JEE Main 2018 — Probability & Statistics

A box ' $A$ ' contanis 2 white, 3 red and 2 black balls. Another box ' $B^{\prime}$ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ' $B^{\prime}$ is

A player $X$ has a biased coin whose probability of showing heads is $p$ and a player $Y$ has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If $X$ starts the game, and the probability of winning the game by both the players is equal, then the value of ' $p$ ' is

If $\sum _{i=1}^{9}({x}_{i}-5)=9$ and $\sum _{i=1}^{9}{({x}_{i}-5)}^{2}=45$, then the standard deviation of the $9$ items ${x}_{1}, {x}_{2},\ldots .,{x}_{9}$ is

If the mean of the data: $7,8,9,7,8,7, \lambda, 8$ is 8 , then the variance of this data is

Let $A, B$ and $C$ be three events, which are pair-wise independent and $\bar{E}$ denotes the complement of an event $E$. If $P(A\cap B\cap C)=0$ and $P(C)>0,$ then $P[(\bar{A}\cap \bar{B})|C]$ is equal to

The mean and the standard deviation $(S.D.)$ of five observations are $9$ and $0$, respectively. If one of the observation is increased such that the mean of the new set of five observations becomes $10$, then their $S.D.$ is

The mean of a set of 30 observations is 75 . If each other observation is multiplied by a nonzero number $\lambda$ and then each of them is decreased by 25 , their mean remains the same. The $\lambda$ is equal to equal to $\{0\}$

The mean of a set of $30$ observation is $75$. If each observations is multiplied by non-zero number $\lambda$ and then each of them is decreased by $25$, their mean remains the same. Then, $\lambda$ is equal to :

Two different families $A$ and $B$ are blessed with equal number of children. There are $3$ tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family $B$ is $\frac{1}{12},$ then the number of children in each family is