MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 2

A ______ experiment is a process whose outcome is undetermined.

0%
variable
0%
random
0%
event
0%
sample

In which experiment outcomes are not predictable?

0%
sample
0%
event
0%
random
0%
essential

Explanation

In random experiment outcomes are not predictable.
Example: toss a coin, sample space

$= \{H, T\}$

A box contains 3 white and 2 black balls. Two balls are drawn at random one after the other. If the balls are not replaced. what is the probability that both the balls are black ?

0%
$\dfrac{2}{5}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{10}$
0%
None of the above

Explanation

Probabilty of drawing black in the first draw

$=\cfrac {\text{total black balls} }{ \text{total balls} } =\cfrac { 2 }{ 5 }$

Probability of drawing black in the second draw

$=\cfrac { 1 }{ 4 }$

Probability that both balls are black

$=$ Probability of first ball being black

$\times$ Probability of second ball being black

$=\cfrac { 2 }{ 5 } \times \cfrac { 1 }{ 4 } \\ =\cfrac { 1 }{ 10 }$

The probability that a leap year will have

$53$ sundays is

0%
$1/7$
0%
$2/7$
0%
$5/7$
0%
$6/7$

Explanation

A leap year has

$52$ weeks and

$2$ days.

The

$53^{rd}$ Sunday will be from these extra two days.

These

$2$ days can be (Sunday,Monday) or (Mon,Tue) or (Tue,Wed).....(Sat,Sun)

There are

$7$ possibilities for these

$2$ days

Out of which Sunday is coming in

$2$ possibilities

$\therefore P(\text{2 sundays in leap year})=\dfrac{2}{7}$

The probability distribution of a discrete random variable X is given in the following table :

$X = x$	$0$	$1$	$2$
$P(x)$	$4C^{3}$	$4C - 13C^{2}$	$7C - 1$

;

$C > 0$ then

$C =$ ________.

0%
$2$
0%
$1$
0%
$\dfrac {1}{4}$
0%
$1$ and $-\dfrac {1}{4}$

Explanation

Probability distribution is given by

$P(X)=P(X=0)+P(X=1)+P(X=2)$

$=4C^{3} + 4C - 13C^{2} + 7C - 1$

Since, total probability is

$1$

$\therefore 4C^{3} + 4C - 13C^{2} + 7C - 1=1$

$\therefore C = 1,2,\dfrac {1}{4}$ .

In the following, find the value of k.

$P(x)=\left\{\begin{matrix} kx & for & x=1, 2, 3\\ 0 & for & otherwise\end{matrix}\right.$ .

0%
$k=\dfrac 16$ .
0%
$k=\dfrac 13$ .
0%
$k=\dfrac 19$ .
0%
$k=\dfrac 15$ .

Explanation

Given

$P(x)=\left\{\begin{matrix} kx & for & x=1, 2, 3\\ 0 & for & otherwise\end{matrix}\right.$ .

$\sum p(x)=1$

Thus,

$k(1)+k(2)+k(3)=1$

$6k=1$

$k=\dfrac{1}{6}$

The p.d.f of a continuous random variable

$X$ is

$f(x)=\begin{cases} \dfrac { x }{ 8 } ,\quad 0<x<4 \\ 0,\ \ \ otherwise \end{cases}$
Then the value of

$P(X>3)$ is

0%
$\dfrac{3}{16}$
0%
$\dfrac{5}{16}$
0%
$\dfrac{7}{16}$
0%
$\dfrac{9}{16}$

Find the c.d.f.

$F(x)$ associated with the following p.d.f.

$f(x)$ :

$f(x)\begin{matrix} =12x^2(1-x), & 0 < x < 1\\ =0, & otherwise\end{matrix}$ . Also, find

$P\left(\dfrac{1}{3} < X < \dfrac{1}{2}\right)$ by using p.d.f. and c.d.f.

0%
$F(x)=4x^3-3x^2, P\left(\dfrac{1}{3} < X < \dfrac{1}{2}\right)=\dfrac{119}{432}$ .
0%
$F(x)=3x^3-5x^2, P\left(\dfrac{1}{3} < X < \dfrac{1}{2}\right)=\dfrac{129}{432}$ .
0%
$F(x)=4x^2-3x^2, P\left(\dfrac{1}{3} < X < \dfrac{1}{2}\right)=\dfrac{119}{432}$ .
0%
$F(x)=4x-3x^3, P\left(\dfrac{1}{3} < X < \dfrac{1}{2}\right)=\dfrac{19}{432}$ .

Two symmetrical dice are thrown

$200$ times. Getting a sum of

$9$ points is considered to be a success. The probability distribution of successes is

0%
$\left( 200,\ \cfrac { 1 }{ 9 } ,\ \cfrac { 8 }{ 9 } \right)$
0%
$\left( 200,\ \cfrac { 2 }{ 9 } ,\ \cfrac { 7 }{ 9 } \right)$
0%
$\left( 200,\ \cfrac { 4 }{ 9 } ,\ \cfrac { 5 }{ 9 } \right)$
0%
$\left( 200,\ \cfrac { 1 }{ 4 } ,\ \cfrac { 3 }{ 4 } \right)$

Explanation

Two symmetrical dice are thrown

$200$ times.

$n = 200$

Getting a sum of

$9$ points is considered to be a success.

Ways of sum

$9$ are

$= (3,6),(6,3),(4,5),(5,4)$ total ways of sum

$9 = 4$

Total ways of when

$2$ die are rolled

$= 36$

Probability of success

$= p =$

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

So,

$q = (1-p) = ( 1-$

$\dfrac {1}{9} ) = \dfrac {8}{9}$

The probability distribution of successes is

$= ( n , p , q )$

$=\left( 200,\ \dfrac {1}{9}, \dfrac {8}{9} \right)$

If the variance of the random variable

$X$ is

$5$ , then the variance of the random variable

$-3X$ is

0%
$15$
0%
$45$
0%
$-45$
0%
$60$

Explanation

We know that

${ \sigma }^{ 2 }_{ a+bx }={ b^{ 2 }\sigma }^{ 2 }$
Here

$a=0, b= -3$

$\Rightarrow { \sigma }^{ 2 }$ for

$-3X = 9*5 = 45$

A die is tossed twice. Getting an odd number is termed a success. The probability distribution of number of successes (X) is formed. Then its mean, variance are

0%
$1,\: \displaystyle\frac{1}{2}$
0%
$\displaystyle\frac{1}{2},\: 1$
0%
$\displaystyle\frac{1}{2},\: \displaystyle\frac{1}{2}$
0%
$1,\: 1$

Explanation

Here,

$n = 2, p = \cfrac{1}{2}, q = 1-p = \cfrac{1}{2}$

$\therefore$ mean

$=np = 1$ and variance

$=npq = \cfrac{1}{2}$

A random variable

$X$ takes values

$-1, 0, +1$ , Its mean is

$0.6$ and if

$P(X=0)=0.2$ , then

$P(X=1)=$

0%
$0.7$
0%
$0.5$
0%
$0.1$
0%
$0.2$

Explanation

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

$P(1)+P(-1) = 1-0.2 = 0.8$

$\dots (i)$

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

$1\times P(1) + (-1)\times P(-1) + 0 = 0.6$

$P(1)-P(-1)=0.6$

$P(-1)=P(1)-0.6$

substitute in

$(i)$

$\Rightarrow P(1)+P(1)-0.6=0.8$

$\Rightarrow 2P(1)=1.4$

$\Rightarrow P(1)=0.7$

The standard deviation

$\sigma$ of

$(q+p)^{16}$ isThe mean of the distribution is

0%
$2$
0%
$8$
0%
$16$
0%
$20$

Explanation

$n =$ number of trials

$= 16$

Standard deviation

$= \sigma = \sqrt { npq } = 2$

$2 = \sqrt { 16p(1-p) }$

$\Rightarrow \displaystyle p (1-p) = \frac { 1 }{ 4 }$

$\Rightarrow \displaystyle 4 { p }^{ 2 } - 4p + 1 = 0$

$\Rightarrow \displaystyle { (p }-\frac { 1 }{ 2 } )^{ 2 } = 0$

$\Rightarrow \displaystyle p = \frac { 1 }{ 2 }$

Mean

$\displaystyle = \mu = np = 16 \times \frac { 1 }{ 2 } = 8$

A coin is tossed successively until for the

$1$ st time head occurs. The expected number of tosses required is

0%
$4$
0%
$2$
0%
$1$
0%
$5$

Explanation

Either the event occurs on the first trial with probability p, or with probability 1-p the expected wait is

$1 + E$ . (E is Expectation)
Therefore

$E = p + ( 1-p)(1 + E)$ , from which

$E = 1/p$ .
(Assuming E is finite)
Therefore the expected wait for a single head to appear

$= 2.$

A random variable

$X$ takes the values

$0, 1$ and

$2$ . If

$P(X=1)=P(X=2)$ and

$P(X=0)=0.4$ , then the mean value of the random variable

$X$ is

0%
$0.2$
0%
$0.5$
0%
$0.7$
0%
$0.9$

Explanation

$P(X=0) + P(X=1) + P(X=2) =1$
Hence,

$P(X=1)=P(X=2) = (1-0.4)/2 = 0.3$
Mean Value

$=\sum { x(i)\times P(x(i)) }$

Therefore, average value

$= 0\times 0.4 +1\times 0.3 +2\times 0.3 = 0.9$ .

A fair die is rolled

$180$ times. The expected number of

$6$ is

0%
$50$
0%
$30$
0%
$10$
0%
$5$

Explanation

When a fair die is rolled once, the probability of getting a six is

$\dfrac{1}{6}$ .
When a fair die is rolled twice, the sample space contains

$6^{2}=36$ outcomes.
The probability of getting a six is

$\dfrac{6}{36}=\dfrac{1}{6}$
Hence the number of expected 6 when a fair die is rolled n times is

$=n\dfrac{1}{6}$

$=\dfrac{180}{6}$

$=30$

The mean or average number of points when we throw a symmetrical die is

0%
$14$
0%
$7$
0%
$7/2$
0%
$14/3$

Explanation

$E[X] = x*P[X]$

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

$=21/6$

$=3.5$

The mathematical expectation of sum of points when

$2$ symmetrical dice are rolled is

0%
$14$
0%
$7$
0%
$7/2$
0%
$14/3$

Explanation

Sample space:

(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

$Sum\quad can\quad be\quad 2,3,4,5,6,7,8,9,10,11,12\\ P(2)=\dfrac { 1 }{ 6\times6 } =\dfrac { 1 }{ 36 } \\$

$P(3)=\dfrac { 2 }{ 6\times6 } =\dfrac { 2 }{ 36 } \\$

$P(4)=\dfrac { 1+1+1 }{ 6\times6 } =\dfrac { 3 }{ 36 } \\$

$P(5)=\dfrac { 2+2 }{ 6\times6 } =\dfrac { 4 }{ 36 } \\$

$P(6)=\dfrac { 2+2+1 }{ 6\times6 } =\dfrac { 5 }{ 36 } \\$
$P(7)=\dfrac { 2+2+2 }{ 6\times6 } =\dfrac { 6 }{ 36 } \\$

$P(8)=\dfrac { 2+2+2+1 }{ 6\times6 } =\dfrac { 5 }{ 36 } \\$

$P(9)=\dfrac { 2+2+2+2 }{ 6\times6 } =\dfrac { 4 }{ 36 } \\$

$P(10)=\dfrac { 2+2+2+2+1 }{ 6\times6 } =\dfrac { 3 }{ 36 } \\$

$P(11)=\dfrac { 2+2+2+2+2 }{ 6\times6 } =\dfrac { 2 }{ 36 } \\$

$P(12)=\dfrac { 1 }{ 6\times6 } =\dfrac { 1 }{ 36 } \\$

$Expectation\quad =\quad \dfrac { 1 }{ 36 } (2+6+12+20+30+42+40+36+30+22+12)=\dfrac { 252 }{ 36 } =7\\ \\ \\ \\ \\$

In a business venture a man can make a profit of Rs.

$2000/-$ with probability of

$0.4$ or have a loss of Rs.

$1000/-$ with probability 0.His expected profit is

0%
Rs. $800/-$
0%
Rs. $600/-$
0%
Rs. $200/-$
0%
Rs. $400/-$

Explanation

Both cases of profit and loss are independent.
Expected profit will be:

$2000\times 0.4+(-1000\times 0.6)$

$800-600=200$

A random variable

$X$ takes values

$-1, 0, +1$ . Its mean is

$0.6$ . If

$P(X=0)=0.2$ , then

$P(X=1)=$

0%
$0.1$
0%
$0.3$
0%
$0.7$
0%
$0.6$

Explanation

Let

$P(X = 1) = a, P(X = 0) = b, P(X = -1) = c$ ,

We have,

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

$\Rightarrow 0.2 + a + c = 1$

$\Rightarrow a + c = 0.8$ ....(i)

Also, mean =

$\sum X.P(X)$

$\Rightarrow 0.6 = a \times 1 + 0 \times b + c \times (-1)$

$\Rightarrow a - c = 0.6$ .....(ii)

From (i) and (ii)

$a= 0.7$ and

$c=0.1$

$4$ bad apples accidentally got mixed up with

$20$ good apples. In a draw of

$2$ apples at random, expected number of bad apples is

0%
$1$
0%
$2/3$
0%
$1/3$
0%
$1/6$

Explanation

Total apples are:

$4$ Bad apples

$(B)$ +

$20$ Apples

$(A)$

$=$

$24$
Selection of

$2$ apples from

$24$ apples

$= ^{24}C_{2}$
Case

$A$ :

$1\ B$ and

$1\ A$

$= \dfrac{^{4}C_{1} \times ^{20}C_{1}}{^{24}C_{2}}=\dfrac{80}{276}$
Case

$B$ :

$2\ B$ and

$0\ A$

$= \dfrac{^{4}C_{2} \times ^{20}C_{0}}{^{24}C_{2}}=\dfrac{6}{120}$
Expected number of Bad apples drawn is

$=1\times Case (A)+ 2\times Case (B)$

$=\dfrac{80}{276}+ 2\times \dfrac{6}{120}$

$=\dfrac{1}{3}$

An urn A contains 4 white and 6 red balls. Three balls are drawn at random the expected number of red balls drawn is

0%
3.0
0%
1.8
0%
1.2
0%
1.6

Explanation

Total number of balls: 4 White and 6 Red

$=$ 10 balls
Cases for three balls are drawn at random:
Selection of 3 balls from 10 balls

$= ^{10}C_{3}$
Case A: 1 Red and 2 White

$= \dfrac{^{6}C_{1} \times ^{4}C_{2}}{^{10}C_{3}}=\dfrac{36}{120}$
Case B: 2 Red and 1 White

$= \dfrac{^{6}C_{2} \times ^{4}C_{1}}{^{10}C_{3}}=\dfrac{60}{120}$
Case C: 3 Red and 0 White

$= \dfrac{^{6}C_{3} \times ^{4}C_{0}}{^{10}C_{3}}=\dfrac{20}{120}$
Expected number of red balls drawn is

$=1\times Case (A)+ 2\times Case (B) + 3\times Case (C)$

$=\dfrac{36}{120}+ 2\times \dfrac{60}{120}+ 3\times\dfrac{20}{120}$

$=\dfrac{9}{5}=1.8$

The probability that there would be

$1, 2$ or

$3$ persons riding a bicycle are

$0.85, 0.12$ and

$0.03$ respectively. The expected number of persons per bicycle is

0%
$2$
0%
$1$
0%
$1.18$
0%
$3$

Explanation

Expected number of persons per bicycle is

$=1\times P(1)+2\times P(2)+3\times P(3)$

$=1 \times 0.85 + 2\times 0.12 + 3 \times 0.03$

$=1.18$

Two cards are drawn simultaneously from a well shuffled pack of

$52$ cards. The expected number of aces is

0%
$\displaystyle \frac{4}{13}$
0%
$\displaystyle \frac{3}{13}$
0%
$\displaystyle \frac{2}{13}$
0%
$\displaystyle \frac{1}{13}$

Explanation

$P(X=0) = \dfrac { 48C_2 }{ 52C_2 } \\ P(X=1) = \dfrac { 4C_1*48C_1 }{ 52C_2 } \\ P(X=2) = \dfrac { 4C_2 }{ 52C_2 } \\ E(X) = \sum _{ 0 }^{ 2 }{ { x }_{ i } } { p }_{ i } = \dfrac { 48C_2 }{ 52C_2 }*0 +\dfrac { 4C_1*48C_1 }{ 52C_2 }*1 +\dfrac { 4C_2 }{ 52C_2 } \\ =\dfrac { 204 }{ 26*51 } = \dfrac { 2 }{ 13 }$

Three coins whose faces are marked

$1$ and

$2$ are tossed. The expected sum of numbers on their faces is

0%
$4$
0%
$4.5$
0%
$5$
0%
$6$

Explanation

Probability of coin when it shows 1

$=P(1)=\dfrac{1}{2}$
Probability of coin when it shows 2

$=P(2)=\dfrac{1}{2}$
Expected sum of numbers for three coins is

$=1\times P(1)+2\times P(2)+1\times P(1)+2\times P(2)+1\times P(1)+2\times P(2)$

$=(1\times \dfrac{1}{2} +2\times \dfrac{1}{2})+(1\times \dfrac{1}{2} +2\times \dfrac{1}{2})+(1\times \dfrac{1}{2} +2\times \dfrac{1}{2})$

$=4.5$

Two unbiased coins whose faces are marked

$1$ and

$2$ are tossed. The mean value of the total of the numbers is

0%
$3$
0%
$4$
0%
$5$
0%
$2$

Explanation

Probability of coin when it shows 1

$=P(1)=\dfrac{1}{2}$
Probability of coin when it shows 2

$=P(2)=\dfrac{1}{2}$
The mean value of the total of the numbers is

$=1\times P(1)+2\times P(2)+1\times P(1)+2\times P(2)$

$=(1\times \dfrac{1}{2} +2\times \dfrac{1}{2})+(1\times \dfrac{1}{2} +2\times \dfrac{1}{2})$

$=3$

An urn A contains 4 white and 6 red balls. Three balls are drawn at random the expected number of white balls drawn is

0%
$3.0$
0%
$1.8$
0%
$1.2$
0%
$1.6$

Explanation

Total number of balls: 4 White and 6 Red

$=$ 10 balls
Cases for three balls are drawn at random:
Selection of 3 balls from 10 balls

$= ^{10}C_{3}$
Case A: 1 White and 2 Red

$= \dfrac{^{4}C_{1} \times ^{6}C_{2}}{^{10}C_{3}}=\dfrac{60}{120}$
Case B: 2 White and 1 Red

$= \dfrac{^{4}C_{2} \times ^{6}C_{1}}{^{10}C_{3}}=\dfrac{36}{120}$
Case C: 3 White and 0 Red

$= \dfrac{^{4}C_{3} \times ^{6}C_{0}}{^{10}C_{3}}=\dfrac{4}{120}$
Expected number of white balls drawn is

$=1\times Case (A)+ 2\times Case (B) + 3\times Case (C)$

$=\dfrac{60}{120}+ 2\times \dfrac{36}{120}+ 3\times\dfrac{4}{120}$

$=\dfrac{6}{5}=1.2$

If it rains a dealer in rain coats can earn Rs.

$500$ /- a day. If it is fair he will lose Rs.

$40$ /- a day. His mean profit if the probability of a fair day is

$0.6$ is:

0%
Rs. $230$ /-
0%
Rs. $460$ /-
0%
Rs. $176$ /-
0%
Rs. $88$ /-

Explanation

$Mean \ Value\quad =\sum { x(i)*P(x(i)) }$
Mean profit

$= 0.4*500+0.6*(-40) = 176$ .

The probability that the value of certain stock will remain the same is

$0.46$ . The probability that its value will increase by Rs.

$0.50$ or Re.

$1$ per share are respectively

$0.17$ and

$0.23$ and the probability that its value will decrease by Rs.

$0.25$ per share is

$0.14$ . The expected gain per share is

0%
Rs. $0.75$
0%
Rs. $0.25$
0%
Rs. $0.28$
0%
Rs. $0.50$

Explanation

The probability that the value of certain stock will remain the same is

$0.46$ .

The probability that its value will increase by Rs.

$0.50$ or Re.

$1$ per share are respectively

$0.17$ and

$0.23$ .

The probability that its value will decrease by Rs.

$0.25$ per share is

$0.14$ .

Expected value

$=\sum { x(i)\times P(x(i)) }$

$= 0.5\times0.17+0\times 0.46+1\times0.23-0.25\times0.14$

$=0.28$

The value of

$K$ , if the probability distribution of a discrete random variable

$x$ is

$X=x_{1}:1 2 3$

$p(X=x_{1}):\dfrac{1}{k^{2}}\dfrac{2}{k^{2}}\dfrac{3}{k^{2}}$

0%
$\sqrt{6}$
0%
$-\sqrt{6}$
0%
$\pm \sqrt{6}$
0%
6

Explanation

$\displaystyle\sum_{1}^{3}P(X=x)=1$

$\dfrac{1}{K^{2}}+\dfrac{2}{K^{2}}+\dfrac{3}{K^{2}}=1$

hence

$K^{2}=6$

$K=\pm \sqrt{6}$

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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