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CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Probability Distribution And Its Mean And Variance
Quiz 3
If the variance of the random variable $$X$$ is $$4$$, then the variance of the random variable $$5X+10$$ is
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0%
$$100$$
0%
$$10$$
0%
$$50$$
0%
$$25$$
Explanation
If a random variable
X
is adjusted by multiplying by the value
b
and adding the value
a
, then the variance is affected as follows: $$ { \sigma }_{ a+bX }^{ 2 }={ b }^{ 2 }{ \sigma }^{ 2 } $$
If the variance of the random variable X = $$ { \sigma }^{ 2 }$$ =
4
For the random variable $$5X+10$$,
$$ { \sigma }_{ 10+5X }^{ 2 } = {b }^{ 2 }{ \sigma }^{ 2 } = { 5 }^{ 2 }{4} = 100 $$
$$ { \sigma }_{ 10+5X }^{ 2 } = 100 $$
From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random without replacement. The expected number of good items is
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$$3$$
0%
$$2.8$$
0%
$$1.2$$
0%
$$1.8$$
Explanation
Probability of good items = $$7/10$$
Total number of items drawn =$$ 4$$
Expected number of good items = $$4*7/10 = 2.8$$
If the probability distribution of a random variable x is
$$X=x_{1}:\quad -2 \quad -1 \quad 0 \quad 1 \quad 2 \quad 3$$
$$p(X=x_{1}): 0.1 \ \ \ \ \ \ k \ \ \ \ 0.2 \ \ \ 2k \ \ 0.3 \ \ \ k$$ then the mean of x is
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0%
$$0.6$$
0%
$$0.8$$
0%
$$1.0$$
0%
$$0.3$$
Explanation
$$\sum _{ i=-2 }^{ 3 }{ { p }_{ i } } =1\\ 0.1+k+0.2+2k+0.3+k=1\\ 4k=1-0.6=0.4\\ k=0.1\\ Mean=\sum _{ i=-2 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =-0.2-k+2k+0.6+3k\\ =4k+0.4=0.4+0.4\\ =0.8\\ $$
The Probability distribution of a random variable $$X$$ is given by $$P(X=x) = 0.1 ,0.1, 0.1 ,0.3 ,0.4$$ for $$(X=x) = 4,3,2,1,0$$. The variance of $$X$$ is
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0%
$$1. 76$$
0%
$$2.45$$
0%
$$3.2$$
0%
$$4.8$$
Explanation
Mean = $$\sum x_{i}p_{i}=1.24$$
variance = $$\sum (x_{i}-\bar{x})^{2}p_{i}$$
$$=1.44*0.4+0.2^{2}*0.3+0.8^{2}*0.1+1.8^{2}*0.1+2.8^{2}*0.1$$
$$=1.76$$
If a random variable $$x$$ has the following probability distribution
$$X=x_{i} : \quad \quad0 \ \quad\quad 1 \ \quad\quad 2 \ \quad 3$$
$$P(X=x_{i}): \ 2K^{2},3K^{2},5K^{2},6K^{2}$$
then the value of $$K$$ is
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0%
$$\displaystyle \frac{1}{4}$$
0%
$$\displaystyle \frac{-1}{4}$$
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$$\displaystyle \pm\frac{1}{4}$$
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$$\displaystyle \frac{1}{2}$$
Explanation
Using the fact , $$\sum P(X) = 1$$
$$\Rightarrow 2K^2+3K^2+5K^2+6K^2=1$$
$$\Rightarrow K^2 = \cfrac{1}{16} \Rightarrow K = \pm \cfrac{1}{4}$$
The value of $$C$$ for which $$P( X=k)= Ck^{2}$$ can serve the probability function of a random variable $$X$$ that takes values $$0, 1, 2, 3, 4$$ is
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$$\displaystyle \frac{1}{30}$$
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$$\displaystyle \frac{1}{10}$$
0%
$$\displaystyle \frac{1}{3}$$
0%
$$\displaystyle \frac{1}{15}$$
Explanation
$$\displaystyle \sum _{ k=0 }^{ 4 }{ P\left( X=k \right) } =1\Rightarrow \sum _{ k=0 }^{ 4 }{ c{ k }^{ 2 } } =1$$
$$\displaystyle \Rightarrow c\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } \right) =1\Rightarrow c=\frac { 1 }{ 30 } $$
If the probability distribution of a random variable $$x$$ is
$$X=x_{1}:\quad -2 \quad -1 \quad 0 \quad 1 \quad 2 \quad 3$$
$$p(X=x_{1}): 0.1 \ \ \ \ \ \ k \ \ \ \ 0.2 \ \ \ 2k \ \ 0.3 \ \ \ k$$, then the variance of $$x$$ is
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$$2.16$$
0%
$$2.8$$
0%
$$\sqrt{2.16} $$
0%
$$
\sqrt{2.8}$$
Explanation
$$\sum _{ i=-2 }^{ 3 }{ { p }_{ i } } =1\\ 0.1+k+0.2+2k+0.3+k=1\\ 4k=1-0.6=0.4\\ k=0.1\\ E[X]=\sum _{ i=-2 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =-0.2-k+2k+0.6+3k\\ =4k+0.4=0.4+0.4\\ =0.8\\ E[{ X }^{ 2 }]=\sum _{ i=-2 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =0.4+k+2k+1.2+9k=2.6\\ Variance = E[{ X }^{ 2 }]-{ E[X] }^{ 2 }=2.6-{ 0.8 }^{ 2 }= 2.16\\ $$
If a random variable $$x$$ has the following probability distribution
$$X=x_{i} : \quad \quad0 \ \quad\quad 1 \ \quad\quad 2 \ \ \quad 3$$
$$P(X=x_{i}): \ 2K^{2}\ \ 3K^{2}\ \ \ 5K^{2}\ \ 6K^{2}$$
Then find the mean.
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$$\displaystyle \frac{33}{16}$$
0%
$$\displaystyle \frac{31}{16}$$
0%
$$\displaystyle \frac{35}{16}$$
0%
$$\displaystyle \frac{29}{16}$$
Explanation
Using $$ \sum P(X) = 1\Rightarrow 16 K^2 = 1\Rightarrow K^2 = \cfrac{1}{16}$$
Thus mean $$ = \sum x_iP(X=x_i) = 0+P(X=1)+2P(X=2)+3P(X=3)$$
$$0+3K^2+10K^2+18K^2$$
$$=31K^2 = \cfrac{31}{16}$$
The probability distribution of a random variable $$X$$ is given below, then $$K =$$
$$X=x_{1}: \quad 1 \quad , 2, \quad 3, \quad 4$$
$$p(X=x_{1}):2k , 4k \ \ ,3k , \ \ \ k$$
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0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{1}{5}$$
0%
$$\dfrac{1}{10}$$
0%
$$\dfrac{1}{15}$$
Explanation
$$\displaystyle\sum_{1}^{4}P(X=x)=1$$
$$K(2+4+3+1)=1$$
$$K=\dfrac{1}{10}$$
A random variable $$X$$ takes the values $$-1, 0, +1$$. Its mean is $$0.6$$. If $$P(X=0)=0.2$$, then find $$P(X=1)$$ and $$P(X=-1)$$
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$$0.2, 0.8$$
0%
$$0.3, 0.7$$
0%
$$0.7, 0.1$$
0%
$$0.4, 0.2$$
Explanation
We know $$\sum P(X)=1\Rightarrow P(X=-1)+P(X=0)+P(X=1)=1\Rightarrow P(X=-1)+P(X=1)=0.8$$ -----(i)
Given mean is $$10.6$$ $$\Rightarrow (-1)(P(X=-1))+(0)(P(X=0))+1(P(X=1))=0.6$$
$$\Rightarrow -P(X=-1)+P(X=1)=0.6$$ ------(ii)
Adding (i) and (ii)
$$2P(x=1)=1.4$$
$$P(x=1)=0.7$$
$$P(x=-1)=0.8-P(x=1)=0.8-0.7=0.1$$
$$\therefore\ P(X=-1) =0.1, P(X=1)=0.7 $$
The value of k, if the probability distribution of a $$X=x:1 , 2 , 3$$ random variable X is$$p(X=x):\dfrac{1}{k} , \dfrac{2}{k} , \dfrac{3}{k}$$ is
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0%
$$\dfrac{1}{6}$$
0%
$$6$$
0%
$$\dfrac{1}{3}$$
0%
$$\dfrac{1}{\sqrt{6}}$$
Explanation
concept : {
sum of probability distribution against random variable always is equal to 1 }
$$\sum_{i=1}^{n}P(x_i)=1$$
$$\frac{1}{k}+\frac{2}{k}+\frac{3}{k}=1$$
$$\frac{6}{k}=1$$
$$k=6$$
value of k is 6.
A random variable $$X$$ has the following probability distribution, then $$C =$$
$$X=x: 1 ,2 , 3 , 4$$, $$P(X=x) :C , 2C, 3C ,4C$$
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$$0.1$$
0%
$$0.2$$
0%
$$10$$
0%
$$20$$
Explanation
$$\displaystyle\sum_{1}^{4}P(X=x)=1$$
$$C(2+4+3+1)=1$$
$$C=0.1$$
A random variable $$X$$ takes the values $$0,\ 1,\ 2,\ 3$$ and its mean is$$1.3$$. If $$P(X=3)=2P(X=1)$$ and $$P(X=2)=0.3$$ , then $$P(X=0)=$$
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0%
$$0.1$$
0%
$$0.2$$
0%
$$0.3$$
0%
$$0.4$$
Explanation
$$E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0*P(X=0)+1*P(X=1)+2*P(X=2)+3*P(X=3)=1.3\\ 1*P(X=1)+2*0.3+3*2*P(X=1)=1.3\\ P(X=1)=0.1\\ Now\quad using\\ \sum _{ i=0 }^{ 3 }{ { p }_{ i } } =1=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ P(X=0)+0.1+0.3+0.2=1\\ P(X=0)=0.4$$
The variance of the random variable $$x$$ whose probability distribution is given by
$$X=x : \quad \quad 0 \quad 1 \quad 2 \quad 3 $$
$$p(X=x): \ \ \ \dfrac{1}{3} \ \ \ \dfrac{1}{2} \ \ \ 0 \ \ \ \dfrac{1}{6}$$
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0%
$$0.5$$
0%
$$1$$
0%
$$1.5$$
0%
$$2.0$$
Explanation
$$E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0+\dfrac { 1 }{ 2 } +0+\dfrac { 1 }{ 2 } =1\\ E[{ X }^{ 2 }]=\sum _{ i=0 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =0+\dfrac { 1 }{ 2 } +0+\dfrac { 9 }{ 6 } =2\\ Variance= E[{ X }^{ 2 }]-{ E[X] }^{ 2 }= 2-{ 1 }^{ 2 }=1$$
If $$X$$ is a random variable with the following probability distribution given below:
$$X=x$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P(X=x)$$
$$k$$
$$3k$$
$$3k$$
$$k$$
Then the value of $$k$$ and its variance are:
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0%
$$\dfrac{1}{8},\dfrac{22}{27}$$
0%
$$\dfrac{1}{8},\dfrac{23}{27}$$
0%
$$\dfrac{1}{8},\dfrac{8}{9}$$
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$$\dfrac{1}{8},\dfrac{3}{4}$$
Explanation
$$\sum _{ i=-2 }^{ 3 }{ { p }_{ i } } =1\\ k+3k+3k+k=1\\ \Rightarrow k=\dfrac { 1 }{ 8 } \\ E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0\times P(X=0)+1\times P(X=1)+2\times P(X=2)+3\times P(X=3)\\ 3k+6k+3k=12k=\dfrac { 12 }{ 8 } \\ E[{ X }^{ 2 }]=\sum _{ i=0 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =3k+12k+9k=24k=3\\ Variance = E[{ X }^{ 2 }]-{ E[X] }^{ 2 } = 3-\left( \dfrac { 3 }{ 2 }\right)^{ 2 } = 3-\dfrac { 9 }{ 4 } =\dfrac { 3 }{ 4 } \\ $$
A random variable X follows the following distribution
$$X=x_{i}: \quad \ \ 1 , \ 2 , \ 3, \ 4 $$
$$p(X=x_{i}):\dfrac{2}{6}, \dfrac{3}{6} , \dfrac{0}{6}, \dfrac{1}{6}$$
, then the mean and variance are
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0%
$$1, 1$$
0%
$$1, 2$$
0%
$$2, 1$$
0%
$$2, 2$$
Explanation
$$E[X]=\sum _{ i=1 }^{ 4 }{ { x }_{ i }{ p }_{ i } } =1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4)\\ =\dfrac { 2 }{ 6 } +\dfrac { 6 }{ 6 } +\dfrac { 0 }{ 6 } +\dfrac { 4 }{ 6 } =2\\ E[{ X }^{ 2 }]=\sum _{ i=0 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =\dfrac { 2 }{ 6 } +\dfrac { 12 }{ 6 } +\dfrac { 0 }{ 6 } +\dfrac { 16 }{ 6 } =5\\ Variance\quad =\quad E[{ X }^{ 2 }]-{ E[X] }^{ 2 }\quad =\quad 5-{ 2 }^{ 2 }\quad =\quad 1$$
A random variable X has its range $${1, 2, 3}$$ with respective probabilities $$P(X=1)=K, \ P(X=2)=2K, P(X=3)=3K,$$ then the value of K is
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0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{1}{5}$$
0%
$$\dfrac{1}{6}$$
0%
$$\dfrac{1}{8}$$
Explanation
$$\displaystyle\sum_{1}^{3}P(X=x)=1$$
$$K(1+2+3)=1$$
$$K=\dfrac{1}{6}$$
Let X be the random variable with the probability distribution function $$f(x)=\dfrac{e^{-4}4^{x}}{x!};x=0,1,2,3,....$$ then the standard deviation of X is
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$$2$$
0%
$$4$$
0%
$$16$$
0%
$$\sqrt{2}$$
Explanation
The above distribution is Poisson Distribution of type
$$P(k)=\dfrac{e^{-\lambda}.\lambda^{k}}{k!}$$
Here $$\lambda=4$$.
Now $$Mean=Variance=\lambda$$ in a Poisson Distribution.
Hence
$$Variance=\lambda=4$$
Or
$$S.D^{2}=Variance=4$$
Or
$$S.D=\sqrt{Variance}=\sqrt{4}$$
$$=2$$.
A random variable $$X$$ has its range $$X = {0, 1, 2}$$ with respective probabilities $$P(X=0)=3K^{3}, P(X=1)=4K-10K^{2}, P(X=2)=5K-1$$ , then the value of $$K$$ is
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0%
$$2$$
0%
$$1$$
0%
$$\dfrac{1}{3}$$
0%
$$2,1,\dfrac{1}{3}$$
Explanation
Given, $$P(X=0)=3K^{3}, P(X=1)=4K-10K^{2}, P(X=2)=5K-1 $$
Now, $$P(X=0)+P(X=1)+P(X=2)=1$$
$$\Rightarrow 3K^{3}-10K^2+9K-2=0$$
$$\Rightarrow (K-1)(3K^2-7K+2)=0$$
$$\Rightarrow (K-1)(K-2)(3K-1)=0$$
$$\Rightarrow K=1,2, \dfrac{1}{3}$$
But $$K=1,2$$ not possible as at these values of $$K$$, the probabilities $$P(X=0)$$ , $$P(X=1)$$ ,$$P(X=2)$$ does not lie between $$0$$ and $$1$$.
If a random variable X takes values$$(-1)^{k}2^{k}/k;k=1,2,3,....$$with probabilities $$P(X=k)=\dfrac{1}{2^{k}}$$then E(X)=
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0%
$$log \ 2$$
0%
$$log \ e$$
0%
$$log \ \left ( \dfrac{1}{2} \right )$$
0%
$$log \ \left ( \dfrac{1}{4} \right )$$
Explanation
$$E[X]=\sum _{ k=1 }^{ \infty }{ \dfrac { { (-1) }^{ k }{ 2 }^{ k } }{ k } \dfrac { 1 }{ { 2 }^{ k } } } =\quad \dfrac { { (-1) }^{ k } }{ k } =\dfrac { -1 }{ 1 } +\dfrac { 1 }{ 2 } +\dfrac { -1 }{ 3 } +\dfrac { 1 }{ 4 } +........\\ =-1+\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 4 } +.........\\ =\quad -\log { 2= } \log { \dfrac { 1 }{ 2 } } $$
If F(x) is the cumulative distributive function of a random variable x whose range is from $$-\alpha $$ to $$+\alpha $$, then $$P(X< -\alpha )$$
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0%
$$1$$
0%
$$\dfrac{1}{2}$$
0%
$$0$$
0%
$$\dfrac{1}{3}$$
Explanation
We have for a random variable $$X$$,
$$F(x)=P(X\le x)= 0$$ for $$x<0$$ (since the random variable cannot assume negative values, so its probability is 0)
So, $$P(X<-\alpha)=0$$
A random variable $$X$$ takes value $$0, 1, 2$$. Its mean is $$1.3$$. If $$P(X=0)=0.2$$, then $$P(X=2)=$$
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0%
$$0.3$$
0%
$$0.4$$
0%
$$0.5$$
0%
$$0.2$$
Explanation
Given Mean $$= 1.3,P(x=0)=0.2$$
$$ 0\times P(X=0)+ 1\times P(X=1)+ 2\times P(X=2)= 1.3 $$
Let $$ P(X=2)= p $$
$$ P(X=1)= 1- P(X=0)- p= 0.8-p$$
Substituting in original equation
$$ P(X=1)+ 2P(X=2)= 1.3 $$
$$ 0.8-p+2p= 1.3$$
$$ p=0.5$$
Let the discrete random variable $$X = x$$ has the probabilities given by $$\displaystyle \frac{x}{6}$$ for $$x=0, 1, 2, 3$$, then its mean is
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0%
$$\displaystyle \frac{1}{3}$$
0%
$$\displaystyle \frac{5}{3}$$
0%
$$\displaystyle\frac{7}{3}$$
0%
$$\displaystyle \frac{9}{3}$$
Explanation
$$E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0*P(X=0)+1*P(X=1)+2*P(X=2)+3*P(X=3)\\ =0+\dfrac { 1 }{ 6 } +\dfrac { 2*2 }{ 6 } +\dfrac { 3*3 }{ 6 } =\dfrac { 14 }{ 6 } $$
If the range of the random variable $$X$$ is from $$-\alpha$$ to $$+\alpha$$ , the limits of $$F(X)$$ are
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0%
$$0$$ to $$\alpha$$
0%
$$-\alpha $$ to $$ 3 $$
0%
$$-1 $$ to $$+1$$
0%
$$0 $$ to $$ 1$$
Explanation
We have for a random variable $$X$$,
$$F(x)=P(X\le x)= 0$$ for $$x<0$$ (since the random variable cannot assume negative values, so its probability is 0)
$$F(x)=P(X\le x)$$ for $$x \ge 0$$
Since, the value of probability lies between 0 and 1.
So, $$0 < F(x) \le 1$$
Hence, limits of $$F(x)$$ are 0 to 1.
The range of a random variable $$X$$ is $$1, 2, 3, 4 ....$$
and the probabilities are given by $$\displaystyle p(X=k)=\frac{c^{k}}{k!}k=1,2,3,4....,$$, then the value of $$C$$ is
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0%
$$2 2$$
0%
$$\log e$$
0%
$$\log_e 2$$
0%
4
Explanation
We know, $$\sum P(X)=1$$
$$\Rightarrow \displaystyle c+\frac{c^2}{2!}+\frac{c^3}{3!}+........... = 1$$
$$\Rightarrow \displaystyle 1+c+\frac{c^2}{2!}+\frac{c^3}{3!}+........... =1+1=2$$
$$\displaystyle \Rightarrow e^c = 2\Rightarrow c = \log_e 2$$
A person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $$X$$ is observed, then the range of $$X$$ is
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0%
$${0,3,6}$$
0%
$${-3,0,3}$$
0%
$${-3,0,3,6}$$
0%
$${-3,3,6}$$
Explanation
$$3$$ coins are tossed . There can be $$4$$ possible outcomes
$$1.\ 3 T , 0 H$$. Score $$= -3 $$
$$2.\ 2 T , 1 H$$. Score $$ = -2 + 2 = 0 $$
$$3.\ 1 T , 2 H$$. Score $$ = -1 +4 = 3 $$
$$4.\ 0 T, 3 H$$. Score $$ = 0 + 6 = 6 $$
Hence, the range of $$X$$ is $$\{ -3, 0,3,6 \}$$
A discrete random variable $$X$$, can take all possible integer values from $$1$$ to $$K$$, each with a probability $$1/K$$. Its mean is
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0%
$$K$$
0%
$$K+1$$
0%
$$K/2$$
0%
$$K/4$$
Explanation
Mean $$=\mu=\sum_{{i=1}}^{K}x_ip_i$$
$$ =1\times 1+2 \times \dfrac{1}{2}+3 \times \dfrac{1}{3}+.....K \times \dfrac{1}{K}$$
$$=1+1+....upto\ K\ times $$
Mean $$\mu=K$$
In a World Cup final match against Srilanka, for six times Sachin Tendulkar hits a six out of 30 balls he plays. What is the probability that in a given throw, the ball does not hit a six?
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0%
$$\frac{1}{4}$$
0%
$$\frac{5}{4}$$
0%
$$\frac{4}{5}$$
0%
$$\frac{3}{4}$$
The probability distribution of random variable X: number of heads , when a fair coin is tossed twice is given by
$$x$$
$$0$$
$$1$$
$$2$$
p(x)
$$p_1$$
$$p_2$$
$$p_3$$
then
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0%
$$\sum p_i=0$$
0%
$$\Pi p_i=1$$
0%
$$\sum p_i=1$$
0%
$$\sum p_i=3$$
Explanation
We know that, the probability distribution of a random variable is a list of probabilities associated with each of its possible values.
Suppose a random variable $$X$$ may take $$k$$ different values, with the probability that $$X=x_i$$ defined to be$$P(X=x_i)=p_i$$.
The probabilities $$p_i$$ must satisfy the following:
$$0\leq p_i \leq 1$$ for each $$i$$
$$p_1+p_2+...+p_k=1. \:That \:is\: \sum p_i=1$$
Thus if the probability distribution of random variable $$X$$: number of heads , when a fair coin is tossed twice is given then $$\sum p_i=1$$
A car hire firm has $$2$$ cars which it hires out day by day. If the number of demands for a car on each day follows poisson distribution with parameter $$1.5$$, then the probability that some demand is refused is
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0%
$$1.12 \times e^{-1.5}$$
0%
$$1.2.5 \times e^{-1.5}$$
0%
$$1-3.625 \times e^{-1.5}$$
0%
$$3.625 \times e^{-1.5}$$
Explanation
let $$X$$ be a random variable denoting the number of demands for a car on any day.
So, $$X$$ is Poisson distributed with the parameter $$\mu =1.5$$
Therefore, the probability mass function is
$$\displaystyle P(X=i)=\frac { { e }^{ \mu }{ \mu }^{ t } }{ i! } ={ e }^{ -1.5 }\frac { { \left( 1.5 \right) }^{ 2 } }{ i! } ;i=0,1,2$$
Now, the proportion of days in which neither car is used is actually the probability of there being no demand of cars which is given by
$$\displaystyle P(X=0)=\frac { { e }^{ -1.5 }{ \left( 1.5 \right) }^{ 2 } }{ 0! } ={ e }^{ -1.5 }$$
and the proportion of days on which some demand is refused, is the probability that the number of demands become more than $$2$$ and is given by
$$P(X>2)=1-P(X\le 2)$$
$$=1-{P(X=0)+P(X=1)+P(X=2)}$$
$$\displaystyle=1-{\frac { { e }^{ 1.5 }{ \left( 1.5 \right) }^{ 0 } }{ 0! } +\frac { { e }^{ -1.5 }{ \left( 1.5 \right) }^{ 1 } }{ 1! } +\frac { { e }^{ -1.5 }{ \left( 1.5 \right) }^{ 2 } }{ 2! } }$$
$$=1-3.625\times { e }^{ -1.5 }$$
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