In the p.d.f. of a random variable of three missing entries are in the ratio $$1:2:3$$. $$X=x$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$P(X=x)$$ $$\dfrac{1}{10}$$ - - - $$\dfrac{3}{20}$$ Then the missing entries are?

$$\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}$$

$$\dfrac{1}{5}, \dfrac{2}{5}, \dfrac{3}{5}$$

$$\dfrac{3}{12}, \dfrac{4}{15}, \dfrac{6}{15}$$

$$\dfrac{1}{10}, \dfrac{2}{10}, \dfrac{3}{10}$$

MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 1

The p.m.f.of a r.v. X is as follows ;

$P(X=0)=3{ K }^{ 3 }\quad ,\quad P(X=1)=4K-10{ K }^{ 2 },\quad P(X=2)=5K-1,\quad P(X-x)=0$ for any other value of x. then k is equal to

0%
1
0%
2
0%
$\frac { 1 }{ 3 }$
0%
$\frac { 2 }{ 3 }$

Variance of the random variable

$X$ is

$4$ . Its mean is

$2$ . Then

$E(X^{2})$ is:

0%
$6$
0%
$8$
0%
$2$
0%
$4$

Explanation

Given: variance

$=4$ , Mean

$=2$

We know that,

$var(x)=E(x^2)-E((x))^2$

Given that,

$var(x)=4,E(x)=2$

$\therefore E(x^2)=var(x)+[E(x)]^2$

$=4+4=8$

Another name for the mean of a probability distribution is expected value.

0%
True
0%
False

The variance of the random variable

$x$ whose probability distribution is given by

$X=x_{i}: \quad-1 \quad, 0, \quad +1$

$p(X=x_{i}): 0.4, \ 0.2, \ \ \ 0.4$ is

0%
$0.4$
0%
$0.6$
0%
$0.8$
0%
$1.0$

Explanation

mean,

$\bar{x} = \cfrac{-1+0+1}{3}=0$
Hence required variance is

$=\sum (x_i-\bar{x})^2P(x_i)=\sum (x_i)^2P(x_i)=(-1)^2(.4)+0(.2)+(1)^2(.4)=.8$

To define probability disribution function we assign to each variable

0%
the respective probabilities
0%
the specific random values
0%
some integers
0%
none

Out of following which are random variables

0%
$x=$ "Number of heads when two coins are tossed.
0%
$x=$ "Sum of digits on uppermost face of two dice"
0%
solution of " $X-4=0"$
0%
$x=$ "Raining"

Explanation

$1.)$

$x=$ Number of heads when two coins are tossed.

The sample space

$=[HH,HT,TH,TT].$ When

$2$ coin are tossed,

Probability Distribution Function.

$2.)$

$X=$ "Sum of digit on uppermost face of two dice".

Two dices are thrown, so the total outcome

$=6^2=36$

Sum pf digits on the dice can be greater than

$2$ and less than or equal to

$12.$

$3.)$ Solution of

$'x-4=0'$

$\rightarrow x=4,$ it is not a random variable.

$4.)$

$x=$ "Raining"

The word consist of repeating words like

$I$ and

$N.$

Given

$E(X + c) = 8$ and

$E(X - c) = 12$ then the value of

$c$ is

0%
$-2$
0%
$4$
0%
$-4$
0%
$2$

Explanation

Considering linearity of expected values

$E(x)$ , we know that

$E[X+c]=E[X]+c$ .
Hence, consider the above equations

$E[X+c]=E[X]+c=8$

$E[X+c]=E[X]-c=12$
Subtracting second equation from i gives us

$2c=-4$

$c=-2$ .

Which of the following is an example of a random experiment?

0%
Selecting a card from a pack of playing cards.
0%
Measuring the weight of a person.
0%
Finding the length of your pencil box.
0%
Throwing two coins together.

To verify Pythagoras theorem is a random experiment.

0%
True
0%
False

The mathematical expectation of sum of points when we throw n symmetrical dice is

0%
7n
0%
$7\displaystyle \times\frac{n}{2}$
0%
$\displaystyle \frac{n}{2}$
0%
$\displaystyle \frac{7n}{3}$

Explanation

$Expected\quad Value\quad =\sum { x(i)*P(x(i)) }$
Expectation

$= n*$ Expectation of one die

$= n*(1/6*(1+2+3+4+5+6))$

$= 7*n/2$

Difference between sample space and subset of sample space is considered as

0%
numerical complementary events
0%
equal compulsory events
0%
complementary events
0%
compulsory events

In the p.d.f. of a random variable of three missing entries are in the ratio

$1:2:3$ .

$X=x$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$\dfrac{1}{10}$	-	-	-	$\dfrac{3}{20}$

Then the missing entries are?

0%
$\dfrac{1}{5}, \dfrac{2}{5}, \dfrac{3}{5}$
0%
$\dfrac{3}{12}, \dfrac{4}{15}, \dfrac{6}{15}$
0%
$\dfrac{1}{10}, \dfrac{2}{10}, \dfrac{3}{10}$
0%
$\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}$

Explanation

$P(2)+P(3)+P(4)=1-\cfrac{1}{10}-\cfrac{3}{20}\\ \quad=\cfrac{20-2-3}{20}\\ \quad\quad=\cfrac{15}{20}=\cfrac{3}{4}$

Ratio of Probability

$=1:2:3$

$P(2)=\cfrac{\cfrac{3}{4}}{1+2+3}\times1=\cfrac{3}{4}\times\cfrac{1}{6}=\cfrac{1}{8}\\P(3)=\cfrac{3}{4}\times\cfrac{2}{6}=\cfrac{3}{4}\times\cfrac{2}{6}=\cfrac{1}{4}\\ P(4)=\cfrac{3}{4}\times\cfrac{3}{6}=\cfrac{3}{8}$

A random variable

$X$ has the probability distribution

$X$ :

$1, 2, 3, 4, 5, 6, 7, 8$

$P(X): 0.15, 0.23, 0.12, 0.10 ,0.20 ,0.08 ,0.07 ,0.05$ . For the events

$\mathrm{E}=$

$\{$ X is a prime number

$\}$ and

$\mathrm{F}=\{\mathrm{X}<4\}$ , the probability

$\mathrm{P}(\mathrm{E}\cup\mathrm{F})$ is:

0%
$0.87$
0%
$0.77$
0%
$0.35$
0%
$0.50$

Explanation

$E$ ={ is a prime number}

$=\left\{ 2,3,5,7 \right\}$

$P\left( E \right) =P\left( X=2 \right) +P\left( X=3 \right) +P\left( X=5 \right) +P\left( X=7 \right) \\ P\left( E \right) =0.23+0.12+0.20+0.07=\quad 0.62\\ F=\left\{ X<4 \right\} =\left\{ 1,2,3 \right\} \\ P\left( F \right) =\quad P\left( X=1 \right) =+P\left( X=2 \right) +P\left( X=3 \right) \\ P\left( F \right) =\quad 0.15+0.23+0.12=0.5$

$E\cap F=$ {X is prime number as well as <4}

$=\left\{ 2,3 \right\}$

$P\left( E\cap F \right) =\quad P\left( X=2 \right) +P\left( X=3 \right) =0.23+0.12=0.35$
Therefore required probability

$P\left( E\cup F \right) =P\left( E \right) +P\left( F \right) -P\left( E\cap F \right) \\ \Rightarrow P\left( E\cup F \right) =0.62+0.5-0.35\\ \Rightarrow P\left( E\cup F \right) =0.77$

Statement 1: The variance of first

$\mathrm{n}$ even natural numbers is

$\displaystyle \frac{\mathrm{n}^{2}-1}{4}$
Statement 2: The sum of first

$\mathrm{n}$ natural numbers is

$\displaystyle \frac{\mathrm{n}(\mathrm{n}+1)}{2}$ and the sum of squares of first

$\mathrm{n}$ natural numbers is

$\displaystyle \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}$

0%
Statement 1 is true, Statement2 is true,Statement 2 is a correct explanation for Statement 1
0%
Statement 1 is true, Statement2 is true;Statement2 is not a correct explanation for statement 1
0%
Statement 1 is true, Statement 2 is false.
0%
Statement 1 is false, Statement 2 is true

Explanation

Sum of first n even natural numbers

$\displaystyle =2+4+6+...+2n=2(1+2+...+n)$

$\displaystyle =2\frac{n(n+1)}{2}=n(n+1)$

Mean

$\displaystyle (\bar{x})=\frac{n(n+1)}{n}=n+1$

variance

$\displaystyle =\frac{1}{n}(\sum x_{1})^{2}-(\bar{x})^{2} =\frac{1}{n}(2^{2}+4^{2}+...+(2n)^{2})-(n+1)^{2}$

$\displaystyle =\frac{1}{n}2^{2}(1^{2}+2^{2}+...+n^{2})-(n+1)^{2}$

$\displaystyle =\frac{4}{n}\frac{n(n+1)(2n+1)}{6}-(n+1)^{2}$

$\displaystyle =\frac{2}{3}(n+1)(2n+1)-(n+1)^{2}$

$\displaystyle =\frac{n+1}{3}[2(2n+1)-3(n+1)]$

$\displaystyle=\frac{(n+1)}{3} (n-1)=\frac{n^{2}-1}{3}$

Hence statement

$1$ is false while statement

$2$ is true

A random variable X has its range

$X = \{3, 2, 1\}$ with the probabilities,

$\dfrac{1}{2},\dfrac{1}{3}$ and

$\dfrac{1}{6}$ respectively. The mean value of X is

0%
$\dfrac{5}{3}$
0%
$\dfrac{7}{3}$
0%
$3$
0%
$4$

Explanation

Mean=np

$=(3.\dfrac{1}{2})+(2.\dfrac{1}{3})+\dfrac{1}{6}$

$=\dfrac{13}{6}+\dfrac{1}{6}$

$=\dfrac{14}{6}$

$=\dfrac{7}{3}$

Four different objects 1,2,3,4 are distributed at random in four places marked 1,2,3,What is the probability that none of the objects occupy the place corresponding to its number ?

0%
$\frac{17}{24}$
0%
$\frac{3}{8}$
0%
$\frac{1}{2}$
0%
$\frac{5}{8}$

Explanation

Let four places be
1 2 3 4
Now object i cannot occupy the place 1. ...(A)
Suppose object 2 occupies the place 1. Then other placements can be done in following 6 ways :
(i) 2 1 3 4
(ii) 2 1 4 3
(iii) 2 3 1 4
(iv) 2 3 4 1
(v) 2 4 1 3
(vi) 2 4 3 1

Here out of six ways, only there are permissible, because (i), (iii) and (vi) are not permissible because of the non-fulfilment of condition (A).

Hence, required probability is

$\frac{3}{6}=\frac{1}{2}$ .

Similarly we can allow objects 3 and 4 to occupy place 1 and in each case we can find that the Probability is

$\frac{1}{2}$

If the range of the random variable X is from a to

$\mathrm{b},\ \mathrm{a}<\mathrm{b},\ \mathrm{F}(\mathrm{X}<\mathrm{a})=$

0%
$0$
0%
$1$
0%
$0.5$
0%
$3$

Explanation

$X$ cannot attain any value less than

$a$
Hence, the probability is 0.

The cumulative distribution function of a random variable

$x$ is defined as

0%
$P(X < x)$
0%
$P(X \ge x)$
0%
$P(X \le x)$
0%
$P(X > x)$

Explanation

The cumulative distribution function of a real-valued random variable X is the function given by

${ F }_{ X }(x) = P (X\le x)$

An urn contains

$5$ red and

$2$ black balls. Two balls are randomly drawn. Let

$X$ represent the number of black balls. What are the possible values of

$X$ ? Is

$X$ a random variable ?

0%
$0,1,2$
0%
$3,5,7$
0%
$7,7,8$
0%
$1,5,7$

Explanation

The two balls selected can be represented as

$BB, BR, RB, RR$ , where

$B$ represents a black ball and

$R$ represents a red ball.

$X$ represents the number of black balls.

$\therefore X \left(BB\right) = 2$

$X \left(BR\right) = 1$

$X \left(RB\right) = 1$

$X \left(RR\right) = 0$
Therefore, the possible values of

$X$ are

$0, 1$ and

$2$ .
Yes,

$X$ is a random variable.

If an unbiased coin is tossed once, then the two events of getting a Head and a Tail are -

0%
Mutually exclusive
0%
Exhaustive
0%
Equally likely
0%
All of these

If a random variable X takes value

$0$ and

$1$ with respective probabilities

$\dfrac{2}{3}$ and

$\dfrac{1}{3}$ , then the expected value of X is

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

Explanation

$E[X]=\sum _{ i=0 }^{ 2 }{ { x }_{ i }{ p }_{ i } } =\quad 0*\dfrac { 2 }{ 3 } +1*\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 }$

In a class of 80 students, 48 are boys and the rest of the students are girls. If 10 students shift to the other class room and a student is selected at random from the remaining class, what is the probability that a girl is selected ?

0%
11/35
0%
27/70
0%
11/40
0%
Cannot be determined

10 different books and 2 different pens are given to 3 boys so that each gets equal number of things. The probability that the same boy does not receive both the pens is

0%
$\frac{5}{11}$
0%
$\frac{7}{11}$
0%
$\frac{2}{3}$
0%
$\frac{8}{11}$

Explanation

Probability of same boy does not receive both pens

$=1-$ probability of any boy receive both pens

$\because$ All the students receives same number of articles, so each on gets

$=(10+2)/3=4$
No. of ways in which

$10$ books and

$2$ pens is distributed to

$3$ students

$=^{ 12 }{ C }_{ 4 }^{ 8 }{ C }_{ 4 }^{ 4 }{ C }_{ 4 }$
Now to find the no. of ways such that any boy receive the both pens
Select one boy from

$3$ and give both pens to him

$=\left( ^{ 3 }{ C }_{ 1 }\times 1 \right)$ ways
Rest two article can be given to this boy in

$=^{ 10 }{ C }_{ 2 }$ ways
Other

$2$ boys will get article in

$=\left( ^{ 8 }{ C }_{ 4 }\times ^{ 4 }{ C }_{ 4 } \right)$ ways

$\therefore$ no. of ways such that any boy receives the both pen

$^{ 3 }{ C }_{ 1 }\times \left( 1\times ^{ 10 }{ C }_{ 2 } \right) \times ^{ 8 }{ C }_{ 4 }\times ^{ 4 }{ C }_{ 4 }$

$\therefore$ Required probability

$=1-\left( ^{ 3 }{ C }_{ 1 }\times 1 \right) \times ^{ 10 }{ C }_{ 2 }\times ^{ 8 }{ C }_{ 4 }\times \cfrac { ^{ 4 }{ C }_{ 4 } }{ ^{ 12 }{ C }_{ 4 } } \quad ^{ 8 }{ C }_{ 4 }^{ 4 }{ C }_{ 4 }$

$=\cfrac { 8 }{ 11 }$

If the range of the random variable X is from

$a$ to

$b$ then

$F(X\leq b)$ is

0%
0
0%
1
0%
0.5
0%
3

Explanation

If the range of the random variable X is from a to b then

$F(X\le b ) = P(X\le b ) = P(X =a ) + .... + P ( X = b) = 1$
Total sum of probability of every variable X is equal to 1.

Expected number of heads when we toss

$n$ unbiased coins is

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

Explanation

Expected number

$=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+.........n$ times

$=\dfrac{n}{2}$

$5$ throws of a die, getting

$1$ or

$2$ is a success. The mean number of successes is

0%
$\dfrac{5}{3}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{5}{9}$
0%
$\dfrac{9}{5}$

Explanation

Clearly the question is based on binomial trial

Here number of trial is

$n=5$

Now in one trial probability of getting 1 or 2 (probability of success ) on throwing a dice is

$p=\dfrac{2}{6}=\dfrac{1}{3}$

Hence mean of success

$= np =5\cdot \dfrac{1}{3}=\dfrac{5}{3}$

Which of the following is NOT a random experiment ?

0%
Rolling an unbiased dice
0%
Tossing a fair coin
0%
Drawing a card from a well shuffled pack of 52 card
0%
None of these

Explanation

Rolling an unbiased dice results in

$6$ outcomes. All have equal probability of showing up.

Tossing a fair coin results in

$2$ outcomes. Both have equal probabilities of showing up.

Drawing a card from from a well shuffled pack of

$52$ cards can have

$52$ outcomes in total with equal probabilities.

Hence, above all are example of random experiment.

The probability distribution of a discrete random variable

$X$ is:

$X = x$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$k$	$2k$	$3k$	$4k$	$5k$

Find

$P (X\leq 4)$

0%
$\cfrac 23$
0%
$\cfrac 34$
0%
$\cfrac 45$
0%
$\cfrac 56$

Explanation

Here, we must have

$k + 2k + 3k + 4k + 5k = 1$

$\Rightarrow 15k = 1$

$\Rightarrow k = \dfrac {1}{15}$
Now,

$P(X \leq 4) = k + 2k + 3k + 4k$

$= \dfrac {1}{15} + \dfrac {2}{15} + \dfrac {3}{15} + \dfrac {4}{15}$

$= \dfrac {10}{15}$

$\therefore P(X\leq 4) = \dfrac {2}{3}$

Which of the following is not a random experiment?

0%
Tossing a coin
0%
Rolling a dice
0%
Choosing a card from a deck of 52 cards
0%
Throw a stone from a roof of a building

Which of the following is an random experiment?

0%
Rolling a pair of dice
0%
Choosing $2$ marbles from a jar
0%
Choosing a number at random from $1$ to $10$
0%
Tossing two coins

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 1 - MCQExams.com

0:0:5

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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