Probability & Statistics PYQ — Page 12

A random variable $X$ has the following probability distribution: $\begin{matrix}\begin{matrix}X: & 1 & \begin{matrix}2 & 3 & \begin{matrix}4 & 5\end{matrix}\end{matrix}\end{matrix} \\ P(X):\begin{matrix}{k}^{2} & \begin{matrix}2k & k & \begin{matrix}2k & 5{k}^{2}\end{matrix}\end{matrix}\end{matrix}\end{matrix}$ Then, $P(X>2)$ is equal to:

The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11,$ then the correct variance is

In a box, there are $20$ cards, out of which $10$ are labelled as $A$ and the remaining $10$ are labelled as $B$ . Cards are drawn at random, one after the other and with replacement, till a second $A$ card is obtained. The probability that the second $A$ card appears before the third $B$ card is:

Consider the data on $x$ taking the values $0,2,4,8,.....,{2}^{n}$ with frequencies $C0n,C1n,C2n,....,Cnn$ respectively. If the mean of this data is $\frac{728}{{2}^{n}}$, then $n$ is equal to ....... .

If the mean and variance of eight numbers $3,7,9,12,13,20,x$ and $y$ be $10$ and $25$ respectively, then $x\cdot y$ is equal to

Let $\mathrm{S}=\{1,2, \ldots . ., 20\}$. A subset $\mathrm{B}$ of $\mathrm{S}$ is said to be "nice", if the sum of the elements of $\mathrm{B}$ is 203 . Than the probability that a randomly chosen subset of $S$ is "nice" is :

A person throws two fair dice. He wins Rs. $15$ for throwing a doublet (same numbers on the two dice), wins Rs $12$ when the throw results in the sum of $9$ , and loses Rs. $6$ for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is:

If the standard deviation of the numbers $-1,0,1,k$ is $\sqrt{5}$ where $k>0,$ then $k$ is equal to

The mean and the median of the following ten numbers in increasing order $10, 22, 26, 29, 34, x, 42, 67, 70, y$ are $42$ and $35$ respectively, then $\frac{y}{x}$ is equal to:

The outcome of each of 30 items was observed; 10 items gave an outcome $\frac{1}{2}-\mathrm{d}$ each, 10 items gave outcome $\frac{1}{2}$ each and the remaining 10 items gave outcome $\frac{1}{2}+\mathrm{d}$ each. If the variance of this outcome data is $\frac{4}{3}$ then $|\mathrm{d}|$ equals:

If the probability of hitting a target by a shooter, in any shot is $\frac{1}{3},$ then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than $\frac{5}{6},$ is:

Four persons can hit a target correctly with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac{1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is

In a class of $60$ students, $40$ opted for NCC, $30$ opted for NSS and $20$ opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :

A data consists of $n$ observations: ${x}_{1}, {x}_{2},\ldots , {x}_{n}.$ If $\sum _{i=1}^{n}{({x}_{i}+1)}^{2}=9n$ and $\sum _{i=1}^{n}{({x}_{i}-1)}^{2}=5n$, then the standard deviation of this data is

Assume that each born child is equally likely to be a boy or girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is:

In a game, a man wins Rs. $100$ if he gets $5$ or $6$ on a throw of a fair die and loses Rs. $50$ for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least $90%$ is:

The mean of five observations is $5$ and their variance is $9.20.$ If three of the given five observations are $1,3$ and $8,$ then a ratio of other two observations is

Two newspapers $A$ and $B$ are published in a city. It is known that $25%$ of the city population reads $A$ and $20%$ reads $B$ while $8%$ reads both $A$ and $B.$ Further, $30%$ of those who read $A$ but not $B$ look into advertisements and $40%$ of those who read $B$ but not $A$ also look into advertisements, while $50%$ of those who read both $A$ and $B$ look into advertisements. Then the percentage of the population who look into advertisements is:

If for some $x\in R$, the frequency distribution of the marks obtained by $20$ students in a test is: <table class="pyq-table"><tbody><tr><td>Marks</td><td>2</td><td>3</td><td>5</td><td>7</td></tr><tr><td>Frequency distribution</td><td>${(x+1)}^{2}$</td><td>$(2x-5)$</td><td>${x}^{2}-3x$</td><td>$x$</td></tr></tbody></table>Then the mean of the marks is :

Let $A$ and $B$ be two non-null events such that $A\subset B.$ Then, which of the following statements is always correct?

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If $X$ be the number of white balls drawn, then $\left(\frac{\text { mean of } \mathrm{X}}{\text { standard deviation of } \mathrm{X}}\right)$ is equal to:

If the mean and standard deviation of $5$ observations ${x}_{1},{x}_{2}, {x}_{3}, {x}_{4}, {x}_{5}$ are $10$ and $3,$ respectively, then the variance of $6$ observations ${x}_{1}, {x}_{2}, \ldots , {x}_{5}$ and $-50$ is equal to

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: