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CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 5 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Probability Distribution And Its Mean And Variance
Quiz 5
The counter part of probability mass function is:
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Probability distribution
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Probability density function
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Distribution function
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None of these
Explanation
We know that,
The probability mass function is defined for a discrete random variable.
And for a continuous random variable we define probability density function.
Thus the counter part of probability mass function is
probability density function.
A variable that can assume any possible value between two points is called:
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Discrete sample space
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Discrete random variable
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Continuous random variable
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Random variable
Explanation
We know that,
A
continuous random variable
is a
random variable
where the variable can take infinitely many values in the given interval.
Thus a
variable that can assume any possible value between two points is called
continuous random variable.
Hence option C is correct.
Amit tosses a fair coin twice, and let $$X$$ be defined as the number of heads he observe. Find probability mass function $$P_x$$.
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{8}$$
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$$\dfrac{1}{4}$$
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None off these
Explanation
Given Amit tosses a fair coin twice
$$\therefore$$ Sample space, $$S=[HH,HT,TH,TT]$$
$$\Rightarrow$$ When $$P(x)=1$$ it is a PMF
i.e, $$\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{4}{4}=1$$
$$\therefore$$ Probability mass function $$Px=\dfrac{1}{4}.$$
Hence, the answer is $$\dfrac{1}{4}.$$
Probability distribution of discrete random variable is classified as
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probability mass function.
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posterior mass function.
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interior mass function.
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continuous mass function.
Explanation
The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values.
It is also sometimes called the probability function or the probability mass function.
Thus option $$(A)$$ is correct.
A random variable $$X$$ has the following p.d.f.
$$X$$
$$0$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$
$$7$$
$$P(X = x)$$
$$0$$
$$k$$
$$2k$$
$$2k$$
$$3k$$
$$k^{2}$$
$$2k^{2}$$
$$7k^{2} + k$$
The value of $$k$$ is
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$$\dfrac {1}{8}$$
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$$\dfrac {1}{10}$$
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$$0$$
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$$-1$$
Explanation
$$\displaystyle \sum_{x=0} ^{x=7} P(X)=1$$
$$\Rightarrow P(0)+P(1)+P(2)...P(7)=1$$
$$\Rightarrow 0+k+2k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1$$
$$\Rightarrow 10k^{2}+9k-1=0$$
$$\Rightarrow 10k^{2}+10k-k-1=0$$
$$\Rightarrow 10k(k+1)-(k+1)=0$$
$$\Rightarrow (k+1)(10k-1)=0$$
$$k=-1$$ or $$k=\dfrac{1}{10}$$.
But $$0\leq P(x)\leq 1$$
Hence $$k=\dfrac{1}{10}$$.
If $$x$$ is a continuous random variable, then $$P(a < x < b) =$$
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$$P(a \leq x \leq b)$$
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$$P(a < x \leq b)$$
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$$P(a \leq x < b)$$
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all the three above
Explanation
Consider there are $$n$$ number of points between $$a$$ and $$b$$ where $$n$$ tends to infinite $$($$given $$x$$ is a continuous random variable$$)$$
$$P(x=a)=P(x=b)=\dfrac{1}{n}$$ . As $$n\rightarrow \infty $$ , $$\dfrac{1}{n}\rightarrow 0$$
Therefore $$P(x=a)=p(x=b)\rightarrow 0$$
$$\Longrightarrow P(a<x<b)=P(a\le x\le b)=P(a<x\le b)=P(a\le x<b)$$
Therefore the correct option is $$D$$.
A random variable X has the following probability distribution
$$X = x$$
$$-2$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P(x)$$
$$0.1$$
$$0.1$$
$$0.2$$
$$0.2$$
$$0.3$$
$$0.1$$
Then $$E(x) =$$
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$$0.8$$
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$$0.9$$
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$$0.7$$
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$$1.1$$
Explanation
$$E(x) = \sum x. P(x)$$
$$X = x$$
$$-2$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P(x)$$
$$0.1$$
$$0.1$$
$$0.2$$
$$0.2$$
$$0.3$$
$$0.1$$
$$x. P(x)$$
$$-0.2$$
$$-0.1$$
$$0$$
$$0.2$$
$$0.6$$
$$0.3$$
$$E(x) = \sum x. P(x) = -0.2 - 0.1 + 0.2 + 0.6 + 0.3$$
Thus $$E(x)= 0.8$$
If $$E(X + C) = 8$$ and $$E(X - C) = 12$$ then the value of $$C$$ is
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$$-2$$
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$$4$$
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$$-4$$
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$$2$$
Explanation
We know that
$$E(X+C)=E(X)+C$$ ...(i)
Therefore
$$E(X+C)=E(X)+C=8$$
$$E(X-C)=E(X)-C=12$$
Subtracting equation ii from i
$$2C=-4$$
$$C=-2$$.
Let the random variable X follow B(6, p). If 16 P(X = 4) = P(X = 2), then what is the value of p?
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{5}$$
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$$\dfrac{1}{6}$$
Explanation
$$16P(X=4)=P(X=2)$$
$$\therefore 16(^6C_4p^4q^2)=^6C_2p^2q^4$$
$$\therefore 16(^6C_4p^4(1-p)^2)=^6C_2p^2(1-p)^4$$
$$\therefore 16p^2=(1-p)^2$$
$$\therefore 15p^2+2p-1=0$$
$$\therefore (3p+1)(5p-1)=0$$
$$\therefore p=-\dfrac{1}{3}$$ or $$p=\dfrac{1}{5}$$
Probability cannot be negative.
So, $$p=\dfrac{1}{5}$$
Answer is option (C)
What is $$P(Z = 5)$$ equal to ?
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{6}$$
Explanation
Given set $$=\{1,2,3,4,5,6,7\}$$
$$Z=X+Y$$
Where $$X$$ is randomly selected from set of odd numbers and
$$Y$$ is randomly selected from set of even numbers
That gives $$Z$$ is odd and greater than one
Now, for
$$Z=5$$
Possible ordered pairs are $$\{(1,4),(2,3)\}$$
Probability of getting these 2 pairs $$=P(1)\times P(4)+P(2)\times P(3)$$
$$=\cfrac{1}{4}\times \cfrac{1}{3}+\cfrac{1}{4}\times \cfrac{1}{3}$$
$$=\cfrac{1}{6}$$
What is $$P(Z = 10)$$ equal to ?
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$$0$$
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$$1/2$$
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$$1/3$$
0%
$$1/5$$
Explanation
Given set $$=\{1,2,3,4,5,6,7\}$$
$$Z=X+Y$$
Where $$X$$ is randomly selected from set of odd numbers and
$$Y$$ is randomly selected from set of even numbers.
Which gives $$Z$$ as odd and greater than one
Given :
$$Z=10$$
But $$10$$ is even
So, $$P(Z=10)=0$$
If $$x$$ is a continuous random variable then $$P(x \geq a) = $$
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$$P (x < a)$$
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$$1 - P(x > a)$$
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$$P(x > a)$$
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$$1 - P (x \leq a - 1)$$
Explanation
Given $$x$$ is continuous random variable , Let us consider that there are $$n$$ points in total, Since $$x$$ is continuous, $$n\rightarrow \infty $$
The probability of $$x=a$$ is $$\dfrac{1}{n}\rightarrow 0$$ , therefore $$P(x=a)=0$$
Which implies $$P(x\ge a) = P(x>a)+P(x=a) = P(x>a)$$
Therefore the correct option is $$C$$
The probability distribution of a random variable is given below :
$$X = x$$
0
1
2
3
4
5
6
7
$$P(X = x)$$
$$0$$
$$K$$
$$2K$$
$$2k $$
$$3K $$
$$K^2$$
$$2K^2 $$
$$7K^2 + k$$
Then $$P(0 < X < 5) = $$
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$$\dfrac{1}{10}$$
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$$\dfrac{3}{10}$$
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$$\dfrac{8}{10}$$
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$$\dfrac{7}{10}$$
Explanation
$$\sum p(X=x_i)=1$$
$$9K+10K^{2}=1$$
$$10K^{2}+9K-1=0$$
$$10K^{2}+10k-k-1=0$$
$$10K(K+1)-1(K+1)=0$$
$$K=\dfrac{1}{10}$$
$$P(0< X< 5)=p(x=1)+p(x=2)+p(x=3)+p(x=4)=K+2K+2K+3K=8K=\dfrac{8}{10}$$
$$\therefore P(0< X< 5)=\dfrac{8}{10}$$
Let $$X$$ and $$Y$$ be two random variables. The relationship $$E(XY) = E(X) \cdot E(Y)$$ holds
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Always
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If $$E(X +Y)= E(X)+ E(Y)$$ is true
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If $$X$$ and $$Y$$ are independent
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If $$X$$ can be obtained from $$Y$$ by a linear transformation
Explanation
We know that the relationship
$$E(XY)=E(X) \cdot E(Y)$$
holds, when $$X$$ and $$Y$$ are independent variables.
If the range of random variable $$X = \left \{0, 1, 2, ...\right \}$$ and $$P(X = k) = \dfrac {c(k + 1)}{2^{k}}$$ for $$k = 0, 1, 2, ...$$ then $$c =$$
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$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{3}$$
0%
$$\dfrac{1}{4}$$
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$$\dfrac{1}{5}$$
Explanation
$$P(X=0)=c$$
$$P(X=1)=\dfrac{2c}{2}$$
$$P(X=2)=\dfrac{3c}{2^{2}}$$
$$\sum P(X=r)=1$$
$$\Rightarrow \dfrac{c}{2^{0}}+\dfrac{2c}{2^1}+\dfrac{3c}{2^2}+........=1$$
$$\Rightarrow \dfrac{1c}{2}+\dfrac{2c}{2^{2}}+.......=\dfrac{1}{2}$$
$$\Rightarrow c+\dfrac{c}{2}+\dfrac{c}{2^{2}}+.......=\dfrac{1}{2}$$
$$\Rightarrow c\left[1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+.....\right]$$
$$\Rightarrow c\left[\dfrac{1}{1-\dfrac{1}{2}}\right]=\dfrac{1}{2}$$
$$\Rightarrow c[2]=\dfrac{1}{2}$$
$$\Rightarrow c=\dfrac{1}{4}$$
A $$r. v. X\sim B(n, p)$$. If values of mean and variance of $$X$$ are $$18$$ and $$12$$ respectively then total number of possible values of $$X$$ are
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$$54$$
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$$55$$
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$$12$$
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$$18$$
Explanation
$$Mean = np = 18$$
$$Variance = npq = 12$$
$$\dfrac {npq}{np} = \dfrac {12}{18}$$
$$\therefore q = \dfrac {2}{3}$$
Now, $$p = 1 - q = 1 - \dfrac {2}{3}$$
$$\therefore p = \dfrac {1}{3}$$
Also, $$np = 18$$
$$\implies n \left (\dfrac {1}{3}\right ) = 18$$
$$\implies n= 54$$
$$\therefore$$ Possible values of $$X$$ are
$$0, 1, 2, ..... 54$$
$$\therefore$$ No. of possible values are $$55$$.
If $$X$$ is a binomial variate with the range $$\left\{ 0,1,2,3,4,5,6 \right\} $$ and $$P(X=2)=4P(X=4)$$, then the parameter of $$X$$ is
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$$\cfrac{1}{3}$$
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$$\cfrac{1}{2}$$
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$$\cfrac{2}{3}$$
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$$\cfrac{3}{4}$$
Explanation
Here n=6
According to the question,
$$^6C_2p^2q^4=4\times ^6C_4p^4q^2\\\implies q^2=4p^2\\\implies (1-p)^2=4p^2\\\implies 3p^2+2p-1=0\\\implies (p+1)(3p-1)=0\\\implies p=\dfrac 13, p=-1$$
p cannot be negative, hence $$p=\dfrac 13$$
If the p.d.f of a r.v.$$X$$ is given as
$$xi$$
$$-2$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$P(X=xi)$$
$$0.2$$
$$0.3$$
$$0.15$$
$$0.25$$
$$0.1$$
then $$F(0)=$$
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$$P(X< 0)$$
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$$P(X> 0)$$
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$$1-P(X>0)$$
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$$1-P(X< 0)$$
Explanation
We know that, probability function $$F$$ for a random variable $$X$$ is given by
$$F(x)=\displaystyle \sum_{u\leq x} f(u)=P(X\leq x)$$
$$\therefore F(0)=P(X\leq 0)$$
$$=P(X=0)+P(X=-1)+P(X=-2)$$
$$=0.15+0.3+0.2$$
$$=0.65$$
$$=1-0.35$$
$$=1-(0.25+0.1)$$
$$=1-(P(X=1)+P(X=2))$$
$$=1-P(X>0)$$
Hence, $$F(0)=1-P(X>0)$$
The value of k when $$f(x) = \frac {1}{\sqrt{x}},0 < x < 4$$, is the p.d.f of r.v.x
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$$-4$$
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$$\dfrac {-1}{4}$$
0%
$$4$$
0%
$$\dfrac{1}{4}$$
Explanation
$$f(x)=\dfrac {k}{\sqrt x}$$ where $$x\in (0,4)$$
So, $$\displaystyle \int_0^4{\dfrac k{\sqrt x}dx}=1$$ (Summation of all probability distribution=1)
i.e, $$2k\sqrt x]_0^4=1\Rightarrow 4k=1$$
So, $$k=\dfrac 14$$
Hence, option D is correct.
If r.v$$X$$: waiting time in minutes for bus and p.d.f of $$X$$ is given by
$$f(x)=\begin{cases} \cfrac { 1 }{ 5 } ,0\le x\le 5 \\ 0,\quad otherwise \end{cases}$$
then probability of waiting time not more than $$4$$ minutes is $$=$$...........
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$$0.3$$
0%
$$0.8$$
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$$0.2$$
0%
$$0.5$$
Explanation
Given :
$$f(x)=\begin{cases} \cfrac { 1 }{ 5 } ,0\le x\le 5 \\ 0,\quad otherwise \end{cases}$$
Probability of waiting time not more than $$4$$ is same as probability of waiting time less than $$4$$ minutes and it is given by
$$P(X<4)=P(X=3)+P(X=2)+P(X=1)+P(X=0)$$
$$=\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}$$ .... [From the definition of $$f$$]
$$=4\times \dfrac { 1 }{ 5 } =0.8$$
The probability distribution of $$X$$ is
$$X$$
0
1
2
3
$$P(x)$$
0.3
k
2k
3k
The value of $$k$$ is
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$$0.116$$
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$$0.7$$
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$$1$$
0%
$$0.3$$
Explanation
$$\displaystyle \sum _{ x=0 }^{ \infty }{ P(x) } =1$$
$$\displaystyle \sum _{ x=0 }^{ 3 }{ P(x) } =1$$
$$P\left( 0 \right) +P\left( 1 \right) +P\left( 2 \right) +P\left( 3 \right) =1$$
$$\Rightarrow0.3+k+2k+3k=1$$
$$\Rightarrow 6k=1-0.3$$
$$\Rightarrow 6k=0.7$$
$$k=\cfrac { 0.7 }{ 6 }; \ $$$$k=0.116$$
For the following distribution function $$F(x)$$ of a r.v $$X$$ is given
$$x$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$
$$F(x)$$
$$0.2$$
$$0.37$$
$$0.48$$
$$0.62$$
$$0.85$$
$$1$$
Then $$P(3 < x\leq 5) =$$
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$$0.48$$
0%
$$0.37$$
0%
$$0.27$$
0%
$$1.47$$
Explanation
$$x$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$
$$f(x)$$
$$0.2$$
$$0.37$$
$$0.48$$
$$0.62$$
$$0.85$$
$$1$$
$$p(x)$$
$$0.2$$
$$0.17$$
$$0.11$$
$$0.14$$
$$0.23$$
$$0.15$$
$$P(3 < x\leq 5) = p(x = 4) + p(x = 5)$$
$$= 0.14 + 0.23$$
$$= 0.37$$
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
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$$\frac{14}{3}$$
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$$\frac{13}{3}$$
0%
$$\frac{12}{3}$$
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$$\frac{11}{3}$$
Explanation
E(X) is nothing but mean of X
$$\rightarrow $$ first sic natural numbers are
$$1,2,3,4,5,6$$
$$\rightarrow $$We can select two positive no.s in
$$6\times 5=30$$ ways
$$\rightarrow 2$$ numbers are selected at random among which X denotes the larger of the two no.s
$$\rightarrow $$But as X is largest among the two ,
it can be $$2,3,4,5$$ or $$6$$
$$\therefore P(X=2)=P$$ (larger no. is $$2$$) $${(1,2),(2,1)}$$
$$=\cfrac { 2 }{ 30 } $$
$$\rightarrow P\left( X=3 \right) =P$$ (Larger no. is $$3$$)
$$=\left\{ \left( 1,3 \right) ,\left( 3,1 \right) ,\left( 2,3 \right) ,\left( 3,2 \right) \right\} $$
$$=\cfrac { 4 }{ 30 } $$
$$\rightarrow P\left( X=4 \right) =P$$ (larger no. is $$4$$)
$$=\left\{ \left( 1,4 \right) ,\left( 4,1 \right) ,\left( 2,4 \right) ,\left( 4,2 \right) ,\left( 3,4 \right) ,\left( 4,3 \right) \right\} $$
$$=\cfrac { 6 }{ 30 } $$
$$\rightarrow P\left( X=5 \right) =P$$ (Larger no. is $$5$$)
$$=\left\{ \left( 1,5 \right) ,\left( 5,1 \right) ,\left( 2,5 \right) ,\left( 5,2 \right) ,\left( 3,5 \right) ,\left( 5,3 \right) ,\left( 4,5 \right) ,\left( 5,4 \right) \right\} $$
$$=\cfrac { 8 }{ 30 } $$
$$\rightarrow P\left( X=6 \right) =P$$ (Larger no. is $$6$$)
$$=\left\{ \left( 1,6 \right) ,\left( 6,1 \right) ,\left( 2,6 \right) ,\left( 6,2 \right) ,\left( 3,6 \right) ,\left( 6,3 \right) ,\left( 4,6 \right) ,\left( 6,4 \right) ,\left( 5,6 \right) ,\left( 6,50 \right) \right\} $$
$$=\cfrac { 10 }{ 30 } $$
$$\therefore E\left( X \right) =$$ mean $$=\sum { P\left( { X }_{ i } \right) } .{ X }_{ i }$$
$$=2\left( \cfrac { 2 }{ 30 } \right) +3\left( \cfrac { 4 }{ 30 } \right) +4\left( \cfrac { 6 }{ 30 } \right) +5\left( \cfrac { 8 }{ 30 } \right) +6\left( \cfrac { 10 }{ 30 } \right) $$
$$=\cfrac { 140 }{ 30 } =\cfrac { 14 }{ 3 } $$
The expected value of the number of points, obtained in a single throw of die, is
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$$\dfrac{3}{2}$$
0%
$$\dfrac{5}{2}$$
0%
$$\dfrac{7}{2}$$
0%
$$\dfrac{9}{2}$$
Explanation
Considering that the dice is a fair dice, the probability of all outcomes is equal.
$$\therefore$$ sample space of outcomes of dice is $$\{1,2,3,4,5,6\}$$
Probability of each outcome = $$\dfrac{1}{6}$$
$$\therefore $$ $$E[\text{outcome of a single throw}]=\dfrac{1}{6}\times(1+2+3+4+5+6)=\dfrac{21}{6}=\dfrac{7}{2}$$
This is the required answer.
$$X$$ has three children in his family. What is the probability that all the three children are boys?
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$$\dfrac{1}{8}$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{3}$$
0%
$$\dfrac{3}{8}$$
Explanation
X has $$3$$ children in his family.
All possible outcomes are if $$G$$ represents girl child and B represent boy child, $$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$$, hence $$n(S)=8$$
Out of which favorable outcomes for all the three chilren being boy, $$n(E)=1$$
Hence, the probability that all the three children are boys is $$\dfrac {n(E)}{n(S)}=\dfrac 18$$
$$X$$ has three children in his family. Probability of atleast two girls in the family is.....
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0%
$$\dfrac{1}{8}$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{3}{4}$$
Explanation
$$X$$ has $$3$$ children in his family.
All possible outcomes are if $$G$$ represents girl child and $$B$$ represent boy child, $$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$$, hence $$n(S)=8$$
Out of which favorable outcomes for at-least two girls are, $$n(E)=4$$
Hence, the probability that there are at-least two girls is $$\dfrac {n(E)}{n(S)}=\dfrac 48=\dfrac 12$$
If $$A$$ and $$B$$ are two non empty sets then $${ \left( A\cup B \right) }^{ C }=$$
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$${ A }^{ C }\cup { B }^{ C }$$
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$${ A }^{ C }\cap { B }^{ C }$$
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$$A\cup { B }^{ C }$$
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$${ A }^{ C }\cup B$$
Explanation
$${ \left( A\cup B \right) }^{ C }={ A }^{ C }\cap { B }^{ C }$$
The probability distribution of a random variable $$X$$ is given below:
$$x$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$
$$P(X=x)$$
$$a$$
$$a$$
$$a$$
$$b$$
$$b$$
$$0.3$$
If mean of $$X$$ is $$4.2$$, then $$a$$ and $$b$$ are respectively equal to
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$$0.3,0.2$$
0%
$$0.1,0.4$$
0%
$$0.1,0.2$$
0%
$$0.2,0.1$$
Explanation
Given, mean of $$x$$ is $$4.2$$
Thus that $$a+2a+3a+4b+5b+1.8=4.2$$
$$\Rightarrow 6a+9b=2.4$$ .....(1)
Also $$a+a+a+b+b+0.3=1$$
$$\Rightarrow 3a+2b=0.7$$ .....(2)
Multiply equation (2) by $$2$$,
$$6a+4b=1.4$$ .....(3)
Subtracting equations (1) and (3), we get
$$5b=1.0\Rightarrow b=0.2$$
Substituting this value in equation (1), we get
$$6a+9(0.2)=2.4$$
$$\Rightarrow 6a+1.8=2.4\Rightarrow a=0.1$$
Therefore, $$a=0.1,b=0.2$$
$$X$$ has three children in his family. What is the probability of two or more boys in the family?
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0%
$$\dfrac{1}{8}$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{3}{8}$$
Explanation
$$X$$ has $$3$$ children in his family.
All possible outcomes are if $$G$$ represents girl child and $$B$$ represent boy child, $$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$$, hence $$n(S)=8$$
Out of which favorable outcomes for two or more boys are, $$n(E)=4$$
Hence, the probability that there are two or more boys in the family is $$\dfrac {n(E)}{n(S)}=\dfrac 48=\dfrac 12$$
$$X$$ has three children in his family. The probability of one girl and two boys is......
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0%
$$\dfrac{1}{8}$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{3}{8}$$
Explanation
$$X$$ has $$3$$ children in his family.
All possible outcomes are if G represents girl child and $$B$$ represent boy child,$$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$$, hence $$n(S)=8$$
Out of which favorable outcomes for one girl and two boys are, $$n(E)=3$$
Hence, the probability that there are one girl and two boys is $$\dfrac {n(E)}{n(S)}=\dfrac 38$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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