Probability
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, $\frac{b+c}{a} = 12$.
The random variable $\mathrm{X}$ follows binomial distribution $\mathrm{B}(\mathrm{n}, \mathrm{p})$, for which the difference of the mean and the variance is 1 . If $2 \mathrm{P}(\mathrm{X}=2)=3 \mathrm{P}(\mathrm{X}=1)$, then $n^{2} \mathrm{P}(\mathrm{X}>1)$ is equal to :
A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $\mathrm{X}$ denotes the number of tosses of the coin, then the mean of $\mathrm{X}$ is :
Two dice A and B are rolled. Let the numbers obtained on A and B be $\alpha$ and $\beta$ respectively. If the variance of $\alpha-\beta$ is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to :
Let $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$ be a sample space and $A=\{M \in S: M$ is invertible $\}$ be an event. Then $P(A)$ is equal to :
Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$, then $\mathrm{k}$ is equal to :
Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that ${2^N} < N!$ is ${m \over n}$, where m and n are coprime, then $4m-3n$ is equal to :
If the probability that the random variable $\mathrm{X}$ takes values $x$ is given by $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 3^{-x}, x=0,1,2,3, \ldots$, where $\mathrm{k}$ is a constant, then $\mathrm{P}(\mathrm{X} \geq 2)$ is equal to :
In a bolt factory, machines $A, B$ and $C$ manufacture respectively $20 \%, 30 \%$ and $50 \%$ of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine $C$ is :
Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then $q-p$ is equal to :
A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3^{11}}$, then $k$ is equal to :
Two dice are thrown independently. Let $\mathrm{A}$ be the event that the number appeared on the $1^{\text {st }}$ die is less than the number appeared on the $2^{\text {nd }}$ die, $\mathrm{B}$ be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and $\mathrm{C}$ be the event that the number appeared on the $1^{\text {st }}$ die is odd and that on the $2^{\text {nd }}$ is even. Then :
In a binomial distribution $B(n,p)$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $6(n+p-q)$ is equal to :
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :
If an unbiased die, marked with $-2,-1,0,1,2,3$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :
Let $\mathrm{S} = \{ {w_1},{w_2},......\} $ be the sample space associated to a random experiment. Let $P({w_n}) = {{P({w_{n - 1}})} \over 2},n \ge 2$. Let $A = \{ 2k + 3l:k,l \in N\} $ and $B = \{ {w_n}:n \in A\} $. Then P(B) is equal to :
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :
Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N-2,\sqrt{3N},N+2$ are in geometric progression be $\frac{k}{48}$. Then the value of k is :
Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = \left\{ {x \in \mathbb{Z}:x(66 - x) \ge {5 \over 9}M} \right\}$ and the event $\mathrm{A = \{ x \in S:x\,is\,a\,multiple\,of\,3\}}$. Then P(A) is equal to :
Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations
$x + y + z = 1$
$2x + \mathrm{N}y + 2z = 2$
$3x + 3y + \mathrm{N}z = 3$
has unique solution is ${k \over 6}$, then the sum of value of k and all possible values of N is :
Let $\Omega$ be the sample space and $\mathrm{A \subseteq \Omega}$ be an event.
Given below are two statements :
(S1) : If P(A) = 0, then A = $\phi$
(S2) : If P(A) = 1, then A = $\Omega$
Then :
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 $
Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :
Let $S=\{1,2,3, \ldots, 2022\}$. Then the probability, that a randomly chosen number n from the set S such that $\mathrm{HCF}\,(\mathrm{n}, 2022)=1$, is :
Let $\mathrm{A}$ and $\mathrm{B}$ be two events such that $P(B \mid A)=\frac{2}{5}, P(A \mid B)=\frac{1}{7}$ and $P(A \cap B)=\frac{1}{9} \cdot$ Consider
(S1) $P\left(A^{\prime} \cup B\right)=\frac{5}{6}$,
(S2) $P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}$
Then :
Out of $60 \%$ female and $40 \%$ male candidates appearing in an exam, $60 \%$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :
Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $P(X>n-3)=\frac{k}{2^{n}}$, then k is equal to :
A six faced die is biased such that
$3 \times \mathrm{P}($a prime number$)\,=6 \times \mathrm{P}($a composite number$)\,=2 \times \mathrm{P}(1)$.
Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :
Let $S$ be the sample space of all five digit numbers. It $p$ is the probability that a randomly selected number from $S$, is a multiple of 7 but not divisible by 5 , then $9 p$ is equal to :
Let $X$ be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$. Then, $54 \,P(X \leq 2)$ is equal to :
The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $\mathrm{P}(X=1)=\frac{4}{243}$, then $\mathrm{P}(X=4$ or 5$)$ is equal to :
Let $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$ be three mutually exclusive events such that $\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2+3 \mathrm{p}}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{2-\mathrm{p}}{8}$ and $\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1-\mathrm{p}}{2}$. If the maximum and minimum values of $\mathrm{p}$ are $\mathrm{p}_{1}$ and $\mathrm{p}_{2}$, then $\left(\mathrm{p}_{1}+\mathrm{p}_{2}\right)$ is equal to :
If $A$ and $B$ are two events such that $P(A)=\frac{1}{3}, P(B)=\frac{1}{5}$ and $P(A \cup B)=\frac{1}{2}$, then $P\left(A \mid B^{\prime}\right)+P\left(B \mid A^{\prime}\right)$ is equal to :
If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :
If the numbers appeared on the two throws of a fair six faced die are $\alpha$ and $\beta$, then the probability that $x^{2}+\alpha x+\beta>0$, for all $x \in \mathbf{R}$, is :
If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then ${{P(X = 2)} \over {P(X = 3)}}$ is equal to :
The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal to :
The probability that a randomly chosen 2 $\times$ 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :
The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) $-$ f(c) = f(d) is :