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CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 6 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Probability Distribution And Its Mean And Variance
Quiz 6
The probability distribution of a random variable $$X$$ is given below
$$X=x$$
$$-1.5$$
$$-0.5$$
$$0.5$$
$$1.5$$
$$2.5$$
$$P[X=X]$$
$$0.05$$
$$0.2$$
$$0.15$$
$$0.25$$
$$0.35$$
The variance of $$X$$ is
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$$1.6$$
0%
$$0.24$$
0%
$$0.84$$
0%
$$0.75$$
Explanation
Now, $$Var(X) = (x-\mu)^2 \ P(X-x)$$
Now, we have
$$\mu =0\times 1+0.4 \times 1+0.3 \times 2+3\times 0.2 + 4 \times 0=1.6$$
Thus, we have $$Var(X) = (0-1.6)^2 \times 0.2 + (1-1.6)^2 \times 0.4 + (2-1.6)^2 \times 0.3 + (3-1.6)^2 \times 0.2 + (4-1.6)^2 \times 0$$
Thus, $$Var(X) = 0.256 + 0.144+ 0.048 + 0.392$$
or, $$\Rightarrow Var(X) = 0.840$$
The probability function of a binomial distribution is $$P(x) = \binom{6}{x} p^{x} q^{6 - x}, x = 0, 1, 2, ..., 6$$. If $$2P(2) = 3P(3)$$, then $$p =$$ __________.
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$$\dfrac {1}{3}$$
0%
$$\dfrac {1}{4}$$
0%
$$\dfrac {1}{2}$$
0%
$$\dfrac {1}{5}$$
Explanation
Given
$$P(x) = \binom{6}{x} p^{x} q^{6 - x}$$ for $$x = 0, 1, 2, ..., 6$$
Also, $$2P(2) = 3P(3)$$
$$2\times {}^{6}C_{2}p^{2}q^{4} = 3\times {}^{6}C_{3}p^{3}q^{3}$$
$$\Rightarrow 2\times \dfrac{6!}{2!4!}p^{2}q^{4}=3\times \dfrac{6!}{3!3!}p^{3}q^{3}$$
$$\Rightarrow q=2p$$
(as $$p + q = 1)$$.
$$\Rightarrow p = \dfrac {1}{3}$$
The probability that an event A occurs in a single trial of an experiment is $$0.3$$. Six independent trials of the experiment are performed. What is the variance of probability distribution of occurrence of event A?
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$$12.6$$
0%
$$0.18$$
0%
$$1.26$$
0%
$$1.8$$
Explanation
X
0
1
2
3
4
5
6
P(X)
$$^6C_{0}(0.7^6)(0.3^0)$$
$$^6C_{1}(0.7^5)(0.3^1)$$
$$^6C_{2}(0.7^4)(0.3^2)$$
$$^6C_{3}(0.7^3)(0.3^3)$$
$$^6C_{4}(0.7^2)(0.3^4)$$
$$^6C_{5}(0.7^1)(0.3^5)$$
$$^6C_{6}(0.7^0)(0.3^6)$$
$$ \mu =\sum X_{i}P(X_{i})$$
$$ = \sum_{0}^{6}i \times ^6C_{i}(0.7^{6-i})(0.3^i)$$
$$ = 1.8$$
$$variance =\sum X_{i}^2P(X_{i})-\mu^2$$
$$ = 4.5 -1.8^2$$
$$ = 1.26$$
Hence the answer is option (C).
What is the mean of $$f(x)=3x+2$$ where x is a random variable with probability distribution.
$$X=x$$
$$1$$
$$2$$
$$3$$
$$4$$
$$P(X=x)$$
$$1_{/6}$$
$$1_{/3}$$
$$1_{/3}$$
$$1_{/6}$$
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$$\displaystyle\frac{19}{2}$$
0%
$$\displaystyle\frac{5}{2}$$
0%
$$\displaystyle\frac{15}{2}$$
0%
$$\displaystyle\frac{5}{3}$$
Explanation
Given, $$f(x) = 3x+2$$
applying expectation on both sides, we get
$$E(f(x))=3E(x)+2$$ (by the properties of expectation)
mean $$E(x)= \sum_{i=1}^{n} x_i P(x_i) $$
Given, number of samples $$n=4$$
Hence, mean = $$\sum_{i=1}^4x_iP(x_i)$$
substituting the values from the table we get,
$$=1\left(\dfrac{1}{6}\right)+2\left(\dfrac{1}{3}\right)+3\left(\dfrac{1}{3}\right)+4\left(\dfrac{1}{6}\right)$$
$$=\dfrac{1+4+6+4}{6} = \dfrac{15}{6}$$
$$E(x)=\dfrac{5}{2}$$
Therefore, $$E(f(x))=3\left(\dfrac{5}{2}\right)+2$$
$$=\dfrac{19}{2}$$
A random variable $$X$$ has the probability distribution,
$$X = x$$
$$-2$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P (x)$$
$$\dfrac{1}{10}$$
$$K$$
$$\dfrac{1}{5}$$
$$2K$$
$$\dfrac{3}{10}$$
$$K$$
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$$\dfrac{2}{10}$$
0%
$$\dfrac{1}{10}$$
0%
$$\dfrac{3}{10}$$
0%
$$\dfrac{7}{10}$$
Explanation
The sum of probability of all the possible outcomes should be $$1$$.
That is $$\sum P(x_i)=1$$
$$\implies \dfrac{1}{10}+K+\dfrac{1}{5}+2K+\dfrac{3}{10}+K=1$$
$$\implies 4K+\dfrac{6}{10}=1$$
$$\implies 4K=1-\dfrac{6}{10}=\dfrac{4}{10}$$
$$\implies K=\dfrac{1}{10}$$
An experiment is known to be random if the results of the experiment -
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Cannot be predicted
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Can be predicted
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Can be split into further experiments
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Can be selected at random
Let $$a={ i }^{ { k }_{ 1 } }+{ i }^{ { k }_{ 2 } }+{ i }^{ { k }_{ 3 } }+{ i }^{ { k }_{ 4 } },(i=\sqrt { -1 } )$$ where each $${k}_{n}$$ is randomly chosen from the set $${1,2,3,4}$$. The probability that $$a=0$$, is
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0%
$$\frac { 7 }{ 64 } $$
0%
$$\frac { 9 }{ 64 } $$
0%
$$\frac { 37 }{ 256 } $$
0%
$$\frac { 39 }{ 256 } $$
Two cards are drawn successively with replacement from a well shuffled deck of $$52$$ cards. Find the probability distribution of the number of aces.
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$$\dfrac{1}{169}$$
0%
$$\dfrac{1}{221}$$
0%
$$\dfrac{1}{265}$$
0%
$$\dfrac{4}{663}$$
Explanation
Total no. of cards = 52
Total no. of aces = 4
$$P(ace) = \cfrac{4}{52} = \cfrac{1}{13}$$
$$P(non-ace) = \cfrac{48}{52} = \cfrac{12}{13}$$
Let X be a random variable, such that
X = no. of aces obtained in 2 draws.
Case I:- X = 0 (No aces are obtained)
$$P(X = 0) = P(\text{non-ace and non-ace})$$
$$= P(non-ace) \times P(non-ace)$$
$$= \cfrac{12}{13} \times \cfrac{12}{13}$$
$$= \cfrac{144}{169}$$
Case II:- X = 1 (1 ace is obtained)
$$P(X = 1) = P(\text{(ace and non-ace) or (non-ace and ace)})$$
$$= \left( P(non-ace) \times P(non-ace) \right) + \left( P(ace) \times P(non-ace) \right)$$
$$= \left( \cfrac{12}{13} \times \cfrac{1}{13} \right) + \left( \cfrac{1}{13} \times \cfrac{12}{13} \right)$$
$$= \cfrac{24}{169}$$
Case III:- X = 2 (two aces are obtained)
$$P(X = 2) = P(\text{ace and ace})$$
$$= P(ace) \times P(ace)$$
$$= \cfrac{1}{13} \times \cfrac{1}{13}$$
$$= \cfrac{1}{169}$$
A box contains $$10$$ items, $$3$$ of which are defective. If $$4$$ are selected at random without replacement, the probability that at least $$2$$ are defective is?
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$$50\%$$
0%
$$33.33\%$$
0%
$$67\%$$
0%
$$100\%$$
Given two independent events A and B such that $$P\left( A \right) = 0.3,$$ $$P\left( A\cap B \right) = 0.1$$. Find
$$P\left( {A \cap B'} \right)$$
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$$0.1$$
0%
$$0.2$$
0%
$$0.15$$
0%
$$0.3$$
Explanation
Given $$P\left(A\right)=0.3,\,\,P\left(A\cap\,B\right)=0.1$$
If $$A$$ and $$B$$ are independent events, $$P\left(A\cap\,B\right)=P\left(A\right)P\left(B\right)$$
$$\Rightarrow\,P\left(B\right)=\dfrac{P\left(A\cap\,B\right)}{P\left(A\right)}=\dfrac{0.1}{0.3}=\dfrac{1}{3}$$
If $$A$$ and $$B$$ are independent events,
$$P\left(A\cap{B}^{\prime}\right)=P\left(A\right)\times P\left({B}^{\prime}\right)$$
$$=P\left(A\right)\left(1-P\left(B\right)\right)$$
$$=0.3\left(1-\dfrac{1}{3}\right)$$
$$=0.3\times\dfrac{3-1}{3}=0.3\times\dfrac{2}{3}=0.1\times 2=0.2$$
If the variance of a random variable X is $$\sigma^2$$, then the variance of the random variable X-$$5$$ is?
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$$5\sigma^2$$
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$$25\sigma^2$$
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$$\sigma^2$$
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$$2\sigma^2$$
Explanation
$${\textbf{Step -1: Solve using the variables given}}{\textbf{.}}$$
$${\text{If variance of }}X = k,{\text{ then variance of }}ax + b = {a^2}k.$$
$${\text{Here }}a = 1,b = - 5$$
$$\therefore {\text{ variance of }}x - 5 = {\sigma ^2}$$
$${\textbf{Hence, the value of variance of the random variable }}\mathbf{x - 5}{\textbf{ is }}\mathbf{{\sigma ^2}.}$$
Consider the word $$W = MISSISSIPPI$$
Number of ways in which the letters of the word W can be arranged if at least one vowel is separated from rest of the vowels
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$$\dfrac { 8\ !.16\ ! }{ 4\ !.4\ !.2\ ! } $$
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$$\dfrac { 8\ !.16\ ! }{ 4.4\ !.2\ ! } $$
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$$\dfrac { 8\ !.16\ !}{ 4\ !.2\ ! } $$
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$$\dfrac { 8\ ! }{ 4\ !.2\ ! } .\dfrac { 165 }{ 4\ ! } $$
Explanation
W = MISSISSIPPI , Letters= 11
1 M. 4I's, 4S's, 2P's
Total number of arranging these letters
$$=\quad \cfrac { 11! }{ 4!4!2! } $$
Suppose all I's are together , then the Number of ways
$$=\quad \cfrac { \left( 1+1+4+2 \right) ! }{ 4!2! } \quad =\quad \cfrac { 8! }{ 4!2! } $$
$$\therefore$$ Number of ways in which atleast one vowel is separated
$$=\quad \cfrac { 11! }{ 4!4!2! } \quad -\quad \cfrac { 8! }{ 4!2! } \\ =\quad \cfrac { 8!16! }{ 4!4!2! } $$
A biased die is tossed and the respective probabilities for various faces to turn up are given below:
Face:
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$
Probability
$$0.1$$
$$0.24$$
$$0.19$$
$$0.18$$
$$0.15$$
$$0.14$$
If an even face has turned up, then probability that it is face $$2$$ or face $$4$$, is
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0%
$$0.25$$
0%
$$0.42$$
0%
$$0.75$$
0%
$$0.9$$
Explanation
Required probability = $$\dfrac{P(A \cap B)}{P(B)} = P$$
$$B =$$ even face
$$A =$$ (face $$= 2$$ or $$4$$)
$$A \cap B =$$ (face $$= 2$$ or $$4$$)
$$\therefore P = \dfrac{P (2 \, or \, 4)}{P (2 \, or \, 4 \, or \, 6)} = \dfrac{0.24 + 0.18}{0.24 + 0.18 + 0.14} = 0.75$$
A special die with number $$1$$, $$-1$$, $$2$$, $$-2$$, $$0$$ and $$3$$ is thrown thrice. The probability that total is $$6$$ is
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$$\dfrac{1}{{108}}$$
0%
$$\dfrac{{25}}{{216}}$$
0%
$$\dfrac{5}{{216}}$$
0%
$$\dfrac{5}{{108}}$$
Explanation
outcomes are 1,-1.2,-2,0,3
For Sum to be 6 -
2,2,2 1 case
3,3,0 3!/2= 3 cases
1,2,3 3!=6 cases
Total favourable outcomes=1+3+6=10
Total outcomes =6$$\times$$6
$$\times$$6=216
Probability=10/216=5/108
A discrete random variable takes
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only a finite number of values
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all possible values between certain given limits
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infinite number of values
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a finite or countable number of values
A random variable $$X$$ has the following probability distribution. Find the value of $$10k$$.
$$X$$
$$-2$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P(X)$$
$$0.1$$
$$k$$
$$0.2$$
$$2k$$
$$0.3$$
$$k$$
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1
0%
2
0%
3
0%
4
Explanation
As we know that $$\displaystyle\sum P(X)=1$$
$$\implies 0.1+k+0.2+2{k}+0.3+k=1\implies k=0.1\implies 10{k}=1$$
A random variable $$X$$ has the probability distribution:
$$X$$:
1
2
3
4
5
6
7
8
$$p(X)$$:
0.2
0.2
0.1
0.1
0.2
0.1
0.1
0.1
For the events $$E=\left\{ X\quad is\quad a\quad prime\quad number \right\} $$ and $$F=\left\{ X<4 \right\} $$, the $$P(E\cup F)$$ is
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0%
$$0.50$$
0%
$$0.77$$
0%
$$0.35$$
0%
$$0.80$$
Explanation
given I = $$\{$$'x' is a prime number $$\}$$
P(E) = p(x=2 or x = 3 or x =5 or x =7)
$$=P(x=2)+p(x=3)+p(x=5)+p(x=7)$$
$$ = 0.2+0.1 +0.2+0.1$$
$$ p(B) = 0.6$$
$$ p(F) = P(x = 1)+p(x=2)+p(x =3)$$
$$ = 0.2+0.2+0.1$$
$$ = 0.5$$
$$ p(EAF) = p(x=2)+p(x=3)$$
$$ = 0.2+0.1$$
$$ 0.3 $$
$$p(E\vee F)=p(E)+p(F)-p(E\wedge F)$$
$$ = 0.6+0.5 = 0.3$$
$$ P(E\vee F) = 0.80 $$
If $$P ( A ) = 0.8,$$ $$P ( B ) = 0.5$$ and
$$P \left( \dfrac { B } { A } \right) = 0.4$$ then $$P \left( \dfrac { A } { B } \right) =?$$
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$$0.32$$
0%
$$0.64$$
0%
$$0.16$$
0%
$$0.25$$
Explanation
$$P(A)=0.8 ; P(B)=0.5$$
$$P(\dfrac{B}{A})=0.4$$
$$P(A\cap B)=P(A)P(\dfrac{B}{A})$$
$$P(A\cap B)=P(B)P(\frac{A}{B})$$
$$\dfrac{P(A\cap B)}{P(A\cap B)}=\dfrac{P(A)\times P(B/A)}{P(B)\times P(A/B)}$$
$$\Rightarrow 1=\dfrac{0.8\times 0.4}{0.5\times P(A/B)}$$
$$\Rightarrow 0.5\times P(A/B)=0.32$$
$$P(A/B)=\dfrac{0.32}{0.5}\Rightarrow 0.64$$
Two cards are drawn successively with replacement from a well-shuffled deck of $$52$$ cards. Let $$X$$ denote the random variable of number of aces obtained in the two drawn cards. Then $$P(X = 1) + P(X = 2)$$ equals
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0%
$$\dfrac {52}{169}$$
0%
$$\dfrac {25}{169}$$
0%
$$\dfrac {49}{169}$$
0%
$$\dfrac {24}{169}$$
Explanation
Two cards are drawn successively with replacement
$$4\ Aces\ 48$$ Non Aces
$$P(x = 1) = \dfrac {^{4}C_{1}}{^{52}C_{1}}\times \dfrac {48C_{1}}{52C_{1}} + \dfrac {48C_{1}}{52C_{1}} \times \dfrac {4C_{1}}{52C_{1}} = \dfrac {24}{169}$$
$$P(x = 2) = \dfrac {^{4}C_{1}}{^{52}C_{1}}\times \dfrac {^{4}C_{1}}{^{52}C_{1}} = \dfrac {1}{169}$$
$$P(x = 1) + P(x = 2) = \dfrac {25}{169}$$.
In the following, find the value of k and find mean and variance of X:
$$X=x$$
$$-2$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P(X=x)$$
$$0.1$$
k
$$0.2$$
$$2k$$
$$0.3$$
k
Find the value of:
1) k
2) E(X)
3) V(X)
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$$0.1 , 0.8, 2.16$$
0%
$$0.8 , 2.16, 2.16$$
0%
$$0.1 , 2, 3$$
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$$2.16, 0.8, 0.1$$
The outcome of each of $$30$$ items was observed; $$10$$ items gave an outcome $$\dfrac{1}{2}$$- d each, $$10$$ items gave outcome $$\dfrac{1}{2}$$ each and the remaining $$10$$ items gave outcome $$\dfrac{1}{2}+$$ d each. If the variance of this outcome data is $$\dfrac{4}{3}$$ then $$|d|$$ equals:-
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0%
$$2$$
0%
$$\dfrac{\sqrt5}{2}$$
0%
$$\dfrac{2}{3}$$
0%
$$\sqrt2$$
Explanation
Variance is independent of origin. So we shift the given data by $$\dfrac{1}{2}$$.
so, $$\dfrac{10d^2+10\times0^2+10d^2}{30}-(0)^2=\dfrac{4}{3}$$
$$\Rightarrow d^2=2\Rightarrow|d|=\sqrt2$$
A perfect die is thrown twice. The expected value of the product of the number of point obtained in two thrown is
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0%
$$7/2$$
0%
$$7$$
0%
$$49/2$$
0%
$$none of these$$
A random variable $$X$$ has the following probability mass function:
$$X$$
$$-2$$
$$3$$
$$1$$
$$P(X = x)$$
$$\dfrac{\lambda}{6}$$
$$\dfrac{\lambda}{4}$$
$$\dfrac{\lambda}{12}$$
Then the value of $$\lambda$$ is:
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$$3$$
0%
$$1$$
0%
$$4$$
0%
$$2$$
Explanation
We know that sum of Probability mass function$$=1$$
$$\sum{P\left(X=x\right)}=P\left(-2\right)+P\left(3\right)+P\left(1\right)$$
$$\Rightarrow 1=\dfrac{\lambda}{6}+\dfrac{\lambda}{4}+\dfrac{\lambda}{12}$$
$$\Rightarrow 1=\dfrac{2\lambda}{6}+\dfrac{3\lambda}{12}+\dfrac{\lambda}{12}$$
$$\Rightarrow \dfrac{2\lambda+3\lambda+\lambda}{12}=1$$
$$\Rightarrow 6\lambda=12$$
$$\therefore \lambda=\dfrac{12}{6}=2$$
For the probability distribution given by $$\left.\begin{matrix} X=x_i & 0 \\ P. & \dfrac{25}{36}\end{matrix}\right|$$ $$\begin{matrix} 1 \\ 5 \\ 18\end{matrix}$$ $$\begin{vmatrix} 2 \\ 1 \\ 36\end{vmatrix}$$ the standard deviation $$(\sigma)$$ is?
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0%
$$\sqrt{\dfrac{1}{3}}$$
0%
$$\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$$
0%
$$\sqrt{\dfrac{5}{36}}$$
0%
None of the above
Explanation
$$M^2+\sigma^2=\sum x^2_iP(x=x_i)$$
$$M=\sum x_iP(x=x_i)=\dfrac{5}{18}+\dfrac{2}{36}=\dfrac{6}{18}=\dfrac{1}{3}$$
$$\therefore\dfrac{1}{9}+\sigma^2=\dfrac{5}{18}+\dfrac{4}{36}=\dfrac{7}{18}$$
$$\therefore \sigma^2=\dfrac{7}{18}-\dfrac{2}{18}=\dfrac{5}{18}$$
$$\therefore S.D=\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$$.
A coin is rolled n times. If the probability of getting head at least once is greater than $$90\%$$ then the minimum value of n is?
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0%
$$4$$
0%
$$3$$
0%
$$5$$
0%
$$6$$
Explanation
$$1-\dfrac{1}{2^n} > \dfrac{9}{10}\Rightarrow \dfrac{1}{10} > \dfrac{1}{2^n}\Rightarrow 2^n > 10$$
$$\therefore$$ minimum value of n is $$4$$.
A random variable X has following probability distribution.
$$X=x$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$
$$P(X=x)$$
K
$$3K$$
$$5K$$
$$7K$$
$$8K$$
K
Then $$P(2\leq x < 5)=$$ _______.
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0%
$$\dfrac{3}{5}$$
0%
$$\dfrac{7}{25}$$
0%
$$\dfrac{23}{25}$$
0%
$$\dfrac{24}{25}$$
Explanation
$$since\quad \sum { P(x)=1 } \\ k+3k+5k+7k+8k+k=1\\ k=1/25\\ P(2\le x<5)=P(2)+P(3)+P(4)\\ \qquad \qquad \quad \quad =3k+5k+7k=15k=15/25=3/5$$
A random variable $$X$$ has the following probability distribution:
$$X$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$P(X)$$
$$k^{2}$$
$$2k$$
$$k$$
$$2k$$
$$5k^{2}$$
Then $$P(X>2)$$ is equal to
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0%
$$\dfrac{1}{6}$$
0%
$$\dfrac{7}{12}$$
0%
$$\dfrac{1}{36}$$
0%
$$\dfrac{23}{36}$$
Explanation
$$\displaystyle \sum Pi=1$$
$$\Rightarrow 6k^2+5k=1$$
$$6k^2+5k-1=0$$
$$6k^2+6k-k-1=0$$
$$6k(k+1)-1(k+1)=0$$
$$(k+1)(6k-1)=0$$
$$k=-1,\dfrac{1}{6}$$
$$\therefore\ k=\dfrac 16$$ as k can't be negative
$$p(x>2)=k+2k+5k^2$$
$$=\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{5}{36}$$
$$=\dfrac{23}{36}$$
If the c.d.f.(cumulative distribution function) is given by $$F(x)=\dfrac{x-25}{10}$$, then $$P(27\leq x\leq 33)=$$ ________.
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0%
$$\dfrac{3}{5}$$
0%
$$\dfrac{3}{10}$$
0%
$$\dfrac{1}{5}$$
0%
$$\dfrac{1}{10}$$
Explanation
By definition, $$F(x) = P(X \leq x)$$.
Therefore, $$P(27 \leq X \leq 33) = P(X \leq 33) - P(X \leq 27)$$
$$= F(33) - F(27)$$
$$= \dfrac{33-25}{10} - \dfrac{27 - 25}{10}$$
$$= \dfrac{6}{10} = \dfrac{3}{5}$$.
That is, $$P(27 \leq X \leq 33) = \dfrac{3}{5}$$.
A fair die is tossed repeatedly until a 6 is obtained.Let X denote the number of tosses required.
The probability that X=3 equals
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0%
25/216
0%
25/36
0%
5/36
0%
125.216
For the following probability distribution.
E(X) is equal to:
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0%
0
0%
-1
0%
-2
0%
-1.8
Explanation
$$E(X)=\sum {XP}(X)$$
$$=4\times (0.1)+(-3\times0.2)+(-2\times0.3)+(-1\times0.2)+(0\times0.2)$$
$$=-0.4-0.6-0.6-0.2=-1.8$$
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