Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively, then $7 \bar{X}+4 \bar{Y}$ is equal to ___________.
Explanation:
| $X$ | 3 | 2 | 1 | 0 |
|---|---|---|---|---|
| $Y$ | 0 | 1 | 2 | 3 |
$\begin{aligned} & \bar{X}=\sum X p(X) \\ & \bar{Y}=\sum Y p(Y) \\ & P(X=3)=P(Y=0)=\frac{{ }^5 C_3 \cdot C_0}{{ }^9 C_3}=\frac{{ }^5 C_2}{{ }^9 C_3}=\frac{5}{42} \\ & P(X=2)=P(Y=1)=\frac{{ }^5 C_2 \cdot C_1}{{ }^9 C_3}=\frac{10}{21} \\ & P(X=1)=P(Y=2)=\frac{{ }^5 C_1 \cdot C_2}{{ }^9 C_3}=\frac{5}{14} \\ & P(X=0)=P(Y=3)=\frac{{ }^5 C_0 \cdot C_3}{{ }^9 C_3}=\frac{4}{84}=\frac{1}{21} \\ & \bar{X}=3 \times \frac{5}{42}+2 \times \frac{10}{21}+\frac{5}{14}+0 \times \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \\ & \bar{Y}=0 \times \frac{5}{42}+1 \times \frac{10}{21}+2 \times \frac{5}{14}+3 \times \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \\ & 7 \bar{X}+4 \bar{Y}=17 \end{aligned}$
From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to _________.
Explanation:
Given a lot of 12 items, 3 are defective.
Good items, $12-3=9$
Let $X$ denote the number of defective items.
So, value of $X=0,1,2,3$
A sample of $S$ items is drawn.
$P(X=0)=G G G G G$
(here $G$ is good item and $d$ is defective)
$\begin{aligned} & \frac{9}{12} \cdot \frac{8}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{5}{8}=\frac{21}{132}=\frac{7}{44} \\ & P(X=1)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 3}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{21}{44} \\ & P(X=2)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{14}{44} \\ & P(X=3)=5\left[\frac{3 \cdot 2 \cdot 1 \cdot 9 \cdot 8}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{2}{44} \\ & P(X=4)=0 \\ & P(X=5)=0 \end{aligned}$
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $P(X)$ | $ \frac{7}{44} $ |
$ \frac{21}{44} $ |
$ \frac{14}{44} $ |
$ \frac{2}{44} $ |
0 | 0 |
| $XP(X)$ | 0 | $ \frac{21}{44} $ |
$ \frac{28}{44} $ |
$ \frac{6}{44} $ |
0 | 0 |
| $X^2P(X)$ | 0 | $ \frac{21}{44} $ |
$ \frac{56}{44} $ |
$ \frac{18}{44} $ |
0 | 0 |
$\begin{aligned} & \sigma_x^2=\sum X^2 P(x)-\left(\sum x P(x)\right)^2 \\ & =\frac{95}{44}-\left(\frac{55}{44}\right)^2 \\ & =\frac{4180-3025}{1936}=\frac{1155}{1936}=\frac{105}{176}=\frac{m}{n} \\ & =n-m=71 \end{aligned}$
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$, then $96 \sigma^2$ is equal to __________.
Explanation:
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(x)$ | $ \frac{{ }^7 C_5}{{ }^{10} C_5}=\frac{1}{12} $ |
$ \frac{C_4 \cdot{ }^3 C_1}{{ }^{10} C_5}=\frac{5}{12} $ |
$ \frac{{ }^7 C_3 \cdot{ }^3 C_2}{{ }^{10} C_5}=\frac{5}{12} $ |
$ \frac{{ }^7 C_2 \cdot{ }^3 C_3}{{ }^{10} C_5}=\frac{1}{12} $ |
| $xP(x)$ | 0 | $\frac{5}{12}$ | $\frac{10}{12}$ | $\frac{3}{12}$ |
$\begin{aligned} & \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\ & \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\ & =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12} \end{aligned}$
$\Rightarrow \sigma^2 \cdot 96=8 \times 7=56$
In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $\mathrm{P}(|x-y| \leq 2)$ is $p$, then $3^9 p$ equals _________.
Explanation:
$\begin{aligned} & x+y=10 \\ & A=x-y \\ & P(|A|<2) \text { is } P \\ & \Rightarrow|A|=2,1,0 \Rightarrow A=0,1,-1,2,-2 \\ & \Rightarrow x=\frac{10+A}{2} \Rightarrow A \in \text { even as } x \in \text { integer } \\ & \Rightarrow A=0,-2,2 \\ & \Rightarrow P(|A| \leq 2)=P(A=0)+P(A=-2)+P(A=2) \end{aligned}$
(1) $A=0 \Rightarrow x=5=y$
$P(A=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5$
(2) $A=-2$
$\Rightarrow x=4$ and $y=6$
$P(A=-2)={ }^{10} C_4 \cdot\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6$ and
$\begin{aligned} & \text { Similarly, } P(A=2)={ }^{10} C_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4 \\ & \Rightarrow P(|A| \leq 2) 3^9=3\left({ }^{10} C_5 \cdot 2^5+{ }^{10} C_4 \cdot 2^6+{ }^{10} C_6 \cdot 2^4\right) \\ & =8288 \end{aligned}$
A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is _________.
Explanation:
$ \begin{aligned} & n(C \cup P \cup M) \leq n(U)=40 . \\\\ & n(C)+n(P)+n(M)-n(C \cap M)-n(P \cap M)-n(C \cap \\\\ & P)+n(C \cap P \cap M) \leq 40 \\\\ & 20+25+16-11-15-15+x \leq 40 \\\\ & x \leq 20 \end{aligned} $
But $11-x \geq 0$ and $15-x \geq 0$
$ \Rightarrow x \geq 11 $
$a=P(X=3), b=P(X \geqslant 3)$ and $c=P(X \geqslant 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to __________.
Explanation:
To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.
Let's begin by defining each of the variables:
- $a = P(X=3)$: This is the probability that the first six appears on the third toss.
- $b = P(X \geqslant 3)$: This is the probability that the first six appears on the third toss or later.
- $c = P(X \geqslant 6 \mid X>3)$: This is the probability that the first six appears on the sixth toss or later, given that it has not appeared in the first three tosses.
Since we're dealing with a fair die, each side has an equal probability of $\frac{1}{6}$ of landing face up. Let's find the probabilities step by step:
Calculating $a$:
The probability of rolling anything other than a six is $\frac{5}{6}$. So for the first six to show up exactly on the third roll, the sequence of rolls must be NN6, where N is anything but a six (i.e., the results of the first two rolls). Thus,
$a = P(X=3) = \left(\frac{5}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right)$
Calculating $b$:
For the first six to appear on the third roll or later, we can think of two cases: when the first six appears on the third roll (which we've already calculated, $a$), and when it appears after the third roll. To combine these probabilities, we can use the fact that $P(X \geqslant 3) = 1 - P(X < 3)$, where $P(X < 3)$ is the probability that the first six appears on either the first or the second roll. So we calculate the latter first:
$ P(X < 3) = P(X=1) + P(X=2) $
$ P(X < 3) = \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) $
Thus,
$ b = P(X \geqslant 3) = 1 - P(X < 3) = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) \right] $
Calculating $c$:
This is the probability that the first six appears on or after the sixth roll, given that it hasn't appeared in the first three rolls. Since $X>3$, the first three outcomes must not be a six, which occurs with probability $\left(\frac{5}{6}\right)^3$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,
$c = P(X \geqslant 6 \mid X>3) = \left(\frac{5}{6}\right)^2$
Notice here, we did not include the probability of rolling a six, because we are looking for the probability that we have not yet rolled a six after the fifth roll.
Now we can calculate $a$, $b$, and $c$:
$a = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$
$b = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right]$
$c = \left(\frac{5}{6}\right)^2$
Now we'll substitute to find $\frac{b+c}{a}$:
$\frac{b+c}{a} = \frac{1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$
Simplifying the numerator:
$1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2$
$= 1 - \left[\frac{1}{6} + \frac{5}{36}\right] + \frac{25}{36}$
$= 1 - \left[\frac{6}{36} + \frac{5}{36}\right] + \frac{25}{36}$
$= 1 - \frac{11}{36} + \frac{25}{36}$
$= \frac{36}{36} - \frac{11}{36} + \frac{25}{36}$
$= \frac{50}{36}$
Now, substitute this back into the expression and solve:
$\frac{b+c}{a} = \frac{\frac{50}{36}}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$
$\frac{b+c}{a} = \frac{50}{36} \cdot \frac{6}{\left(\frac{5}{6}\right)^2}$
$\frac{b+c}{a} = \frac{50 \cdot 6}{25}$
$\frac{b+c}{a} = \frac{300}{25}$
$\frac{b+c}{a} = 12$
Therefore, $\frac{b+c}{a} = 12$.
A fair $n(n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$, then $n$ is equal to ____________.
Explanation:
$ \text { Mean }=\sum\limits_{i=1}^{\infty} p_i x_i=1 \cdot \frac{n-1}{n}+\frac{2}{n} \cdot\left(\frac{n-1}{n}\right)+\frac{3}{n^2}\left(\frac{n-1}{n}\right)+\ldots $
$ \frac{n}{9}=\left(1-\frac{1}{n}\right) S $ ......(1)
where
$ \begin{aligned} & S=1+\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^3}+\ldots \\\\ & \frac{1}{n} S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\ldots \\\\ & ----------------\\\\ & \left(1-\frac{1}{n}\right) S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\ldots \end{aligned} $
$ \begin{aligned} & \Rightarrow \left(1-\frac{1}{n}\right) S=\frac{1}{1-\frac{1}{n}} \\\\ & \Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right) \times \frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n}{n-1} \end{aligned} $
$ \Rightarrow $ n = 10
Let the probability of getting head for a biased coin be $\frac{1}{4}$. It is tossed repeatedly until a head appears. Let $\mathrm{N}$ be the number of tosses required. If the probability that the equation $64 \mathrm{x}^{2}+5 \mathrm{Nx}+1=0$ has no real root is $\frac{\mathrm{p}}{\mathrm{q}}$, where $\mathrm{p}$ and $\mathrm{q}$ are coprime, then $q-p$ is equal to ________.
Explanation:
This gives us :
$(5N)^2 - 4\times64\times1 < 0$
$\Rightarrow 25N^2 < 256$
$\Rightarrow N^2 < \frac{256}{25}$
$\Rightarrow N < \sqrt{\frac{256}{25}} = \frac{16}{5}$
Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3.
The probability of getting the first head on the $n$-th toss (given the probability of getting a head is $1/4$) is given by the geometric distribution formula, $(1 - p)^{n-1}\times p$.
So, the probability for our specific values of $N$ is:
$P(N=1) = (1 - 1/4)^{1-1}\times(1/4) = 1/4$
$P(N=2) = (1 - 1/4)^{2-1}\times(1/4) = 3/4 \times 1/4 = 3/16$
$P(N=3) = (1 - 1/4)^{3-1}\times(1/4) = (3/4)^2 \times 1/4 = 9/64$
Therefore, the total probability (p/q) is :
$p/q = P(N=1) + P(N=2) + P(N=3)$
$= 1/4 + 3/16 + 9/64$
$= 16/64 + 12/64 + 9/64$
$= 37/64$
So, $p = 37$, $q = 64$ and $q-p = 64 - 37 = 27$.
Therefore, $q-p$ is equal to $27$.
Explanation:
$ \begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{\operatorname{ar}(\mathrm{OACDEG})}{(\mathrm{OBDF})} \\\\ & =\frac{\operatorname{ar}(\mathrm{OBDF})-\operatorname{ar}(\mathrm{ABC})-\operatorname{ar}(\mathrm{EFG})}{\operatorname{ar}(\mathrm{OBDF})} \\\\ & \Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60-\mathrm{a})^2-\frac{1}{2}(60-\mathrm{a})^2}{3600} \\\\ & \Rightarrow 1100=3600-(60-\mathrm{a})^2 \\\\ & \Rightarrow (60-\mathrm{a})^2=2500 \Rightarrow 60-\mathrm{a}=50 \\\\ & \Rightarrow \mathrm{a}=10 \end{aligned} $
Explanation:
$p = {6 \over {36}} = {1 \over 6}$
$q = {{{}^6{C_1} \times {}^5{C_1} \times {{4!} \over {3!}}} \over {{6^4}}} = {{120} \over {1296}} = {5 \over {54}}$
${p \over q} = {{{1 \over 6}} \over {{5 \over {54}}}} = {{54} \over {6 \times 5}} = {9 \over 5} = {m \over n}$
$m + n = 14$
25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}%$. Then the value of k is __________.
Explanation:
Probability of a person being non-smoker $=\frac{3}{4}$
$P\left(\frac{\text { Person is smoker }}{\text { Person diagonsed with cancer }}\right)=\frac{\frac{1}{4} \cdot 27 P}{\frac{1}{4} \cdot 27 P+\frac{3 P}{4}}$
$=\frac{9}{10}=\frac{k}{10}$
$\Rightarrow k=9$
Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and $\lambda$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^2=\lambda x$ with one vertex at the vertex of the parabola, is :
Explanation:
$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$
$E_{3}$ : Ball is drawn from urn $C(\lambda R+4 B)$
$A \rightarrow$ Ball drawn is red.
Required probability $=P\left(\frac{E_{3}}{A}\right)$
$=\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{\lambda}{\lambda+4}}=\frac{2}{5}$
$\Rightarrow\frac{10 \lambda}{19 \lambda+36}=\frac{2}{5}$
$\Rightarrow \lambda=6$
So, parabola $y^2=6 x$
Let side length of the triangle be $l$.
$ \begin{aligned} & \tan 30^{\circ}=\frac{3 t} {\frac{3}{2} t^2} \\\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \\\\ & \therefore t=2 \sqrt{3} \\\\ & \text { So, }\left(\frac{3}{2} t^2, 3 t\right) \\\\ & =(18,6 \sqrt{3}) \end{aligned} $
$ \text { Now, } l^2=18^2+(6 \sqrt{3})^2=324+108=432 $
The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is ____________.
Explanation:
Given $n p+n p q=82.5$
and $n p(n p q)=1350$
$ \therefore $ Quadratic equation is
$ x^{2}-82.5 x+1350=0$
$\Rightarrow x^{2}-22.5 x-60 x+1350=0$
$\Rightarrow x-(x-22.5)-60(x-22.5)=0$
Mean $=60$ and Variance $=22.5$
$ n p=60, n p q=22.5 $
$ \Rightarrow q=\frac{9}{24}=\frac{3}{8}, p=\frac{5}{8} $
$ \therefore \quad n \frac{5}{8}=60 \quad \Rightarrow n=96 $
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $\mathrm{X}$ be the number of white balls, among the drawn balls. If $\sigma^{2}$ is the variance of $\mathrm{X}$, then $100 \sigma^{2}$ is equal to ________.
Explanation:
$X = $ Number of white ball drawn
$P(X = 0) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = {1 \over 6}$
$P(X = 1) = {{{}^6{C_2} \times {}^4{C_1}} \over {{}^{10}{C_3}}} = {1 \over 2},$
$P(X = 2) = {{{}^6{C_1} \times {}^4{C_2}} \over {{}^{10}{C_3}}} = {3 \over {10}}$
and $P(X = 3) = {{{}^6{C_0} \times {}^4{C_3}} \over {{}^{10}{C_3}}} = {1 \over {30}}$
Variance $ = {\sigma ^2} = \sum {{P_i}X_i^2 - {{\left( {\sum {{P_i}{X_i}} } \right)}^2}} $
${\sigma ^2} = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {\left( {{1 \over 2} + {6 \over {10}} + {1 \over {10}}} \right)^2}$
$ = {{56} \over {100}}$
$100{\sigma ^2} = 56.$
The probability distribution of X is :
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | ${{1 - d} \over 4}$ | ${{1 + 2d} \over 4}$ | ${{1 - 4d} \over 4}$ | ${{1 + 3d} \over 4}$ |
For the minimum possible value of d, sixty times the mean of X is equal to _______________.
Explanation:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(x)$ | ${{1 - d} \over 4}$ | ${{1 + 2d} \over 4}$ | ${{1 - 4d} \over 4}$ | ${{1 + 3d} \over 4}$ |
We know, $0 \le P(x) \le 1$
$\therefore$ $0 \le {{1 - d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 - d \le 4$
$ \Rightarrow - 1 \le - d \le 3$
$ \Rightarrow 1 \ge d \ge - 3$
Also,
$0 \le {{1 + 2d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 + 2d \le 4$
$ \Rightarrow - 1 \le 2d \le 3$
$ \Rightarrow - {1 \over 2} \le d \le {3 \over 2}$
Also,
$0 \le {{1 - 4d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 - 4d \le 4$
$ \Rightarrow - 1 \le - 4d \le 3$
$ \Rightarrow 1 \ge 4d \ge - 3$
$ \Rightarrow {1 \over 4} \ge d \ge - {3 \over 4}$
And,
$0 \le {{1 + 3d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 + 3d \le 4$
$ \Rightarrow - 1 \le 3d \le 3$
$ \Rightarrow - {1 \over 3} \le d \le 1$
Common range is $ = - {1 \over 3}$ to ${1 \over 4}$
$\therefore$ $d\, \in \left[ { - {1 \over 3},{1 \over 4}} \right]$
$\therefore$ Minimum value of $d = - {1 \over 3}$
We know, mean
$E(x) = \sum {x\,.\,P(x)} $
$ = 0 \times {{1 - d} \over 4} + 1 \times {{1 + 2d} \over 4} + 2 \times {{1 - 4d} \over 4} + 3 \times {{1 + 3d} \over 4}$
$ = {{1 + 2d + 2 - 8d + 3 + 9d} \over 4}$
$ = {{6 + 3d} \over 4}$
For $d = - {1 \over 3}$, $E(x) = {{6 + 3 \times - {1 \over 3}} \over 4} = {5 \over 4}$
$\therefore$ $60E(x) = 60 \times {5 \over 4} = 75$
Let S = {E1, E2, ........., E8} be a sample space of a random experiment such that $P({E_n}) = {n \over {36}}$ for every n = 1, 2, ........, 8. Then the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is ___________.
Explanation:
Here $P({E_n}) = {n \over {36}}$ for n = 1, 2, 3, ......, 8
Here $P(A) = {{Any\,possible\,sum\,of\,(1,2,3,\,...,\,8)( = a\,say)} \over {36}}$
$\because$ ${a \over {36}} \ge {4 \over 5}$
$\therefore$ $a \ge 29$
If one of the number from {1, 2, ......, 8} is left then total $a \ge 29$ by 3 ways.
Similarly by leaving terms more 2 or 3 we get 16 more combinations.
$\therefore$ Total number of different set A possible is 16 + 3 = 19
If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to _______________.
Explanation:
Total number of numbers from given
Condition = n(s) = 26.
Every required number is of the form
A = 7 . (10a1 + 10a2 + 10a3 + .......) + 111111
Here 111111 is always divisible by 21.
$\therefore$ If A is divisible by 21 then
10a1 + 10a2 + 10a3 + ....... must be divisible by 3.
For this we have 6C0 + 6C3 + 6C6 cases are there
$\therefore$ n(E) = 6C0 + 6C3 + 6C6 = 22
$\therefore$ Required probability = ${{22} \over {{2^6}}} = p$
$\therefore$ ${{11} \over {32}} = p$
$\therefore$ $96p = 33$
In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability ${3 \over 4}$ and the remaining 6 questions correctly with probability ${1 \over 4}$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is ${{{{27}k}} \over {{4^{10}}}}$, then k is equal to ___________.
Explanation:
Student guesses only two wrong. So there are three possibilities.
(i) Student guesses both wrong from 1st section
(ii) Student guesses both wrong from 2nd section
(iii) Student guesses two wrong one from each section
Required probabilities
$ = {}^4{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^2}{\left( {{1 \over 6}} \right)^6} + {}^6{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^4}{\left( {{3 \over 4}} \right)^4} + {}^4{C_1}\,.\,{}^6{C_1}\left( {{3 \over 4}} \right)\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}{\left( {{1 \over 4}} \right)^5}$
$ = {1 \over {{4^{10}}}}\left[ {6 \times 9 + 15 \times {9^4} + 24 \times {9^2}} \right]$
$ = {{27} \over {{4^{10}}}}\left[ {2 + 27 \times 15 + 72} \right]$
$ = {{27 \times 479} \over {{4^{10}}}}$
| x | $ - $2 | $ - $1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | ${1 \over 5}$ | a | ${1 \over 3}$ | ${1 \over 5}$ | b |
If the mean of X is 2.3 and variance of X is $\sigma$2, then 100 $\sigma$2 is equal to :
Explanation:
| x | $ - $2 | $ - $1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | ${1 \over 5}$ | a | ${1 \over 3}$ | ${1 \over 5}$ | b |
$\overline X $ = 2.3
$-$a + 6b = ${9 \over {10}}$ ..... (1)
$\sum {{P_i} = {1 \over 5} + a + {1 \over 3} + {1 \over 5} + b = 1} $
$a + b = {4 \over {15}}$ .... (2)
From equation (1) and (2)
$a = {1 \over {10}},b = {1 \over 6}$
${\sigma ^2} = \sum {{p_i}x_i^2 - {{(\overline X )}^2}} $
${1 \over 5}(4) + a(1) + {1 \over 3}(9) + {1 \over 5}(16) + b(36) - {(2.3)^2}$
$ = {4 \over 5} + a + 3 + {{16} \over 5} + 36b - {(2.3)^2}$
$ = 4 + a + 3 + 36b - {(2.3)^2}$
$ = 7 + a + 36b - {(2.3)^2}$
$ = 7 + {1 \over {10}} + 6 - {(2.3)^2}$
$ = 13 + {1 \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$
$ = {{131} \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$
$ = {{1310 - {{(23)}^2}} \over {100}}$
$ = {{1310 - 529} \over {100}}$
${\sigma ^2} = {{781} \over {100}}$
$100{\sigma ^2} = 781$
Explanation:
I2 = second unit is functioning
P(I1) = 0.9, P(I2) = 0.8
P($\overline {{I_1}} $) = 0.1, P($\overline {{I_2}} $) = 0.2
$P = {{0.8 \times 0.1} \over {0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8}} = {8 \over {28}}$
$98P = {8 \over {28}} \times 98 = 28$
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X) | K | 2K | 2K | 3K | K |
Let p = P(1 < X < 4 | X < 3). If 5p = $\lambda$K, then $\lambda$ equal to ___________.
Explanation:
$ \Rightarrow k = {1 \over 9}$
Now, $p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$
$ \Rightarrow p = {2 \over 3}$
Now, $5p = \lambda k$
$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$
$ \Rightarrow \lambda = 30$
Explanation:
1 $-$ P(All tail) $\ge$ 0.9
$1 - {\left( {{1 \over 2}} \right)^n}$ $\ge$ 0.9
$ \Rightarrow {\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$
$\Rightarrow$ nmin = 4
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ and ($\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $ is equal to _____________.
Explanation:
$\alpha $ = P$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$ = $P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$
$ \Rightarrow $ $\alpha $ = x(1 $-$ y) (1 $-$ z) ......(i)
Similarly
β = (1 – x).y(1 – z) ...(ii)
$\gamma $ = (1 – x)(1 – y).z ...(iii)
p = (1 – x)(1 – y)(1 – z) ...(iv)
From (i) and (iv)
${x \over {1 - x}} = {\alpha \over p}$
$ \Rightarrow $ x = ${\alpha \over {\alpha + p}}$
From (iii) and (iv)
${z \over {1 - z}} = {\gamma \over p}$
$ \Rightarrow $ z = ${\gamma \over {\gamma + p}}$
$ \therefore $ ${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$ $ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$ ..(v)
Also given,
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ $ \Rightarrow $ $\alpha $p = ($\alpha $ + 2p)$\beta $ ....(vi)
$\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$ $ \Rightarrow $ 3$\gamma $p = (p - 2$\gamma $)$\beta $ .....(vii)
From (vi) and (vii),
${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$
$ \Rightarrow $ p$\alpha $ - 6p$\gamma $ = 5$\gamma $$\alpha $
$ \Rightarrow $ ${p \over \gamma } - {{6p} \over \alpha } = 5$
$ \Rightarrow $ ${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$ ....(viii)
Now from (v) and (viii),
${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$ = 6
Then ${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$ is equal to ________.
Explanation:
$\alpha $ = P(B1 $ \cap $ $\overline {{B_2}} \cap \overline {{B_3}} $) = $P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$
$ \Rightarrow $ x(1 $-$ y)(1 $-$ z) = $\alpha$
Similarly, y(1 $-$ x)(1 $-$ z) = $\beta$
z(1 $-$ x)(1 $-$ y) = $\gamma$
and (1 $-$ x)(1 $-$ y)(1 $-$ z) = p
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$
(x(1 $-$ y)(1 $-$ z) $-$2y(1 $-$ x)(1 $-$ z)) (1 $-$ x)(1 $-$ y)(1 $-$ z) = xy(1 $-$ x)(1 $-$ y)(1 $-$ z)
x $-$ xy $-$ 2y + 2xy = xy
x = 2y ...... (1)
Similarly ($\beta$ $-$ 3$\gamma $)p = 2$\beta$$\gamma $
$ \Rightarrow $ y = 3z .... (2)
From (1) & (2)
x = 6z
Now
${x \over z} = 6$
Explanation:
at least 2 bombs should hit.
P(x > 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - p(x < 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - (p(x = 0) + p(x = 1)) $ \ge $ 0.99
$ \Rightarrow $ 1 - nC0${\left( {{1 \over 2}} \right)^0}{\left( {{1 \over 2}} \right)^n}$ - nC1.${\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{n - 1}}$ $ \ge $ 0.99
$ \Rightarrow $ 1 - ${1 \over {{2^n}}}$ - ${n \over {{2^n}}}$ $ \ge $ ${{99} \over {100}}$
$ \Rightarrow $ ${1 \over {100}}$ $ \ge $ ${{n + 1} \over {{2^n}}}$
$ \Rightarrow $ 2n $ \ge $ 100(n + 1)
Now checking for value of n, we get
n = 11
Explanation:
$ \Rightarrow 1 - {\left( {{9 \over {10}}} \right)^n} > {1 \over 4}$
$ \Rightarrow {3 \over 4} > {\left( {{9 \over {10}}} \right)^n} \Rightarrow n \ge 3$
STATEMENT-1:
$P\left( {{H_1}|E} \right) > P\left( {E|{H_1}} \right).P\left( {{H_1}} \right)$ for $i=1,2,....,n$ because
STATEMENT-2: $\sum\limits_{i = 1}^n {P\left( {{H_i}} \right)} = 1.$
Three students $S_1, S_2,$ and $S_3$ are given a problem to solve. Consider the following events:
U: At least one of $S_1, S_2,$ and $S_3$ can solve the problem,
V: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem,
W: $S_2$ can solve the problem and $S_3$ cannot solve the problem,
T: $S_3$ can solve the problem.
For any event $E$, let $P(E)$ denote the probability of $E$. If
$P(U) = \dfrac{1}{2}$ , $P(V) = \dfrac{1}{10}$ , and $P(W) = \dfrac{1}{12}$,
then $P(T)$ is equal to
$\dfrac{13}{36}$
$\dfrac{1}{3}$
$\dfrac{19}{60}$
$\dfrac{1}{4}$
A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\frac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is :
Box-I contains 8 red, 3 blue and 5 green balls,
Box-II contains 24 red, 9 blue and 15 green balls,
Box-III contains 1 blue, 12 green and 3 yellow balls,
Box-IV contains 10 green, 16 orange and 6 white balls.
A ball is chosen randomly from Box-I; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to
Two players, $P_{1}$ and $P_{2}$, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let $x$ and $y$ denote the readings on the die rolled by $P_{1}$ and $P_{2}$, respectively. If $x>y$, then $P_{1}$ scores 5 points and $P_{2}$ scores 0 point. If $x=y$, then each player scores 2 points. If $x < y$, then $P_{1}$ scores 0 point and $P_{2}$ scores 5 points. Let $X_{i}$ and $Y_{i}$ be the total scores of $P_{1}$ and $P_{2}$, respectively, after playing the $i^{\text {th }}$ round.
| List-I | List-II |
|---|---|
| (I) Probability of $\left(X_{2} \geq Y_{2}\right)$ is | (P) $\frac{3}{8}$ |
| (II) Probability of $\left(X_{2}>Y_{2}\right)$ is | (Q) $\frac{11}{16}$ |
| (III) Probability of $\left(X_{3}=Y_{3}\right)$ is | (R) $\frac{5}{16}$ |
| (IV) Probability of $\left(X_{3}>Y_{3}\right)$ is | (S) $\frac{355}{864}$ |
| (T) $\frac{77}{432}$ |
The correct option is:
Let G2 = G1 $\cup$ S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.
Let E3 = E2 $\cup$ S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is
(There are two questions based on Paragraph "A", the question given below is one of them)
The probability that, on the examination day, the student S1 gets the previously allotted seat R1, and NONE of the remaining students gets the seat previously allotted to him/her is
(There are two questions based on Paragraph "A", the question given below is one of them)
For i = 1, 2, 3, 4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event ${T_1} \cap {T_2} \cap {T_3} \cap {T_4}$ is
$\,\,\,\,P\,\left( {X > Y} \right)$ is
$P\,\left( {X = Y} \right)$ is
$ = 10P$ (computer turns out to be defective given that it is produced in plant ${T_2}$),
where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant ${T_2}$ is
The probability that ${x_1} + {x_2} + {x_3}$ is odd, is
The probability that ${x_1},$, ${x_2},$ ${x_3}$ are in an arithmetic progression, is
If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from box ${B_2}$ is
If $1$ ball is drawn from each of the boxex ${B_1},$ ${B_2}$ and ${B_3},$ the probability that all $3$ drawn balls are of the same colour is
Given that the drawn ball from ${U_2}$ is white, the probability that head appeared on the coin is