JEE Main Mathematics — Probability & Statistics previous year questions with solutions.
The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2$. then their new variance is equal to:
Two dice are thrown independently. Let $A$ 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, $B$ be the event that the number appeared on the ${1}^{\text{st }}$ die is even and that on the second die is odd, and $C$ 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
Let $N$ denote the sum of the numbers obtained when two dice are rolled. If the probability that ${2}^{N}<N!$ is $\frac{m}{n}$ where $m$ and $n$ are coprime, then $4m-3n$ is equal to
If the probability that the random variable $X$ takes values $x$ is given by $P(X=x)=k(x+1){3}^{-x}$, $x=0,1,2,3,\ldots \ldots$, where $k$ is a constant, then $P(X\geq 2)$ is equal to
The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is
The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and ${\sigma }^{2}$ respectively. If the variance of all the $30$ numbers in the two sets is $13$, then ${\sigma }^{2}$ is equal to
Let the mean and variance of $8$ numbers $x,y,10,12,6,12,4,8$ be $9$ and $9.25$ respectively. If $x>y,$ then $3x-2y$ is equal to $_______$
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
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
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
Five numbers ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$ are randomly selected from the numbers$1,2,3,\ldots \ldots ,18$ and are arranged in the increasing order $({x}_{1}<{x}_{2}<{x}_{1}<{x}_{4}<{x}_{2})$. The probability that ${x}_{2}=7$ and ${x}_{4}=11$ is
A six faced die is biased such that $3\times P$(a prime number)$=6\times P$(a composite number)$=2\times P(1)$. 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
If P(A)=0.4, P(B)=0.5 and P(A∩B)=0.2 then P(A|B) is:
If the mean deviation about the mean of the numbers $1,2,3,\ldots \ldots ,n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to ______.
Let $A$ and $B$ be two events such that $P(B\mid A)=\frac{2}{5}$, $P(A\mid B)=\frac{1}{7}$ and $P(A\cap B)=\frac{1}{9}$. Consider $(S1)P({A}^{'}\cup B)=\frac{5}{6}$, $(S2)P({A}^{'}\cap {B}^{'})=\frac{1}{18}$. Then
Let the mean and the variance of $5$ observations ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $(4a+{x}_{5})$ is equal to
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 $x\in R$, is
Let ${E}_{1}$ and ${E}_{2}$ be two events such that the conditional probabilities $P({E}_{1}\mid {E}_{2})=\frac{1}{2}$, $P({E}_{2}\mid {E}_{1})=\frac{3}{4}$ and $P({E}_{1}\cap {E}_{2})=\frac{1}{8}$. Then
Suppose a class has $7$ students. The average marks of these students in the mathematics examination is $62$, and their variance is $20$. A student fails in the examination if he/she gets less than $50$ marks, then in worst case, the number of students can fail is
The mean and variance of the data $4,5,6,6,7,8,x,y$ where $x<y$ are $6$ and $\frac{9}{4}$ respectively. Then ${x}^{4}+{y}^{2}$ is equal to
If the mean deviation about median for the number $3,5,7,2k,12,16,21,24$ arranged in the ascending order, is $6$ then the median is
A bag contains $4$ white and $6$ black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If ${\sigma }^{2}$ is the variance of $X$, then $100{\sigma }^{2}$ is equal to
The number of values of $a\in N$ such that the variance of $3,7,12,a,43-a$ is a natural number is:
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 $\frac{3}{4}$ and the remaining $6$ questions correctly with probability $\frac{1}{4}$. If the probability that the student guesses the answers of exactly $8$ questions correctly out of $10$ is $\frac{27k}{{4}^{10}}$, then $k$ is equal to