We fix $n \in \mathbf { N } ^ { * }$ and draw successively and with replacement two integers $p$ and $q$ according to a uniform distribution on $\llbracket 1 , n \rrbracket$. We define the events:

• $E _ { n }$: "We obtain $( p , q ) \in E _ { 1 } \cup E _ { 2 } \cup E _ { 3 }$".
• $A _ { n }$: "We obtain $p = q$".
• $B _ { n }$: "We obtain $q > p$ and $q$ is divisible by $p$".
• $C _ { n }$: "We obtain $p > q$".

where $E_1 = \{(p,q)\in(\mathbf{N}^*)^2: p=q\}$, $E_2 = \{(p,q)\in(\mathbf{N}^*)^2: pq\}$.
Justify that the set $\left\{ A _ { n } , B _ { n } , C _ { n } \right\}$ forms a partition of $E _ { n }$.

We fix $n \in \mathbf { N } ^ { * }$ and draw successively and with replacement two integers $p$ and $q$ according to a uniform distribution on $\llbracket 1 , n \rrbracket$. We define the events:

• $A _ { n }$: "We obtain $p = q$".
• $C _ { n }$: "We obtain $p > q$".

Calculate $\mathbf { P } \left( A _ { n } \right)$ then $\mathbf { P } \left( C _ { n } \right)$.

We fix $n \in \mathbf { N } ^ { * }$ and draw successively and with replacement two integers $p$ and $q$ according to a uniform distribution on $\llbracket 1 , n \rrbracket$. We define the events:

• $A _ { n }$: "We obtain $p = q$".
• $B _ { n }$: "We obtain $q > p$ and $q$ is divisible by $p$".

Show that $$\mathbf { P } \left( B _ { n } \right) = \frac { 1 } { n ^ { 2 } } \sum _ { p = 1 } ^ { n } \left\lfloor \frac { n } { p } \right\rfloor - \frac { 1 } { n } ,$$ and deduce $\mathbf { P } \left( A _ { n } \cup B _ { n } \right)$.

We fix $n \in \mathbf { N } ^ { * }$ and draw successively and with replacement two integers $p$ and $q$ according to a uniform distribution on $\llbracket 1 , n \rrbracket$. The event $E_n$ is defined as "We obtain $(p,q) \in E_1 \cup E_2 \cup E_3$".
Using the result $\mathbf{P}(A_n \cup B_n) \sim \dfrac{\ln n}{n}$ as $n \to +\infty$, deduce $$\lim _ { n \rightarrow + \infty } \mathbf { P } \left( E _ { n } \right) .$$

In a win-or-lose game, the winner gets 2 points whereas the loser gets 0. Six players A, B, C, D, E and F play each other in a preliminary round from which the top three players move to the final round. After each player has played four games, A has 6 points, B has 8 points and C has 4 points. It is also known that E won against F. In the next set of games D, E and F win their games against A, B and C respectively. If A, B and D move to the final round, the final scores of E and F are, respectively,
(A) 4 and 2
(B) 2 and 4
(C) 2 and 2
(D) 4 and 4

In a win-or-lose game, the winner gets 2 points whereas the loser gets 0. Six players A, B, C, D, E and F play each other in a preliminary round from which the top three players move to the final round. After each player has played four games, A has 6 points, B has 8 points and C has 4 points. It is also known that E won against F. In the next set of games D, E and F win their games against A, B and C respectively. If A, B and D move to the final round, the final scores of E and F are, respectively,
(A) 4 and 2
(B) 2 and 4
(C) 2 and 2
(D) 4 and 4

A party is attended by twenty people. In any subset of four people, there is at least one person who knows the other three (we assume that if $X$ knows $Y$, then $Y$ knows $X$). Suppose there are three people in the party who do not know each other. How many people in the party know everyone?
(A) 16
(B) 17
(C) 18
(D) Cannot be determined from the given data.

Let $j$ be a number selected at random from $\{1, 2, \ldots, 2024\}$. What is the probability that $j$ is divisible by 9 and 15?
(A) $\frac{1}{23}$
(B) $\frac{1}{46}$
(C) $\frac{1}{44}$
(D) $\frac{1}{253}$

Consider the system of equations $a x + b y = 0 , c x + d y = 0$, where $a , b , c , d \in \{ 0,1 \}$. STATEMENT-1 : The probability that the system of equations has a unique solution is $\frac { 3 } { 8 }$. and STATEMENT-2 : The probability that the system of equations has a solution is 1.
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True

Four persons independently solve a certain problem correctly with probabilities $\frac { 1 } { 2 } , \frac { 3 } { 4 } , \frac { 1 } { 4 } , \frac { 1 } { 8 }$. Then the probability that the problem is solved correctly by at least one of them is
(A) $\frac { 235 } { 256 }$
(B) $\frac { 21 } { 256 }$
(C) $\frac { 3 } { 256 }$
(D) $\frac { 253 } { 256 }$

A box $B _ { 1 }$ contains 1 white ball, 3 red balls and 2 black balls. Another box $B _ { 2 }$ contains 2 white balls, 3 red balls and 4 black balls. A third box $B _ { 3 }$ contains 3 white balls, 4 red balls and 5 black balls.
If 1 ball is drawn from each of the boxes $B _ { 1 } , B _ { 2 }$ and $B _ { 3 }$, the probability that all 3 drawn balls are of the same colour is
(A) $\frac { 82 } { 648 }$
(B) $\frac { 90 } { 648 }$
(C) $\frac { 558 } { 648 }$
(D) $\frac { 566 } { 648 }$

Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{2}{3}$
(D) $\frac{3}{4}$

Box 1 contains three cards bearing numbers $1,2,3$; box 2 contains five cards bearing numbers $1,2,3,4,5$; and box 3 contains seven cards bearing numbers $1,2,3,4,5,6,7$. A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i^{\text{th}}$ box, $i = 1,2,3$.
The probability that $x_1 + x_2 + x_3$ is odd, is
(A) $\frac{29}{105}$
(B) $\frac{53}{105}$
(C) $\frac{57}{105}$
(D) $\frac{1}{2}$

One of the two boxes, box I and box II, 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 II is $\frac { 1 } { 3 }$, then the correct option(s) with the possible values of $n _ { 1 } , n _ { 2 } , n _ { 3 }$ and $n _ { 4 }$ is(are)
(A) $n _ { 1 } = 3 , n _ { 2 } = 3 , n _ { 3 } = 5 , n _ { 4 } = 15$
(B) $n _ { 1 } = 3 , n _ { 2 } = 6 , n _ { 3 } = 10 , n _ { 4 } = 50$
(C) $n _ { 1 } = 8 , n _ { 2 } = 6 , n _ { 3 } = 5 , n _ { 4 } = 20$
(D) $n _ { 1 } = 6 , n _ { 2 } = 12 , n _ { 3 } = 5 , n _ { 4 } = 20$

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is $\frac { 1 } { 3 }$, then the correct option(s) with the possible values of $n _ { 1 }$ and $n _ { 2 }$ is(are)
(A) $\quad n _ { 1 } = 4$ and $n _ { 2 } = 6$
(B) $\quad n _ { 1 } = 2$ and $n _ { 2 } = 3$
(C) $n _ { 1 } = 10$ and $n _ { 2 } = 20$
(D) $n _ { 1 } = 3$ and $n _ { 2 } = 6$

Let $S = \left\{x \in (-\pi, \pi) : x \neq 0, \pm\frac{\pi}{2}\right\}$. The sum of all distinct solutions of the equation $\sqrt{3}\sec x + \operatorname{cosec} x + 2(\tan x - \cot x) = 0$ in the set $S$ is equal to
(A) $-\frac{7\pi}{9}$
(B) $-\frac{2\pi}{9}$
(C) $0$
(D) $\frac{5\pi}{9}$

A computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20\%$ and plant $T_2$ produces $80\%$ of the total computers produced. $7\%$ of computers produced in the factory turn out to be defective. It is known that $P$(computer turns out to be defective given that it is produced in plant $T_1$) $= 10P$(computer turns out to be defective given that it is produced in plant $T_2$), where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is
(A) $\frac{36}{73}$
(B) $\frac{47}{79}$
(C) $\frac{78}{93}$
(D) $\frac{75}{83}$

Three randomly chosen nonnegative integers $x , y$ and $z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is
[A] $\frac { 36 } { 55 }$
[B] $\frac { 6 } { 11 }$
[C] $\frac { 1 } { 2 }$
[D] $\frac { 5 } { 11 }$

Let $|X|$ denote the number of elements in a set $X$. Let $S = \{1,2,3,4,5,6\}$ be a sample space, where each element is equally likely to occur. If $A$ and $B$ are independent events associated with $S$, then the number of ordered pairs $(A, B)$ such that $1 \leq |B| < |A|$, equals\_\_\_\_

Let $E$, $F$ and $G$ be three events having probabilities $$P(E) = \frac{1}{8}, \quad P(F) = \frac{1}{6}, \quad P(G) = \frac{1}{4},$$ and let $P(E \cap F \cap G) = \frac{1}{10}$.
For any event $H$, if $P(H^c)$ denotes its complement, then which of the following statements is(are) TRUE?
(A) $P(E \cap F \cap G^c) \leq \frac{1}{40}$
(B) $P(E^c \cap F \cap G) \leq \frac{1}{15}$
(C) $P(E \cup F \cup G) \leq \frac{13}{24}$
(D) $P(E^c \cap F^c \cap G^c) \leq \frac{5}{12}$

A number is chosen at random from the set $\{ 1,2,3 , \ldots , 2000 \}$. Let $p$ be the probability that the chosen number is a multiple of 3 or a multiple of 7 . Then the value of $500 p$ is $\_\_\_\_$.

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.
List-I (I) Probability of $\left( X _ { 2 } \geq Y _ { 2 } \right)$ is (II) Probability of $\left( X _ { 2 } > Y _ { 2 } \right)$ is (III) Probability of $\left( X _ { 3 } = Y _ { 3 } \right)$ is (IV) Probability of $\left( X _ { 3 } > Y _ { 3 } \right)$ is
List-II (P) $\frac { 3 } { 8 }$ (Q) $\frac { 11 } { 16 }$ (R) $\frac { 5 } { 16 }$ (S) $\frac { 355 } { 864 }$ (T) $\frac { 77 } { 432 }$
The correct option is:
(A) (I) → (Q); (II) → (R); (III) → (T); (IV) → (S)
(B) (I) → (Q); (II) → (R); (III) → (T); (IV) → (T)
(C) (I) → (P); (II) → (R); (III) → (Q); (IV) → (S)
(D) (I) → (P); (II) → (R); (III) → (Q); (IV) → (T)

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\frac { 1 } { 3 }$, then the probability that the experiment stops with head is
(A) $\frac { 1 } { 3 }$
(B) $\frac { 5 } { 21 }$
(C) $\frac { 4 } { 21 }$
(D) $\frac { 2 } { 7 }$

Consider the $6 \times 6$ square in the figure. Let $A _ { 1 } , A _ { 2 } , \ldots , A _ { 49 }$ be the points of intersections (dots in the picture) in some order. We say that $A _ { i }$ and $A _ { j }$ are friends if they are adjacent along a row or along a column. Assume that each point $A _ { i }$ has an equal chance of being chosen.
Two distinct points are chosen randomly out of the points $A _ { 1 } , A _ { 2 } , \ldots , A _ { 49 }$. Let $p$ be the probability that they are friends. Then the value of $7p$ 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:
(1) $P ( A ) + P ( B ) = 1$
(2) $P \left( A \cap B ^ { \prime } \right) = 0$
(3) $A \& B$ are equally likely
(4) $P \left( A ^ { \prime } \cap B \right) = 0$