Mathematics Probability & Statistics questions from JEE Main 2014.
A number $\mathrm{x}$ is chosen at random from the set $\{1$, $2,3,4, \ldots ., 100\}$. Define the event: $\mathrm{A}=$ the chosen number $x$ satisfies $\frac{(x-10)(x-50)}{(x-30)} \geq 0$ Then $\mathrm{P}(\mathrm{A})$ is:
A set $\mathrm{S}$ contains 7 elements. A non-empty subset $A$ of $S$ and an element $x$ of $S$ are chosen at random. Then the probability that $\mathrm{x} \in \mathrm{A}$ is:
If $A$and $B$ are two events such that $P(A\cup B)=P(A\cap B)$, then the incorrect statement amongst the following statements is :
If $X$ has a binomial distribution, $B(n, p)$ with parameters $n$ and $p$ such that $P(X=2)=P(X=3)$, then $\mathrm{E}(\mathrm{X})$, the mean of variable $\mathrm{X}$, is
In a set of $2n$ distinct observations, each of the observation below the median of all the observations is increased by $5$ and each of the remaining observations is decreased by $3$. Then, the mean of the new set of observations :
Let $A$ and $E$ be any two events with positive probabilities Statement I: $P(E/A)\geq P(A/E)P(E).$ Statement II: $P(A/E)\geq P(A\cap E).$
Let $\bar{x}$, $M$ and ${\sigma }^{2}$ be respectively the mean, mode and variance of $n$ observations ${x}_{1},{x}_{2},....,{x}_{n}$ and ${d}_{i}=-{x}_{i}-a,i=1,2,....,n,$ where $a$ is any number. Statement I: Variance of ${d}_{1},{d}_{2},...,{d}_{n}$ is ${\sigma }^{2}$. Statement II: Mean and mode of ${d}_{1},{d}_{2},....,{d}_{n}$ are $-\bar{x}-a$ and $-M-a$, respectively.
Let $A$ and $B$ be two events such that $P(\bar{A\cup B})=\frac{1}{6},P(A\cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{1}{4},$ where $\bar{A}$ stands for the complement of the event $A$. Then the events $A$ and $B$ are
Let $\bar{X}$ and M.D. be the mean and the mean deviation about $\bar{X}$ of $n$ observations $\mathrm{x}_{\mathrm{i}}, \mathrm{i}=1,2$,n. If each of the observations is increased by 5 , then the new mean and the mean deviation about the new mean, respectively, are :
The variance of the first $50$ even natural numbers is :