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) $\frac{1}{2}$

The probability that $x_1, x_2, x_3$ are in an arithmetic progression, is
(A) $\frac{9}{105}$
(B) $\frac{10}{105}$
(C) $\frac{11}{105}$
(D) $\frac{7}{105}$

Let $X$ and $Y$ be two events such that $P(X) = \frac{1}{3}$, $P(X \mid Y) = \frac{1}{2}$ and $P(Y \mid X) = \frac{2}{5}$. Then
[A] $P(Y) = \frac{4}{15}$
[B] $P(X' \mid Y) = \frac{1}{2}$
[C] $P(X \cap Y) = \frac{1}{5}$
[D] $P(X \cup Y) = \frac{2}{5}$

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] $\frac { 5 } { 11 }$

There are three bags $B _ { 1 } , B _ { 2 }$ and $B _ { 3 }$. The bag $B _ { 1 }$ contains 5 red and 5 green balls, $B _ { 2 }$ contains 3 red and 5 green balls, and $B _ { 3 }$ contains 5 red and 3 green balls. Bags $B _ { 1 } , B _ { 2 }$ and $B _ { 3 }$ have probabilities $\frac { 3 } { 10 } , \frac { 3 } { 10 }$ and $\frac { 4 } { 10 }$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
(A) Probability that the chosen ball is green, given that the selected bag is $B _ { 3 }$, equals $\frac { 3 } { 8 }$
(B) Probability that the chosen ball is green equals $\frac { 39 } { 80 }$
(C) Probability that the selected bag is $B _ { 3 }$, given that the chosen ball is green, equals $\frac { 5 } { 13 }$
(D) Probability that the selected bag is $B _ { 3 }$ and the chosen ball is green equals $\frac { 3 } { 10 }$

Let $S$ be the sample space of all $3 \times 3$ matrices with entries from the set $\{ 0,1 \}$. Let the events $E _ { 1 }$ and $E _ { 2 }$ be given by $$\begin{aligned} & E _ { 1 } = \{ A \in S : \operatorname { det } A = 0 \} \text { and } \\ & E _ { 2 } = \{ A \in S : \text { sum of entries of } A \text { is } 7 \} . \end{aligned}$$ If a matrix is chosen at random from $S$, then the conditional probability $P \left( E _ { 1 } \mid E _ { 2 } \right)$ equals $\_\_\_\_$

The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is $\_\_\_\_$

Two fair dice, each with faces numbered $1, 2, 3, 4, 5$ and $6$, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number, then the value of $14p$ is $\_\_\_\_$

Consider three sets $E_1 = \{1,2,3\}$, $F_1 = \{1,3,4\}$ and $G_1 = \{2,3,4,5\}$. Two elements are chosen at random, without replacement, from the set $E_1$, and let $S_1$ denote the set of these chosen elements. Let $E_2 = E_1 \setminus S_1$ and $F_2 = F_1 \cup S_1$. Now two elements are chosen at random, without replacement, from the set $F_2$ and let $S_2$ denote the set of these chosen elements.
Let $G_2 = G_1 \cup S_2$. The value of $P(E_2 = F_2)$ is
(A) $\frac{1}{7}$
(B) $\frac{3}{7}$
(C) $\frac{1}{5}$
(D) $\frac{2}{7}$

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.
Let $p _ { i }$ be the probability that a randomly chosen point has $i$ many friends, $i = 0,1,2,3,4$. Let $X$ be a random variable such that for $i = 0,1,2,3,4$, the probability $P ( X = i ) = p _ { i }$. Then the value of $7 E ( X )$ is

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

A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For $i = 1,2,3$, let $W _ { i } , G _ { i }$, and $B _ { i }$ denote the events that the ball drawn in the $i ^ { \text {th } }$ draw is a white ball, green ball, and blue ball, respectively. If the probability $P \left( W _ { 1 } \cap G _ { 2 } \cap B _ { 3 } \right) = \frac { 2 } { 5 N }$ and the conditional probability $P \left( B _ { 3 } \mid W _ { 1 } \cap G _ { 2 } \right) = \frac { 2 } { 9 }$, then $N$ equals $\_\_\_\_$ .

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) $P \left( A ^ { \prime } \cap B \right) = 0$

If two different numbers are taken from the set $\{ 0 , 1 , 2 , 3 , \ldots , 10 \}$; then the probability that their sum as well as absolute difference are both multiple of 4, is:
(1) $\frac { 6 } { 55 }$
(2) $\frac { 12 } { 55 }$
(3) $\frac { 14 } { 45 }$
(4) $\frac { 7 } { 55 }$

For three events $A$, $B$ and $C$, $P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs}) = P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \dfrac{1}{4}$ and $P(\text{All the three events occur simultaneously}) = \dfrac{1}{16}$. Then the probability that at least one of the events occurs, is:
(1) $\dfrac{7}{32}$
(2) $\dfrac{7}{16}$
(3) $\dfrac{1}{64}$
(4) $\dfrac{3}{16}$

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is:
(1) $\frac { 127 } { 128 }$
(2) $\frac { 63 } { 64 }$
(3) $\frac { 255 } { 256 }$
(4) $\frac { 1 } { 2 }$

If two different numbers are taken from the set $\{0, 1, 2, 3, \ldots, 10\}$; then the probability that their sum as well as absolute difference are both multiples of 4, is:
(1) $\dfrac{6}{55}$
(2) $\dfrac{12}{55}$
(3) $\dfrac{14}{45}$
(4) $\dfrac{7}{55}$

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is:
(1) $\frac { 3 } { 4 }$
(2) $\frac { 3 } { 10 }$
(3) $\frac { 2 } { 5 }$
(4) $\frac { 1 } { 5 }$

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is:
(1) $\frac{21}{49}$
(2) $\frac{26}{49}$
(3) $\frac{32}{49}$
(4) $\frac{27}{49}$

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered $1,2,3 , \ldots , 9$ is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is
(1) $\frac { 13 } { 36 }$
(2) $\frac { 19 } { 72 }$
(3) $\frac { 15 } { 72 }$
(4) $\frac { 19 } { 36 }$

Four persons can hit a target correctly with probabilities $\frac { 1 } { 2 } , \frac { 1 } { 3 } , \frac { 1 } { 4 }$ and $\frac { 1 } { 8 }$ respectively. If all hit at the target independently, then the probability that the target would be hit, is
(1) $\frac { 25 } { 192 }$
(2) $\frac { 7 } { 32 }$
(3) $\frac { 1 } { 192 }$
(4) $\frac { 25 } { 32 }$