Probability · Mathematics · JEE Advanced

A factory has a total of three manufacturing units, $M_1, M_2$, and $M_3$, which produce bulbs independent of each other. The units $M_1, M_2$, and $M_3$ produce bulbs in the proportions of $2: 2: 1$, respectively. It is known that $20 \%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1, 15 \%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$.

If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is __________.

Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in \mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is _____________.

Then $${{\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_1}} \over {\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_3}}}$$

Three students $S_1, S_2,$ and $S_3$ are given a problem to solve. Consider the following events:

U: At least one of $S_1, S_2,$ and $S_3$ can solve the problem,

V: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem,

W: $S_2$ can solve the problem and $S_3$ cannot solve the problem,

T: $S_3$ can solve the problem.

For any event $E$, let $P(E)$ denote the probability of $E$. If

$P(U) = \dfrac{1}{2}$ , $P(V) = \dfrac{1}{10}$ , and $P(W) = \dfrac{1}{12}$,

then $P(T)$ is equal to

A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\frac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is :

Box-I contains 8 red, 3 blue and 5 green balls,

Box-II contains 24 red, 9 blue and 15 green balls,

Box-III contains 1 blue, 12 green and 3 yellow balls,

Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

Two players, $$P_{1}$$ and $$P_{2}$$, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let $$x$$ and $$y$$ denote the readings on the die rolled by $$P_{1}$$ and $$P_{2}$$, respectively. If $$x>y$$, then $$P_{1}$$ scores 5 points and $$P_{2}$$ scores 0 point. If $$x=y$$, then each player scores 2 points. If $$x < y$$, then $$P_{1}$$ scores 0 point and $$P_{2}$$ scores 5 points. Let $$X_{i}$$ and $$Y_{i}$$ be the total scores of $$P_{1}$$ and $$P_{2}$$, respectively, after playing the $$i^{\text {th }}$$ round.

The correct option is:

$$P\,\left( {X = Y} \right)$$ is

$$\,\,\,\,P\,\left( {X > Y} \right)$$ is

The probability that $${x_1} + {x_2} + {x_3}$$ is odd, is

The probability that $${x_1},$$, $${x_2},$$ $${x_3}$$ are in an arithmetic progression, is

If $$1$$ ball is drawn from each of the boxex $${B_1},$$ $${B_2}$$ and $${B_3},$$ the probability that all $$3$$ drawn balls are of the same colour is

If $$2$$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $$2$$ balls are drawn from box $${B_2}$$ is

The probability of the drawn ball from $${U_2}$$ being white is

Given that the drawn ball from $${U_2}$$ is white, the probability that head appeared on the coin is

The conditional probability that $$X\ge6$$ given $$X>3$$ equals :

The probability that $$X\ge3$$ equals :

An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent is :

STATEMENT-2: $$\sum\limits_{i = 1}^n {P\left( {{H_i}} \right)} = 1.$$

If $$P\left( {{u_i}} \right) = c,$$ (a constant) then $$P\left( {{u_n}/w} \right) = $$

If $$P\left( {{u_i}} \right) \propto i,\,$$ where $$i=1,2,3,.......,n,$$ then $$\mathop {\lim }\limits_{n \to \infty } P\left( w \right) = $$

Let $$P\left( {{u_i}} \right) = {1 \over n},$$ if $$n$$ is even and $$E$$ denotes the event of choosing even numbered urn, then the value of $$P\left( {w/E} \right)$$ is

One of the two boxes, box $${\rm I}$$ and box $${\rm I}{\rm I},$$ was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $${\rm I}{\rm I}$$ is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ $${n_2},$$ $${n_3}$$ and $${n_4}$$ is (are)

A ball is drawn at random from box $${\rm I}$$ and transferred to box $${\rm I}$$$${\rm I}.$$ If the probability of drawing a red ball from box $${\rm I},$$ after this transfer, is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ and $${n_2}$$ is(are)

If $$P\left( {A \cup B \cup C} \right) \ge 0.75,$$ then show that $$P$$ $$(BC)$$ lies in the interval $$0.23 \le x \le 0.48$$

This die tossed and you are told that either face $$1$$ or face $$2$$ has turned up. Then the probability that it is face $$1$$ is ...............