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CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Probability Distribution And Its Mean And Variance
Quiz 7
The probability distribution of a discrete random variable X is given below.
The value of K is
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0%
8
0%
16
0%
32
0%
48
Explanation
We know that, $$\sum { p }(x) =1$$
$$\Rightarrow \dfrac{5}{k}+\dfrac{7}{k}+\dfrac{9}{k}+\dfrac{11}{k}=1$$
$$\Rightarrow \dfrac{32}{k}=1$$
$$\therefore K = 32$$
For the following probability distribution
$$E(X^2)$$
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0%
3
0%
5
0%
7
0%
10
Explanation
$$E(X^2)=\sum X^2P(X)=1\cdot\dfrac{1}{10}+4\cdot\dfrac{1}{5}+9\cdot\dfrac{3}{10}+16\cdot\dfrac{2}{5}$$
$$=\dfrac{1}{10}+\dfrac{4}{5}+\dfrac{27}{10}+\dfrac{32}{5}$$
$$\dfrac{1+8+27+64}{10}=10$$
Two cards are drawn randomly from a well-shuffled deck of $$52$$ cards. If $$X$$ denotes the number of aces, then find mean of $$X$$:
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$$\cfrac{5}{13}$$
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$$\cfrac{1}{13}$$
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$$\cfrac{37}{221}$$
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$$\cfrac{2}{13}$$
Explanation
Two cards are drawn randomly from $$52$$ cards, firstly number of total ways that two cards are not an ace.
$$={ _{ }^{ 48 }{ C } }_{ 2 }=\cfrac { 48\times 47 }{ 2 } =1128$$
Number of ways drawing $$2$$ cards out of $$52$$ cards
$$={ _{ }^{ 52 }{ C } }_{ 2 }=\cfrac { 52\times 51 }{ 2 } =1326$$
$$P(X=0)$$. Probably that no ace is not drawn $$=\cfrac { 1128 }{ 1326 } $$
Secondly number of ways that there is an ace and there is no ace out of $$={ _{ }^{ 4 }{ C } }_{ 1 }\times { _{ }^{ 48 }{ C } }_{ 1 }=4\times 48=192$$
$$P(X=1)$$, Probably that there is an ace and there is no ace $$=\cfrac { 192 }{ 1326 } $$
$$P(X=2)$$, Probably of getting two aces $$=\cfrac { 6 }{ 1326 } $$
Probability distribution is:
$$X$$
0
1
2
$$P(X)$$
$$\cfrac { 1128 }{ 1326 } $$
$$\cfrac { 192 }{ 1326 } $$
$$\cfrac { 6 }{ 1326 } $$
$$\sum { { X }_{ i } } =\sum { { p }_{ i }{ x }_{ i } } $$
$$=\cfrac { 1128 }{ 1326 } \times 0+\cfrac { 192 }{ 1326 } \times 1+\cfrac { 6 }{ 1326 } \times 2=\cfrac { 192 }{ 1326 } +\cfrac { 12 }{ 1326 } =\cfrac { 2 }{ 13 } $$
A random variable $$X$$, takes the values $$0,1,2$$ and $$3$$. Mean of $$X$$ is $$P(x=3)=2P(x=1)$$ and $$P(x=2)=0.3$$, then $$P(x=0)$$ is{
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0%
$$0.2$$
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$$0.4$$
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$$0.3$$
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$$0.1$$
Explanation
Let $$P(X=3)=2p(X=1)=p$$
$$P(X=3)=p$$
and $$P(X=1)=\cfrac{p}{2}$$
Also $$P(X=2)=0.3$$ (given)
Let $$P(X=0)=x$$
Probability distribution is:
$$X$$
0
1
2
3
$$P(X)$$
$$x$$
$$\cfrac{p}{2}$$
$$0.3$$
$$p$$
From table $$\sum { { X }_{ i } } =Mean=\cfrac { \sum { { p }_{ i }{ x }_{ i } } }{ \sum { { p }_{ i } } } \Rightarrow 1.3=x\times 0+1\times \cfrac { p }{ 2 } +2\times 0.3+3\times p$$
$$\Rightarrow 1.3=\cfrac { p }{ 2 } +0.6+3p\Rightarrow 2.6=p+6p+1.2\Rightarrow 7p=1.4\Rightarrow p=\cfrac { 1.4 }{ 7 } =0.2$$
Sum of probabilities $$=1$$
$$x+\cfrac { p }{ 2 } +0.3+p=1\Rightarrow x+0.1+0.3+0.2=1\Rightarrow x=1-0.6\Rightarrow x=0.4$$
$$P(X=0)=0.4$$
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, $$E = \{X$$ is a prime number$$\}$$ and $$F = {X < 4}$$, the probability $$\displaystyle P\left ( E\cup F \right )$$ is
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$$0.87$$
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$$0.77$$
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$$0.35$$
0%
$$0.50$$
Explanation
From the given values, prime numbers and the values less than $$ 4 $$ are $$ 1,2,3,5,7 $$
So, $$ P(E \cup F) $$ will be the sum of probabilities of each of the above selected values of $$ X $$
So, $$ P( E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77 $$
Suppose $$X$$ is a random variable which takes values $$0, 1, 2, 3, ...$$ and $$P(X=r)=pq^r$$ where $$0 < p < 1, q=1-p$$ and $$r=0, 1, 2, ...$$, Then
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$$P(X\geq n)=q^n$$
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$$P(X\geq m+n|X\geq m+n|X\geq m)=P(X\geq n)$$
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$$P(X= m+n|X\geq m+n|X\geq m)=P(X= n)$$
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None of these
Explanation
$$\displaystyle P(X\geq n)=\sum_{k=n}^{\infty}P(X=k)=\frac {pq^n}{1-q}=q^n$$
$$P(X\geq m+n|X\geq m)=\dfrac {P(X\geq m+n)}{P(X\geq m)}=\dfrac {q^{m+n}}{q^m}=q^n$$
$$P(X=m+n|X\geq m)=\dfrac {P(X=m+n)}{P(X\geq m)}=pq^n$$.
In a series of 3 independent trials, the probability of exactly 2 success is 12 times as large as the probability of 3 successes. The probability of a success in each trail is
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$$\dfrac{1}{5}$$
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$$\dfrac{2}{5}$$
0%
$$\dfrac{3}{5}$$
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$$\dfrac{4}{5}$$
Explanation
For a binomial experiment consisting of n trials, the probability of exactly
k successes is
$$P(k \text{ successes}) = ^nC_kp^k(1-p)^{n-k}$$
where the probability of success on each trial is p.
According to the question,
$$n=3,, P(\text{2 successes})=12\times P(\text{3 successes})$$
To find:
The probability of a success in each trail, p
$$P(\text{2 successes})=12\times P(\text{3 successes})\\\implies ^3C_2p^2(1-p)^{3-2}=12 \times ^3C_3p^3(1-p)^{3-3}\\\implies 3\times p^2(1-p)^1=12\times 1\times p^3(1-p)^0\\\implies 3p^2-3p^3=12p^3$$
Divide throughout by $$3p^2$$, we get
$$1-p=4p\\\implies 5p=1\\\implies p=\dfrac 15$$
A random variable $$x$$ assumes values which are numbers of the form $$\displaystyle \frac{n}{n+1}$$ and $$\displaystyle \frac{n+1}{n}$$, where
$$n=1, 2, 3,...$$. lf $$P\left(x=\displaystyle \frac{n}{n+1}\right)=P\left(x=\dfrac{n+1}{n}\right)=\left(\dfrac{1}{2} \right)^{n+1}$$, then
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$$\mathrm{P}(\mathrm{x}<1)=\mathrm{P}(\mathrm{x}>1)$$
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$$\mathrm{P}(1/2<\mathrm{x}<1)<\mathrm{P}(\mathrm{x}>1)$$
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$$\mathrm{P}(\mathrm{x}>3/2)<\mathrm{P}(\mathrm{x}<1)$$
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$$\mathrm{P}(\mathrm{x}>3/2)=0$$
Explanation
The numbers of the form $$ \dfrac{n}{n+1} $$ are less than 1 and $$ \dfrac{n+1}{n} $$ are greater than 1.
$$P(x<1) = P(\dfrac{1}{2}) +P(\dfrac{2}{3}) + ... $$
$$\Rightarrow P(x<1) = \dfrac{1}{4} +\dfrac{1}{8} + ... $$
$$\Rightarrow P(x<1) = \dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}} $$
$$\Rightarrow P(x<1) = \dfrac{1}{2}$$
$$P(x>1) = \dfrac{1}{2} $$
$$P\left(\dfrac{1}{2} <x<1\right) = P\left(\dfrac{2}{3} \right) + P\left(\dfrac{3}{4} \right) + ... $$
$$P\left(\dfrac{1}{2} <x<1\right) = \dfrac18 +\dfrac{1}{16} + ... $$
$$P\left(\dfrac{1}{2} <x<1\right) =\dfrac{1}{4} $$
$$P\left(x> \dfrac32\right) =P(x=2) $$
$$P\left(x> \dfrac32\right) =\dfrac{1}{4} $$
$$P\left(x> \dfrac32\right) = \dfrac{1}{4} $$]
$$\therefore$$$$\mathrm{P}(\mathrm{x}<1)=\mathrm{P}(\mathrm{x}>1)$$
$$\mathrm{P}(1/2<\mathrm{x}<1)<\mathrm{P}(\mathrm{x}>1)$$
$$\mathrm{P}(\mathrm{x}>3/2)<\mathrm{P}(\mathrm{x}<1)$$
If the range of a random variable $$X$$ is $$\{0, 1, 2, 3, \ldots\ldots\}$$ with $$\displaystyle P(X=k)=\dfrac{(k+1)(a)}{3^{k}}$$ for sample of $$4$$ items is drawn at random without $$\mathrm{k}\geq 0$$, then $$\mathrm{a}=$$
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$$2/3$$
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$$4/9$$
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$$8/27$$
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$$16/81$$
Explanation
Use the concept of probability distribution function,
$$\sum_{k=0}^{\infty}(P(X=k))=1$$
So,
$$\sum_{k=0}^{\infty}\dfrac{(k+1)a}{3^k}=a\sum_{k=0}^{\infty}(\dfrac{k}{3^k}+\dfrac{1}{3^k})$$
$$\sum_{k=0}^{\infty}\dfrac{k}{3^k}=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+.......=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+.......=\dfrac{1}{3}(1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+.....)$$
Use the Taylor series
$$\dfrac{1}{(1-x)^2}=1+2x+3x^2+4x^3+...........$$
Let us assume that $$x=\dfrac{1}{3}$$
$$\sum_{k=0}^{\infty}\dfrac{k}{3^k}=\dfrac{\dfrac{1}{3}}{(1-\dfrac{1}{3})^2}=\dfrac{3}{4}$$
The second Summation will be infinite Geometric Sequence where $$a=\dfrac{1}{3}$$ and $$r=\dfrac{1}{3}$$
$$\sum_{k=0}^{\infty}\dfrac{1}{3^k}=1+\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{3}{2}$$
Therefore,
$$\sum_{k=0}^{\infty}(P(X=k))=1$$
$$(\dfrac{3}{4}+\dfrac{3}{2})\times a=1$$
Hence, $$a=\dfrac{4}{9}$$
If $$\mathrm{P}(\mathrm{u}_{\mathrm{i}})\propto \mathrm{i}$$, where $$\mathrm{i}=1,2,3,\ \ldots \mathrm{n}$$, then $$\displaystyle \lim_{\mathrm{n}\rightarrow\infty}\mathrm{P}(\mathrm{w})$$ is equal to
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1
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$$\displaystyle \frac{2}{3}$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{1}{4}$$
Explanation
Here $$\mathrm{P}(\mathrm{u}_{\mathrm{i}})=\mathrm{k}\mathrm{i}$$
$$\Sigma \mathrm{P}(\mathrm{u}_{\mathrm{i}})=1$$
$$\displaystyle \Rightarrow \mathrm{k}=\frac{2}{\mathrm{n}(\mathrm{n}+1)}$$
$$\displaystyle \lim_{\mathrm{n}\rightarrow\infty}\mathrm{P}(\mathrm{w})=\lim_{n\rightarrow\infty}\sum_{{i}=1}^{n}P(\frac{w}{u_i})p(u_i)=\lim_{\mathrm{n}\rightarrow\infty}\sum_{\mathrm{i}=1}^{\mathrm{n}}\frac{2\mathrm{i}^{2}}{\mathrm{n}(\mathrm{n}+1)^{2}}=\lim_{\mathrm{n}\rightarrow\infty}\frac{2\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{\mathrm{n}(\mathrm{n}+1)^{2}6}=\frac{2}{3}$$
A fair coin is tossed $$99$$ times. If X is the number of times heads occur then P(X = r) is maximum when r is
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$$49$$
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$$50$$
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$$51$$
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none of these
Explanation
Notice that x has binomial distribution with n = 99, p = q = 0.5.
Therefore the probability
$$P(x=r)=^nC_rP^rq^{n-r}\\\implies P(x=r)=^nC_r(0.5)^r(0.5)^{n-r}=^{n}C_r(0.5)^n$$
Thus maximum of P(x = r) corresponds to the value of r with maximal
$$^nC_r=\dfrac {n!}{r!(n-r)!}$$
Notice that binomial coefficients are symmetric:
$$^nC_r=^{n-r}C_r=\dfrac {n!}{r!(n-r)!}$$
hence the maximum corresponds to the points most closest to $$\dfrac n2$$
Since n is odd, there are two such values:
$$r_1=\dfrac {n-1}2=\dfrac {98}2=49, r_2=\dfrac {n+1}2=\dfrac {100}2=50$$
For these numbers the function P(x = r) has maximum.
Let numbers $$1,2,3...4n$$ be pasted on $$4n$$ blocks. The probability of drawing a number is proportional to $$r$$, then the probability of drawing an even number in one draw is $$\left ( n\in N \right )$$
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$$\displaystyle\frac{n+1}{2n+1}$$
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$$\displaystyle \frac{2n+1}{4n+1}$$
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$$\displaystyle\frac{n+2}{n+3}$$
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$$\displaystyle\frac{2n+3}{4n+1}$$
Explanation
If $$P(r)$$ be the probability that a number $$r$$ is drawn in one draw, it is given that $$P(r) = kr;$$
where $$k$$ is a constant.
$$\therefore P(1) + P(2) + P(3) + \cdots + p(4n) = k \cdot 1 + k \cdot 2 + ...+ k\cdot (4n) = k\left [ 1 + 2 + 3 + \cdots+4n\right ]$$
$$1= \displaystyle \frac{4n(4n + 1)k}{2}$$
$$\displaystyle\therefore \sum_{r=0}^{n}P(r)=1$$ or
$$P(S)= 1$$ $$\displaystyle\Rightarrow K=\frac{1}{2n\left ( 4n+1 \right )}$$
Now $$'A'$$ event of drawing $$2, 4, 6, ...4n$$
$$P(A)= P ( 2$$ or $$4$$ or $$6 ....$$ or $$4n )$$
$$\therefore P(A)= P(2) + P(4) + \cdots + P(4n)$$
$$=2k\left (1 + 2 + ... + 2n \right )$$ $$P(A)= k(2n) \left (2n + I \right )$$
$$P(A)= (2n)\left (2n + I \right )$$
$$\therefore P(A)= \displaystyle \frac{2n+1}{4n+1}$$
Let $$p$$ be the probability that a man aged $$x$$ years will die in a year time. The probability that out of $$n$$ men $$\displaystyle A_{1},A_{2},A_{3},....,A_{n}$$ each aged $$x$$ years, $$\displaystyle A_{1}$$ will die & will be the first to die, is
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$$\displaystyle \frac{1-p^{n}}{n}$$
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$$\displaystyle \frac{p}{n}$$
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$$\displaystyle \frac{p\left ( 1-p \right )^{n-1}}{n}$$
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$$\displaystyle \frac{1-\left ( 1-p \right )^{n}}{n}$$
Explanation
Let $$E_i$$ be the event that $$A_i$$ will die in a year where i=1, 2, 3, 4, ..., n
The probability that none of $$A_1, A_2, A_3, ..., A_n $$ dies in a year
=$$(1-p_{A_1}).(1-p_{A_2}), ..., (1-p_{A_n})$$
Where $$p_{A_1}, p_{A_2}, ..., p_{A_n}$$ are the probabilities that $$A_1, A_2, A_3, ..., A_n $$ dies in a year respectively.
=$$(1-p).(1-p), ...(1-p)=(1-p)^n$$
Now the probability that at least one of $$A_1, A_2, A_3, ..., A_n $$
=$$1-(1-p)^n$$
The probability that among n men $$A_1$$ is the first one to die is $$\dfrac 1n$$
Therefore, the required probability is
$$=\dfrac 1n[1-(1-p)^n]$$
A continuous random variable $$X$$ has p.d.f $$f(x)$$, then:
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$$0\le f(x)\le 1$$
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$$f(x)\ge 0$$
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$$f(x)\le 1$$
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$$0< f(x)< 1$$
Explanation
Let $$X$$ be a continuous random variable.
Let $$f(x)$$ be the p.d.f of $$X$$.
We know that the p.d.f $$f(x)$$ of a continuous random variable $$X$$ is always positive. That is $$f(x) \geq 0, \forall x\in \mathbb{R}$$
Thus a continuous random variable $$X$$ has p.d.f $$f(x)$$ then $$f(x)\geq 0$$
A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. If $$p$$ satisfies the inequality $$ 625p^{2} - 175p + 12 < 0$$ then value of $$n$$ is
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$$1$$
0%
$$2$$
0%
$$3$$
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$$4$$
Explanation
Let p
(
n
)
=
Hitting for the 1st time at the n-th trial
$$\implies p(n)=0.8^{n-1}\times0.2$$
The boundary
of the inequality $$62p^2-175p+12=0$$
is the solution of a quadratic equation in
p
:
or, $$175^2-4\times12\times625=625=25^2$$
$$\implies p=\dfrac {175\pm25}{1250}=\dfrac 3{25}$$ or $$\dfrac 4{25}$$
So
p
(
n
)
is negative between those two values.
$$p(n)=\dfrac 3{25}=0.8^{n-1}\times 0.2\\\implies \dfrac 35=0.8^{n-1}\\\implies \log\left(\dfrac 35\right)=(n-1)\log (0.8)\\\implies n=1+\dfrac {\log \left(\dfrac 35\right)}{\log (0.8)}=3.289$$
$$p(n)=\dfrac 4{25}=0.8^{n-1}\times 0.2\\\implies \dfrac 45=0.8^{n-1}\\\implies \log\left(\dfrac 45\right)=(n-1)\log (0.8)\\\implies n=1+\dfrac {\log \left(\dfrac 45\right)}{\log (0.8)}=2$$
Hence, $$2<n<3.289\implies n=3$$
If $$X$$ is a discrete random variable then which of the following is correct?
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$$0\le F(x)<1$$
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$$F(-\infty )=0;F(\infty )\le 1$$
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$$P\left[ X={ x }_{ n } \right] =F({ x }_{ n })-F({ x }_{ n-1 })$$
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$$F(x)$$ is a constant function
Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of $$52$$ cards. Find the mean and variance of the number of red cards.
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Mean $$= 0.1$$ and Variance $$= 0.7$$
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Mean $$= 0.6$$ and Variance $$= 0.3$$
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Mean $$= 0.49$$ and Variance $$= 0.37$$
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Mean $$= 0$$ and Variance $$= 0.45$$
Explanation
Since we have to draw two red cards(without replacement) from $$52$$ cards, therefore the probability is:
$$p=\dfrac { 26 }{ 52 } \times \dfrac { 25 }{ 51 } =\dfrac { 25 }{ 102 }$$
And, $$q=1-\dfrac { 25 }{ 102 } =\dfrac { 77 }{ 102 }$$
Since $$n=2$$, therefore,
$$Mean=np=2\times \dfrac { 25 }{ 102 } =\dfrac { 50 }{ 102 } =0.49\\$$
$$Variance=npq=2\times \dfrac { 25 }{ 102 } \times \dfrac { 77 }{ 102 } =0.37$$
$$f(x)=\dfrac { A }{ \pi } .\dfrac { 1 }{ 16+{ x }^{ 2 } } ,-\infty <x<\infty $$ is a p.d.f of a continuous random variable $$X$$, then the value of $$A$$ is:
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$$16$$
0%
$$8$$
0%
$$4$$
0%
$$1$$
Explanation
For a probability distribution function $$f(x), a<x<b$$ of a continuous random variable $$X$$, it is required that $$\displaystyle \int_a^b f(x)dx = 1$$
$$\therefore \displaystyle \int_{-\infty}^{\infty} \cfrac{A}{\pi}\cfrac{1}{16+x^2}dx = 1$$
$$\Rightarrow \displaystyle \cfrac{A}{\pi}\int_{-\infty}^{\infty}\cfrac{1}{4^2 + x^2}dx = 1$$
$$\Rightarrow \displaystyle \cfrac{A}{\pi} \cfrac14\tan^{-1}\left(\cfrac x4\right)\big|_{-\infty}^{\infty} =1$$
$$\Rightarrow \cfrac {A}{4\pi}\left[\cfrac{\pi}2 - \left(-\cfrac{\pi}2\right) \right] = 1$$
$$\Rightarrow \cfrac{A}{4} =1$$
$$\Rightarrow A = 4$$
What is P($$Z$$ is the product of two prime numbers) equal to ?
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$$0$$
0%
$$\dfrac{1}{4}$$
0%
$$\dfrac{1}{6}$$
0%
$$\dfrac{1}{12}$$
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
$$X=\{1,3,5,7\}$$
$$Y=\{2,4,6\}$$
$$Z=\{3,5,7,9,11,13\}$$
To find probability of Z being product of two prime numbers-
Among the above obtained values of Z, none can be expressed in the form of product of two different prime numbers
$$\therefore P=0$$
$$NOTE$$- If we consider that Z is to be expressed as product of two different prime numbers, then the answer is 0
But if we assume that the prime numbers used in the products can be same, then
9 can be expressed as product of 2 prime numbers
$$Z=X+Y=9$$
$$\{X,Y\}=\{\{7,2\},\{3,6\},\{5,4\}\}$$
$$P=\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 3 } $$
$$P=\cfrac{1}{4}$$
The distribution of a random variable X is given below:
X = x
2
-1
0
1
2
3
P(X = x)
$$\frac{1}{10}$$
k
$$\frac{1}{5}$$
2k
$$\frac{3}{10}$$
k
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$$\frac{1}{10}$$
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$$\frac{2}{10}$$
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$$\frac{3}{10}$$
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$$\frac{7}{10}$$
The probability that the bag contains 2 balls of each colour, is
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$$\dfrac{1}{3}$$
0%
$$\dfrac{1}{5}$$
0%
$$\dfrac{1}{10}$$
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$$\dfrac{1}{4}$$
Let the p.m.f. of a random variable $$X$$ be -
$$P(x) = \dfrac {3 - x}{10}$$ for $$x = -1, 0, 1, 2 $$ otherwise
Then $$E(X)$$ is _________.
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$$1$$
0%
$$2$$
0%
$$0$$
0%
$$-1$$
Explanation
$$x$$
$$-1$$
$$0$$
$$1$$
$$2$$
$$P(x)$$
$$\dfrac{4}{10}$$
$$\dfrac{3}{10}$$
$$\dfrac{2}{10}$$
$$\dfrac{1}{10}$$
$$x\cdot P(x)$$
$$-\dfrac{4}{10}$$
$$0$$
$$\dfrac{2}{10}$$
$$\dfrac{2}{10}$$
We have to find $$E(X)$$
We know that $$E(X)=\sum x\cdot P(x)$$
$$\therefore E(X)=-\dfrac{4}{10}+0+\dfrac{2}{10}+\dfrac{2}{10}=0$$
Hence $$E(X)=0$$
The range of a random variable X = $${ 1,2,3,4,...}$$ and the probabilities are given by $$ P (X = k) = \frac{C^k}{ k! } $$ ; $$k = 1,2,3,4...,$$ then the value of C is
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$$2$$
0%
$$log_2 e$$
0%
$$log_e 2$$
0%
4
A random variable $$X$$ has the following probability distribution:
$$X$$
0
1
2
3
4
5
$$P(X=x)$$
$$\cfrac{1}{4}$$
$$2a$$
$$3a$$
$$4a$$
$$5a$$
$$\cfrac{1}{4}$$
Then $$P(1\le X\le 4)$$ is:
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$$\cfrac { 10 }{ 21 } $$
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$$\cfrac { 2 }{ 7 } $$
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$$\cfrac { 1 }{ 14 } $$
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$$\cfrac { 1 }{ 2 } $$
Explanation
We know that $$\sum P(X=x)=1$$
$$\Rightarrow \dfrac{1}{4}+2a+3a+4a+5a+\dfrac{1}{4}=1$$
$$\Rightarrow \dfrac{1}{2}+14a=1$$
$$\Rightarrow 14a=1-\dfrac{1}{2}$$
$$\Rightarrow 14a=\dfrac{1}{2}$$
$$\Rightarrow a=\dfrac{1}{28}$$
Now we have to find $$P(1\leq x \leq 4)$$
$$P(1\leq x \leq 4)=P(x=1)+P(x=2)+P(x=3)+P(x=4)$$
$$=2a+3a+4a+5a$$
$$=14a$$
$$=14 \left(\dfrac{1}{28}\right) \quad \left[\because a=\dfrac{1}{28}\right]$$
$$=\dfrac{1}{2}$$
$$\therefore P(1\leq x \leq 4)=\dfrac{1}{2}$$
The random variable $$X$$ follows normal distribution
$$f(x)=c{ e }^{ \cfrac { -\cfrac { 1 }{ 2 } { \left( x-100 \right) }^{ 2 } }{ 25 } }$$. Then the value of $$c$$ is:
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$$\sqrt { 2\pi } $$
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$$\cfrac { 1 }{ \sqrt { 2\pi } } $$
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$$5\sqrt { 2\pi } $$
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$$\cfrac { 1 }{5 \sqrt { 2\pi } } $$
Explanation
The normal density function of the random variable $$X$$ is given by $$f(x)=ce^{\dfrac{-\dfrac{1}{2}(x-100)^2}{25}}$$ $$...(1)$$
We have to find the value of $$c$$.
We know that, if $$X \sim N(\mu ,\sigma^2)$$, the normal density function is $$f(x;\mu,\sigma^2)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{\dfrac{-(x-\mu)^2}{2\sigma^2}}$$ $$...(2)$$
Comparing $$(1)$$ and $$(2)$$ we get
$$c=\dfrac{1}{\sigma\sqrt{2\pi}}, \mu=100, \sigma^2=25 \Rightarrow \sigma=5$$
$$\therefore c=\dfrac{1}{5\sqrt{2\pi}}$$
If the range of random variable X is {0, 1, 2, 3, 4...} with p(X = k) = $$\frac{(k + 1)a}{3^k}$$ for $$k>=$$0, then a =
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0%
$$\frac{2}{3}$$
0%
$$\frac{4}{9}$$
0%
$$\frac{8}{27}$$
0%
$$\frac{16}{81}$$
There are $$5$$ men and $$5$$ women in a party. In how many can $$5$$ dancing pairs be selected ? (each dancing pair consists of $$1$$ man and $$1$$ women )
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$$120$$
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$$44$$
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$$32$$
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$$10$$
Two probability distributions of the discrete random variable $$X$$ and $$Y$$ are given below.
$$X$$
$$0$$
$$1$$
$$2$$
3
$$P\left( X
\right)$$
$$\dfrac { 1 }{ 5
} $$
$$\dfrac { 2 }{ 5
} $$
$$\dfrac { 1 }{ 5
}$$
$$\dfrac { 1 }{ 5 }$$
$$Y$$
$$0$$
$$1$$
$$2$$
$$3$$
$$P\left( Y
\right)$$
$$\dfrac { 1 }{ 5 }$$
$$\dfrac { 3 }{ 10
} $$
$$\dfrac { 2 }{ 5
} $$
$$\dfrac { 1 }{ 10 }$$
Then
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$$E\left( { Y }^{ 2 } \right) =2E\left( X \right)$$
0%
$$E\left( { Y }^{ 2 } \right) =E\left( X \right)$$
0%
$$E\left( Y \right) =E\left( X \right)$$
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$$E\left( { X }^{ 2 } \right) =2E\left( Y \right) $$
Explanation
$$E(X)=\sum { XP( } X)=0*\frac { 1 }{ 5 } +1*\frac { 2 }{ 5 } +2*\frac { 1 }{ 5 } +3*\frac { 1 }{ 5 } =7/5\\ E({ Y }^{ 2 })=\sum { { y }^{ 2 }P(Y) } ={ 0 }^{ 2 }*\frac { 1 }{ 5 } +1^{ 2 }*\frac { 3 }{ 10 } +{ 2 }^{ 2 }*\frac { 4 }{ 10 } +{ 3 }^{ 2 }*\frac { 1 }{ 10 } =28/10=14/5\\ E({ Y }^{ 2 })=2E(X)$$
The total numbers of outcomes when three coins tossed once is .....
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$$2$$
0%
$$8$$
0%
$$6$$
0%
$$4$$
Explanation
When three coins tossed ones then total number of outcomes is : $$8$$
$$\{ TTT,HHH,TTH,THT,HTT,HHT,HTH,THH\} $$
The random variable X has the following probability massfunction $$P[x=x]=k.\dfrac{2x}{x!}, x=0,1,2,3=0 $$, otherwise, then the value of K is
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0%
$$\dfrac{1}{5}$$
0%
$$\dfrac{2}{5}$$
0%
$$\dfrac{3}{5}$$
0%
$$\dfrac{4}{5}$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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