{10}}{10!}$

Explanation

Here P.D parameter

$\lambda = \cfrac{5}{2000}\times 10000=25$
Hence probability that exactly 10 houses will get damaged

$=P(X = 10) = \cfrac{e^{-25}(25)^{10}}{10!}$

The probability of getting a total of

$9$ exactly twice in

$6$ tosses of a pair of dice is

0%
$_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9} \right) }^{ 2}+_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 3 }$
0%
$_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
0%
$1-_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
0%
${ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$

Explanation

We get a sum of

$9$ for

$(5,4),(4,5)$ and

$(6,3),(3,6)$
Hence probability of getting a sum

$9$ ,

$=\dfrac{4}{36}$

$=\dfrac{1}{9}$
Now for

$6$ tosses, getting a sum of

$9$ exactly twice is

$\:^{6}C_{2}(\dfrac{1}{9})^{2}(1-\dfrac{1}{9})^{6-2}$

$=\:^{6}C_{2}(\dfrac{1}{9})^{2}(\dfrac{8}{9})^{4}$

$X$ is a binomial variable with

$E(X)=2$ and

$V(X)=\cfrac { 4 }{ 3 }$ ,
the probability of

$x$ successes is

0%
${ ^{ 6 } }C_{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6-x }$
0%
${ ^{ 6 } }C_{ x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 6-x }$
0%
${ ^{ 9 } }C_{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6-x }$
0%
${ ^{ 9 } }C_{ x }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 6-x }{ \left( \cfrac { 2 }{ 3 } \right) }^{ x }$

Explanation

Given E(X)=2 and V(X)=

$\dfrac{4}{3}$
E(X) = Mean = np = 2
V(X) = Variance = np(1-p) =

$\dfrac{4}{3}$

$\Rightarrow 2 (1-p) = \dfrac{4}{3}$

$\Rightarrow 1-p = \dfrac{2}{3}$

$\Rightarrow p = \dfrac{1}{3}$

$\Rightarrow np= 2$

$\Rightarrow n = \dfrac{2}{p} = 2 \times 3 = 6$

The probability of x successes is

$\Rightarrow P(X=x) = ^{n}C_{x} {p}^{x} ({1-p})^{n-x}$

$=\,{} ^{6}C_{x} ({\dfrac{1}{3}})^{x} ({1-\dfrac{1}{3}})^{6-x}$

$=\,{} ^{6}C_{x} ({\dfrac{1}{3}})^{x} (\dfrac{2}{3})^{6-x}$

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

0%
$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }\left( \frac { 1 }{ 5 } \right)$
0%
$\displaystyle _{ 0 }^{ 5 }{ C{ \left( \frac { 4 }{ 5 } \right) }^{ 5 }+ }\quad _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }\left( \frac { 1 }{ 5 } \right)$
0%
$\displaystyle _{ 0 }^{ 5 }{ C{ \left( \frac { 1 }{ 5 } \right) }^{ 5 }+ }\quad _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }$
0%
$\displaystyle _{ 1 }^{ 5 }{ C }{ \left( \frac { 4 }{ 5 } \right) }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }$

Explanation

The probability for a person being a swimmer is

$0.2$ .
Out of

$5$ boys the probability of at least

$4$ being a swimmer is,

$P(4) =$

$5{C}_{4} \times {(\dfrac{4}{5})}^{4} \times (\dfrac{1}{5})$

$P(5) =$

${(\dfrac{4}{5})}^{4}$
Hence, the probability is

$P(4) + P(5) =$

$5{C}_{4} \times {(\dfrac{4}{5})}^{4} \times (\dfrac{1}{5})$ +

${(\dfrac{4}{5})}^{4}$

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

0%
$\displaystyle \frac{22}{125}$
0%
$\displaystyle \frac{112}{625}$
0%
$\displaystyle \frac{116}{625}$
0%
$\displaystyle \frac {110}{625}$

Explanation

Probability of the team winning is

$\dfrac{2}{5}$
Hence the probability of the team losing is

$\dfrac{3}{5}$
Now the probability of A playing 4 games and winning more that half of the games is

$\:^{4}C_{3}(\displaystyle \frac{2}{5})^{3}(\frac{3}{5})^{1}+\:^{4}C_{4}(\frac{2}{5})^{4}(\frac{3}{5})^{0}$

$=\dfrac{1}{5^{4}}[4(8\times3)+1(16)]$

$=\dfrac{112}{625}$

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

0%
$_{ }^{ 8 }{ C_2 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }$
0%
$_{ }^{ 8 }{ C_2}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }$
0%
${ \left( \cfrac { 19 }{ 20 } \right) }^{ 8 }+\,_{ }^{ 8 }{ C_1 }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 7 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$
0%
${ \left( \cfrac { 19 }{ 20 } \right) }^{ 8 }+\,_{ }^{ 8 }{ C_1 }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 7 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }+\,_{}^{ 8 }{ C_2}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 2 }$

Explanation

A production process is supposed to contain 5% defective items.
Probability of defective items =

$\dfrac {5}{100} = \dfrac{1}{20}$
Given

$n = 8$
The probability that a sample contain less than 2 defective items = Probability of sample containing no defective items + Probability of sample containing one defective items
=

$^{8}C_{0} ({\dfrac{1}{20}})^{0} ({1-\dfrac{1}{20}})^{8-0} + ^{8}C_{1} ({\dfrac{1}{20}})^{1} ({1-\dfrac{1}{20}})^{8-1}$

$({\dfrac{19}{20}})^{8} + ^{8}C_{1} ({\dfrac{19}{20}})^{7} ({\dfrac{1}{20}})^{1}$

A fair die is tossed

$1620$ times. If getting a face with

$6$ points is considered as a success, then

$E(x) = ........, V(x) = .........$

0%
$270, 270$
0%
$270, 225$
0%
$270, 25$
0%
$27, 250$

Explanation

The probability of getting a face with six is

$\dfrac{1}{6}$ .
Thus the mean will be

$np$

$=1620(\dfrac{1}{6})$

$=270$
Variance will be

$np(1-p)$

$=270(\dfrac{5}{6})$

$=225$

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

0%
$1-_{ 4 }^{ 5 }{ C }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 1 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 4 }-{ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$
0%
${ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$
0%
$_{ 4 }^{ 5 }{ C }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 1 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 4 }+{ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$
0%
$_{ 4 }^{ 5 }{ C }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 1 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 4 }$

Explanation

Probability of a vessel reacing safely is

$0.9$ .
The question asks for the probability of at least

$4$ arriving safely. i.e. either

$4$ arrive or all the

$5$ arrive
P(4) =

$5{C}_{4} \times {(\dfrac{9}{10})}^{4} \times \dfrac{1}{10}$
P(5) =

${(\dfrac{9}{10})}^{5}$
So the probability of at least

$4$ reaching safely is

$P(4) + P(5)$
=

$5{C}_{4} \times {(\dfrac{9}{10})}^{4} \times \dfrac{1}{10}$ +

${(\dfrac{9}{10})}^{5}$

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

0%
$\displaystyle ^{ 7 }C_{ 5 }{ \left( \frac { 1 }{ 4 } \right) }^{ 6 }{ \left( \frac { 3 }{ 4 } \right) }$
0%
$\displaystyle ^{ 7 }C_{ 6 }{ \left( \frac { 3 }{ 4 } \right) }^{ 6 }{ \left( \frac { 1 }{ 4 } \right) }$
0%
$\displaystyle ^{ 7 }C_{ 6 }{ \left( \frac { 1 }{ 4 } \right) }^{ 6 }{ \left( \frac { 3 }{ 4 } \right) }$
0%
$\displaystyle ^{ 7 }C_{ 6 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }{ \left( \frac { 1 }{ 2 } \right) }$

Explanation

The probability of a man hitting the target is

$\dfrac {1}{4}$
For fires 7 times,

$n= 7$
The probability of hitting the target exactly six times is =

$P(X=k) = ^{n}C_{k} {p}^{k} ({1-p})^{n-k}$

$P(X=6) = {}^{7}C_{6} \left({\dfrac{1}{4}}\right)^{6} \left({1-\dfrac {1}{4}}\right)^{7-6}$

$= {}^{7}C_{6} \left({\dfrac{1}{4}}\right)^{6} \left(\dfrac {3}{4}\right)$

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

0%
$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }$
0%
$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }$
0%
$_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }+{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }\\$
0%
$1-_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }-_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }-{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }\\$

Explanation

Probability of an experiment which succeeds twice as often as it fails:

$\dfrac{2}{3}$
Probability of its fail is:

$\dfrac{1}{3}$
Selection of success four out of six event is:

$^{4}C_{3}$
Probability for at least for success out of six events is

$=P(4)+P(5)+P(6)$

$=_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+_{ 5 }^{ 6 }{ C }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 }+{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }\\$

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

0%
$^{ 10 }C_{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }$
0%
$^{ 10 }C_{ 6 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 4 }$
0%
$^{ 10 }C_{ 6 }{ \left( \cfrac { 1 }{ 4 } \right) }^{ 6 }{ \left( \cfrac { 3 }{ 4 } \right) }^{ 4 }$
0%
$^{ 10 }C_{ 6 }{ \left( \cfrac { 3 }{ 4 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 4 } \right) }^{ 4 }$

Explanation

Probability of success

$= \dfrac{2}{3}$
Probability of failure

$=1-\dfrac{2}{3}= \dfrac{1}{3}$
Probability of getting 6 successes in 10 trials is

$=^{ 10 }C_{ 6 }{ \left( \dfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \dfrac { 1 }{ 3 } \right) }^{ 4 }$

For a binomial distribution the mean and variance are respectively

$4$ and

$3$ . The probability of getting a non-zero success is

0%
$1-{ \left( \cfrac { 3 }{ 4 } \right) }^{ 16 }$
0%
${ \left( \cfrac { 1 }{ 4 } \right) }^{ 16 }$
0%
${ \left( \cfrac { 3 }{ 4 } \right) }^{ 16 }$
0%
$1-{ \left( \cfrac { 1 }{ 4 } \right) }^{ 16 }$

Explanation

$Mean=np=4$

$Variance=np(1-p)=3$
Hence,

$4(1-p)=3$

$(1-p)=\dfrac{3}{4}$

$p=\dfrac{1}{4}$
Therefore

$n=16$
Where

$n$ is the no. of trials.
Hence probability of a non zero success.

$=1-(\dfrac{3}{4})^{16}$

The probability of getting a total of

$9$ at least twice in

$6$ tosses of a pair of dice is

0%
$_{ }^{ 6 }{ C_2 }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
0%
${ \left( \cfrac { 8 }{ 9 } \right) }^{ 6 }+\ _{ }^{ 6 }{ C_1 }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 1 }$
0%
$1-{ \left( \cfrac { 8 }{ 9 } \right) }^{ 6 }-\ _{ }^{ 6 }{ C_1 }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 1 }$
0%
$1-{ \left( \cfrac { 8 }{ 9 } \right) }^{ 6 }$

Explanation

For getting a total of

$9$ in a pair of dice

$= (3,6), (6,3), (4,5), (5,4)$
Probability of sum of

$9$ in a pair of dice =

$\dfrac {4}{36} = \dfrac {1}{9}$

$n = 9$
The probability of getting a total of

$9$ at least twice

$= 1 -$ [ the probability of not getting a total of

$9\ +$ The probability of getting a total of

$9$ at once ]
=

$1 - \,{} ^{6}C_{0}{(\dfrac {1}{9})}^{ 0 }{ (1-\dfrac {1}{9}) }^{6-0} - \,{}^{6}C_{1}{(\dfrac {1}{9})}^{ 1 }{ (1-\dfrac {1}{9}) }^{6-1}$

$=$

$1 - { (\dfrac {8}{9}) }^{6} - \,{}^{6}C_{1}{(\dfrac {1}{9})}^{ 1 }{ (\dfrac {8}{9}) }^{5}$

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

0%
$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$
0%
${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$
0%
$1-{ \left( \cfrac { 1 }{ 20 } \right) }^{ 5 }$
0%
${ \left( \cfrac { 1 }{ 20 } \right) }^{ 5 }$

Explanation

Probability that bulb will fuse after 100 days

$=0.05=\dfrac{1}{20}$
Probability that bulb will not fuse after 100 days

$=1-\dfrac{1}{20}=\dfrac{19}{20}$
Probability that out of 5 such bulbs none of them fuse

$=^{5}C_{0}\times (\dfrac{1}{20})^0 \times (\dfrac{19}{20})^5=(\dfrac{19}{20})^5$

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

$400$ .... & ....

0%
$40, 36$
0%
$40, 6$
0%
$20, 6$
0%
$10, 6$

Explanation

Mean will

$np$

$=n(0.1)$

$=400(0.1)$

$=40$
And variance will be

$np(1-p)$

$=40(1-0.1)$

$=40(0.9)$

$=36$
Therefore the standard deviation is

$\sqrt{36}$

$=6$

A random variable

$X$ is binomially distributed with mean

$12$ and variance

$8$ . The parameters of the distribution are ...... & .......

0%
$\displaystyle 36, \frac{2}{3}$
0%
$\displaystyle 36, \frac{1}{3}$
0%
$\displaystyle 24, \frac{1}{3}$
0%
$\displaystyle 24, \frac{2}{3}$

Explanation

Mean

$= \mu = np$ = 12;

$n =$ number of trials ;

$p =$ probability of success ;

$q = (1-p) =$ probability of failure
Variance

$={ \sigma }^{ 2 } = npq = 8$

$8 = np(1-p) = 12(1-p)$ ;

$p = \dfrac {1 }{ 3 }$

$\mu = np = 12$

$n = \dfrac { 12}{ p } = 12 \times \dfrac { 3 }{ 1 } = 36$
Parameters are

$n =36$ and

$p = \dfrac {1 }{ 3 }$

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

0%
$^{ 5 }{ C }_1{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$
0%
${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }+^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$
0%
$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$
0%
$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }-^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$

Explanation

Probability that bulb will fuse after 100 days

$=0.5=\dfrac{1}{20}$
Probability that bulb will not fuse after 100 days

$=1-\dfrac{1}{20}=\frac{19}{20}$
Probability that out of 5 such bulbs at least one will fuse

$= 1-$ Probability that out of 5 such bulbs none of them fuse

$1- ^{5}C_{0}\times (\dfrac{1}{20})^0 \times (\dfrac{19}{20})^5=1-(\dfrac{19}{20})^5$

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

0%
${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$
0%
$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }-_{ 1 }^{ 5 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$
0%
$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$
0%
${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }+_{ 1 }^{ 5 }{ C }{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$

Explanation

Probability that bulb will fuse after 100 days

$=0.5=\dfrac{1}{20}$
Probability that bulb will not fuse after 100 days

$=1-\dfrac{1}{20}=\dfrac{19}{20}$
Probability that out of 5 such bulbs at least one will fuse

$= 1-$ Probability that out of 5 none bulb will fuse

$1- ^{5}C_{0}\times (\dfrac{1}{20})^0 \times (\dfrac{19}{20})^5=1-(\dfrac{19}{20})^5$

In a binomial distribution

$n=9$ and

$p=\cfrac { 1 }{ 3 }$ .The mode of the binomial distribution is

0%
$3$
0%
$4$
0%
$3$ and $4$
0%
$3.5$

Explanation

$(n+1)p$

$={10}(\dfrac{1}{3})$

$=3.33$
Hence, no integral.
Therefore mode is

$[3.33]$ where,

$[x]$ is the greatest integer function.
Hence mode is

$3$ .

Let X be a binomially distributed variate with mean 10 and varianceThen p(x>10) is

0%
$\dfrac{1}{2^{20}}\sum_{k=11}^{20}$ $^{20}$ $C_{k}$
0%
$\dfrac{1}{2^{20}}\sum_{k=1}^{11}$ $^{20}C_{k}$
0%
$\dfrac{1}{2^{20}}\sum_{k=11}^{20}$ $^{10}C_{k}$
0%
$\sum_{k=11}^{20}$ $^{20}C_{k}\left ( \dfrac{2}{3} \right )^{30-k}$

Explanation

Mean

$= \mu = np = 10$ and Variance =

${ \sigma }^{ 2 } = np(1-p) = 5$
So,

$10 \times (1-p) = 5 => p= \dfrac {1}{2}$

$n = \dfrac {10}{p} = 10 \times 2 = 20$
p(x=k)=

$^{ n }C_{ k }\quad { p }^{ k }{ (1-p) }^{ n-k }$

p(x>10) =

$^{ 20 }C_{ 11 } { \dfrac {1}{2}}^{ 11 }{ (1 - \dfrac {1}{2}) }^{ 20-11} + ^{ 20 }C_{ 12 } { \dfrac {1}{2}}^{ 12 }{ (1 - \dfrac {1}{2}) }^{20-12} +.....+ ^{ 20 }C_{20} { \dfrac {1}{2}}^{20}{ (1 - \dfrac {1}{2}) }^{20-20}$

$= ^{ 20 }C_{ 11 } ({ \dfrac {1}{2}})^{20} + ^{ 20 }C_{12 } ({ \dfrac {1}{2}})^{20} + ^{ 20 }C_{ 13 } ({ \dfrac {1}{2}})^{20} + ....+ ^{ 20 }C_{20} ({ \dfrac {1}{2}})^{20}$

$= \dfrac{1}{2}^{20} \sum_{k=11}^{20} \ ^{20}C_{k}$

In an experiment success is twice that of failure. If the experiment is repeated

$6$ times, the probability that atleast

$4$ times success is :

0%
$\dfrac{64}{729}$
0%
$\dfrac{192}{729}$
0%
$\dfrac{240}{729}$
0%
$\dfrac{496}{729}$

Explanation

Probability of success

$= \dfrac{2}{3}$
Probability of failure

$=1-\dfrac{2}{3}= \dfrac{1}{3}$
Probability of getting atleast 4 successes in 6 trials is

$= P(4)+P(5)+P(6)$

$=^{ 6 }C_{ 4 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 } + ^{ 6 }C_{ 5 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 1 } + ^{ 6 }C_{ 6 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 6 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 0 }$

$=\dfrac{240}{729}+\dfrac{192}{729}+\dfrac{64}{729}=\dfrac{496}{729}$

For a

$B.D.$

$\bar{x}=4,\sigma =\sqrt{3}$ , then

$P(X=r)$ =

0%
$^{16}C_{r}\left ( \dfrac{1}{4} \right )^{r}\left ( \dfrac{3}{4} \right )^{16-r}$
0%
$^{12}C_{r}\left ( \dfrac{1}{4} \right )^{r}\left ( \dfrac{3}{4} \right )^{12-r}$
0%
$^{12}C_{r}\left ( \dfrac{2}{3} \right )^{r}\left ( \dfrac{1}{3} \right )^{12-r}$
0%
$^{12}C_{r}\left ( \dfrac{3}{4} \right )^{r}\left ( \dfrac{1}{4} \right )^{12-r}$

Explanation

$Mean=np=4$

$Variance=(\sigma)^{2}=3$

$np(1-p)=3$

$4(1-p)=3$

$1-p=\dfrac{3}{4}$

$p=\dfrac{1}{3}$

$n=4(4)$

$=16$
Therefore

$P(X=r)$ implies

$\:^{n}C_{r}(\dfrac{1}{4})^{r}(\dfrac{3}{4})^{n-r}$

$=\:^{16}C_{r}(\dfrac{1}{4})^{r}(\dfrac{3}{4})^{16-r}$
Here

$n$ no. f trials, and

$p$ is the probability of success.

A symmetrical die is thrown

6 times. If getting an odd number is a success, the probability of at the most

$5$ successes is

0%
$\displaystyle \frac{5}{64}$
0%
$\displaystyle \frac{15}{64}$
0%
$\displaystyle \frac{63}{64}$
0%
$\displaystyle \frac{36}{64}$

Explanation

In a die there are

$3$ odd numbers:

$1, 3, 5$
Probability of getting an odd number in a single throw of a die is

$P=\dfrac{3}{6}=\dfrac{1}{2}$
Probability of at the most

$5$ successes is

$=1-P(X>5)$ (

$X$ is number of success)

$=1-P(X=6)$

$=1- ^{6}C_{6} (\dfrac{1}{2})^6= 1-\dfrac{1}{64}= \dfrac{63}{64}$

The binomial distribution for which,
mean

$+$

$2\times$ variance

$=$ 4, mean

$+$ variance

$= 3$ is

0%
$\left( 6,\ \dfrac { 1 }{ 3 } ,\ \dfrac { 2 }{ 3 } \right)$
0%
$\left( 6,\ \dfrac { 1 }{ 2 } ,\ \dfrac { 1 }{ 2 } \right)$
0%
$\left( 4,\ \dfrac { 1 }{ 2 } ,\ \dfrac { 1 }{ 2 } \right)$
0%
$\left( 4,\ \dfrac { 1 }{ 3 } ,\ \dfrac { 2 }{ 3 } \right)$

Explanation

Mean

$= \mu = np$ and Variance

$={ \sigma }^{ 2 } = npq$
mean

$+ 2\times$ variance

$=4$ , mean

$+$ variance

$=3$

$np + 2 npq = 4$ ;

$np + npq = 3$

$np + npq + npq = 4$ ;

$np + 1 = 3$

$3 + npq = 4$ ;

$np = 2$

$npq = 1$ ;

$2q = 1$

$q= \dfrac {1}{2}$ and

$p = 1 -q = \dfrac {1}{2}$

$np = 2$

$n = \dfrac {2}{p} = 2 \times \dfrac {2}{1} = 4$

Binomial distribution

$= ( 4, \dfrac {1}{2}, \dfrac {1}{2} )$

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

0%
$\displaystyle \frac{5}{16}$
0%
$\displaystyle \frac{7}{16}$
0%
$\displaystyle \frac{11}{16}$
0%
$\displaystyle \frac{9}{16}$

Explanation

${\textbf{Step - 1: Finding n, p and q}}$

${\text{We know that, mean = np and variance = np(1 - p)}}$

$\therefore {\text{ np = 2 and np(1 - p) = 1}}$

${\text{Dividing, }}\dfrac{{{\text{np(1 - p)}}}}{{{\text{np}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}$

$\Rightarrow {\text{ 1 - p = }}\dfrac{{\text{1}}}{{\text{2}}}$

$\Rightarrow {\text{ p = }}\dfrac{{\text{1}}}{{\text{2}}}$

$\because {\text{ q = 1 - p}}$

$\Rightarrow {\text{ q = 1 - }}\dfrac{{\text{1}}}{{\text{2}}}$

$\Rightarrow {\text{ q = }}\dfrac{{\text{1}}}{{\text{2}}}$

${\text{Substituting value of p in np = 2}}$

$\Rightarrow {\text{ n }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = 2}}$

$\Rightarrow {\text{ n = 4}}$

${\textbf{Step - 2: Calculating probability}}$

${\text{Probability of value greater than 1, P(X > 1) = 1 - P(X = 0) - P(X = 1)}}$

$\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{1}}}{{{{\text{2}}^{\text{4}}}}}{\text{ - }}{{\text{ }}^{\text{4}}}{{\text{C}}_{\text{1}}}\dfrac{{\text{1}}}{{{{\text{2}}^{\text{4}}}}}$

$\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{1}}}{{{\text{16}}}}{\text{ - }}\dfrac{{\text{4}}}{{{\text{16}}}}$

$\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{5}}}{{{\text{16}}}}$

$\Rightarrow {\text{ P(X > 1) = }}\dfrac{{{\text{11}}}}{{{\text{16}}}}{\text{ }}$

$\textbf{Hence option C is correct}$

$X$ follows a binomial distribution with parameters

$n = 6$ and

$P$ . If

$4P(x=4)=P(x=2)$ , then

$P=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$

Explanation

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

$4\times^{6}C_{4}\times(P^4)\times(1-P)^2=^{6}C_{2}\times(P^2)\times(1-P)^4$

$4\times P^2=(1-P)^2$

$(3P-1)(P+1)=0$

$P=\dfrac{1}{3}$

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

0%
$1/3$
0%
$1/4$
0%
$1/5$
0%
$1/6$

Explanation

In a binomial distribution consisting of 5 independent trials, n = 5
The probability of 1 and 2 successes are 0.4096 and 0.2048 respectively.

$^{5}C_{1}{ p }^{ 1 }{ (1-p) }^{5-1 } = 0.4096$

$^{5}C_{2}{ p }^{2 }{ (1-p) }^{5-2 } = 0.2048$

$5 { p }{ (1-p) }^{4} = 0.4096$ _____ (1)

$10 { p }^{ 2 }{ (1-p) }^{3} = 0.2048$ ______ (2)
Divide equation (1) by (2), we get

$\dfrac {(1-p)}{2p} = 2$

$5p = 1$

$p = \dfrac {1}{5}$

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

$\sigma ^{2}=$

0%
$n$
0%
$\dfrac{n}{2}$
0%
$\dfrac{n}{4}$
0%
$\dfrac{n}{6}$

Explanation

The probability of success, thus become

$\dfrac{1}{2}$

$\sigma^{2}=Variance=np(1-p)$

$=n(\dfrac{1}{2})(1-\dfrac{1}{2})$

$=n\dfrac{1}{4}$

$=\dfrac{n}{4}$

For a binomial distribution if

$P=\dfrac{1}{4}, n=20$ , the probability of mode is

0%
$^{20}C_{5}\left ( \dfrac{3}{4} \right )^{5}$
0%
$^{20}C_{5}\left ( \dfrac{1}{4} \right )^{5}\left ( \dfrac{3}{4} \right )^{15}$
0%
$^{10}C_{10}\left ( \dfrac{1}{4} \right )^{10}\left ( \dfrac{3}{4} \right )^{10}$
0%
$^{10}C_{10}\left ( \dfrac{3}{4} \right )^{10}$

Explanation

Given:

$p = \dfrac14, n=20$

For binomial distribution, mode

$= (n+1)(p)$

$=(20+1)\cdot \dfrac{1}{4}$

$=\dfrac{21}{4}$
Hence,

$(n+1)(p)$ is a non-integral value.

Therefore,
mode

$=\left[\dfrac{21}{4}\right]$ , where

$[\cdot]$ denotes greatest integer function.
mode

$=[5.25] = 5$

Hence, the required value of probability of mode is

$^{20}C_{5}\left(\dfrac{1}{4}\right)^{5}\left(\dfrac{3}{4}\right)^{15}$

Hence, option A.

Out of

$800$ families with

$4$ children each, the expected number of families having atleast one boy is

0%
$550$
0%
$50$
0%
$750$
0%
$300$

Explanation

The probability of having a girl

$=$ probability of having a boy

$=$

$\dfrac{1}{2}$
The probability of having no girl at all is equal to the probability of having at-least one boy

$=1-\:^{4}C_{0}(\dfrac{1}{2})^{0}(\dfrac{1}{2})^{4}$

$=1-\dfrac{1}{2^{4}}$

$=\dfrac{15}{16}$
Hence, the required expectancy

$=800(\dfrac{15}{16})$

$=15(50)$

$=750$

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

0%
$15(0.1)^{2}(0.9)^{5}$
0%
$20(0.1)^{2}(0.9)^{5}$
0%
$21(0.1)^{2}(0.9)^{5}$
0%
$23(0.1)^{2}(0.9)^{5}$

Explanation

Probability that a candidate secure a seat is

$=p=\dfrac{1}{10}=0.1$
Probability that a candidate couldn't secure a seat is

$=q=(1-\dfrac{1}{10})=0.9$
Selection of

$2$ out of

$7$ seats is

$={}^{7}C_{2}=21$

$P(X=r)={}^nC_rp^rq^{n-r}$
Probability that exactly two will get seats out of

$7$ is

$=21(0.1)^{2}(0.9)^{5}$

$X$ follows a binomial distribution with parameters

$n = 8$ and

$P=\frac{1}{2},$ than

$P(\left | x-4 \right |\leq 2)=$

0%
$\dfrac{119}{128}$
0%
$\dfrac{9}{128}$
0%
$\dfrac{101}{128}$
0%
$\dfrac{11}{128}$

Explanation

$|x-4|\leq 2$

$-2\leq x-4 \leq 2$

$2\leq x\leq 6$

$x\epsilon[2,6]$
Hence the required probability

$\dfrac{1}{2^{8}}[\:^{8}C_{2}+\:^{8}C_{3}+\:^{8}C_{4}+\:^{8}C_{5}+\:^{8}C_{6}]$

$=\dfrac{1}{2^{8}}[2(\:^{8}C_{2}+\:^{8}C_{3})+\:^{8}C_{4}]$ ...

(

$\:^{n}C_{r}=\:^{n}C_{n-r}$ )

$=\dfrac{1}{256}[2(28+56)+70]$

$=\dfrac{238}{256}$

$=\dfrac{119}{128}$

4 unbiased coins are tossed

$256$ times. The expected frequency of

$x$ heads

0%
$16^{4}C_{x}$
0%
$12^{4}C_{x}$
0%
$12^{4}C_{0}$
0%
$8^{4}C_{0}$

Explanation

In an unbiased coin, the probability of head=probability of tails =

$\dfrac{1}{2}$
Hence for 4 trials, probability of x heads is

$P(X)$

$=\:^{4}C_{x}(\dfrac{1}{2})^{x}(\dfrac{1}{2})^{4-x}$

$=\dfrac{1}{16}[\:^{4}C_{x}]$
Hence the expected frequency will be

$256\times(\dfrac{1}{16}[\:^{4}C_{x}])$

$=16\:^{4}C_{x}$

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

$=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$1$

Explanation

$P(x=r)=\:^{n}C_{r}p^{r}(1-p)^{n-r}$

$P(x=n-r)=\:^{n}C_{n-r}p^{n-r}(1-p)^{r}$
Now,

$\:^{n}C_{r}=\:^{n}C_{n-r}$

Hence,

$\dfrac{P(x=r)}{P(x=n-r)}$

$=\dfrac{p^{r}(1-p)^{n-r}}{p^{n-r}(1-p)^{r}}$

Is independent of

$n$ .
The above expression is true for

$x=\dfrac{1}{2}$ only.

Out of

$800$ families with

$4$ children each, the expected number of families having

$2$ boys and

$2$ girls

0%
$100$
0%
$200$
0%
$300$
0%
$400$

Explanation

${\textbf{Step - 1: Finding probability of 2 boys and 2 girls}}$

${\text{Probabilty of having a girl = }}\dfrac{{\text{1}}}{{\text{2}}}$

${\text{Probabilty of having a boy = }}\dfrac{{\text{1}}}{{\text{2}}}$

${\text{Probability of 2 boys and 2 girls , P = }}{{\text{ }}^{\text{4}}}{{\text{C}}_{\text{2}}}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}$

$\Rightarrow {\text{ P = 6 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{4}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{4}}}$

$\Rightarrow {\text{ P = }}\dfrac{{\text{3}}}{{\text{8}}}$

${\textbf{Step - 2: Calculating expectancy}}$

${\text{Expectancy, E = 800 }} \times {\text{ }}\dfrac{{\text{3}}}{{\text{8}}}$

$\Rightarrow {\text{ E = 100 }} \times {\text{ 3}}$

$\Rightarrow {\text{ E = 300}}$

${\textbf{Hence option C is correct.}}$

The probability of expected number of boys in a family with 8 children assuming that the sex distributions are equally probable is

0%
$^{8}C_{2}\left ( \dfrac{1}{2} \right )^{8}$
0%
$^{8}C_{3}\left ( \dfrac{1}{2} \right )^{8}$
0%
$^{8}C_{4}\left ( \dfrac{1}{2} \right )^{8}$
0%
$^{8}C_{5}\left ( \dfrac{1}{2} \right )^{8}$

Explanation

The probability of having a boy=probability of having a girl=

$\dfrac{1}{2}$
Hence for 8 children, the probability of having a boy, is the middle term of the binomial expansion.
This implies

$\:^{8}C_{4}(p^{4})(q^{4})$

$p=q=\dfrac{1}{2}$
Hence the required probability is

$\:^{8}C_{4}(\dfrac{1}{2})^8$

If X be B.V. with

$E(X)=5$ and

$E(X^{2})-{E(X)}^{2}=4$ then the parameters of distribution are

0%
$\dfrac{1}{4}, 20$
0%
$\dfrac{1}{5}, 20$
0%
$\dfrac{1}{5}, 25$
0%
$\dfrac{4}{5}, 25$

Explanation

Here

$E(X)=$ Mean

$=np$ and

$E(x^2)-E(x)^{2}=$ Variance

$=np(1-p)$
Here

$n=$ no.of trials while p is the probability of success.
Hence

$np=5$

$np(1-p)=4$

$5(1-p)=4$

$1-p=\dfrac{4}{5}$

$p=\dfrac{1}{5}$
Therefore

$n=5(5)$

$=25$
Hence the parameters are

$\dfrac{1}{5},25$

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

0%
$^{10}C_{8}(0.51)^{8}(0.49)^{2}$
0%
$^{10}C_{8}(0.49)^{8}(0.51)^{2}$
0%
$^{10}C_{8}(0.49X0.51)^{8}$
0%
$^{10}C_{8}(0.49X0.51)^{2}$

Explanation

Probability of having a girl

$=\dfrac{49}{100}=0.49$

Probability of having a boy

$=\dfrac{51}{100}=0.51$

Selection of 8 girls out of 10 babies is

$=^{10}C_{8}$
10 babies in which 8 are girls and 2 are boys, so probability of 8 girls amongst 10 babies born

$=^{10}C_{8}(0.49)^{8}(0.51)^{2}$

If x is $B\left ( x,n,\dfrac{1}{3} \right ),P(x\geq 1)> 0.8$ n the least value of n is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

$P(X\ge 1)>0.8\\ P(X\ge 1)=\sum _{ x=1 }^{ n }{ (\begin{matrix} n \\ x \end{matrix}){ (\dfrac { 1 }{ 3 } ) }^{ x }{ (\dfrac { 2 }{ 3 } ) }^{ n-x } } =\quad { (\dfrac { 1 }{ 3 } +\dfrac { 2 }{ 3 } ) }^{ n }-(\begin{matrix} n \\ 0 \end{matrix}){ (\dfrac { 1 }{ 3 } ) }^{ 0 }{ (\dfrac { 2 }{ 3 } ) }^{ n }\\ 1-(\begin{matrix} n \\ 0 \end{matrix}){ (\dfrac { 1 }{ 3 } ) }^{ 0 }{ (\dfrac { 2 }{ 3 } ) }^{ n }>0.8\\ { (\dfrac { 2 }{ 3 } ) }^{ n }\quad <\quad 0.2\\ \quad n=4\\$

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

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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