Positive integers $a$ and $b$, possibly equal, are chosen randomly from among the divisors of 400. The numbers $a, b$ are chosen independently, each divisor being equally likely to be chosen. Find the probability that $\gcd(a, b) = 1$ and $\text{lcm}(a, b) = 400$.

A bag contains 7 marbles, each labeled with a natural number from 2 to 8. When 2 marbles are drawn simultaneously from the bag, what is the probability that the two natural numbers on the drawn marbles are coprime? [3 points]
(1) $\frac { 8 } { 21 }$
(2) $\frac { 10 } { 21 }$
(3) $\frac { 4 } { 7 }$
(4) $\frac { 2 } { 3 }$
(5) $\frac { 16 } { 21 }$

5. From 7 integers from 2 to 8, two different numbers are randomly selected. The probability that these two numbers are coprime is
A. $\frac { 1 } { 6 }$
B. $\frac { 1 } { 3 }$
C. $\frac { 1 } { 2 }$
D. $\frac { 2 } { 3 }$

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 $\mathbf { P } \left( A _ { n } \cup B _ { n } \right)$.

148. From the set of consecutive integers $\{300, 301, \ldots, 51\}$... wait: $\{300, 301, \ldots, 51\}$, i.e., $\{300, 301, \ldots, 451\}$, a number is chosen at random. What is the probability that this number is divisible by 6 but not by 7 and not by 42?
(1) $0/24$ (2) $0/26$ (3) $0/28$ (4) $0/31$

145- A two-digit natural number is chosen at random. What is the probability that the chosen number is a multiple of 3 or 5?
\[ (1)\quad \frac{2}{5} \qquad (2)\quad \frac{3}{5} \qquad (3)\quad \frac{7}{15} \qquad (4)\quad \frac{8}{15} \]

130- We write the numbers 1 to 21 each on a card and place them in a bag. We then randomly draw two cards one after another without replacement and place them aside. The set of all possible outcomes forms set $A$. We select one number from set $A$. What is the probability that the selected number is divisible by 6?
(1) $\dfrac{13}{84}$ (2) $\dfrac{65}{417}$ (3) $\dfrac{11}{70}$ (4) $\dfrac{67}{417}$

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) $\frac{1}{253}$

25. If the integers $m$ and $n$ are chosen at random between 1 and 100 , then the probability that a number of the form $7 m + 7 n$ is divisible by 5 equals :
(A) $1 / 4$
(B) $1 / 7$
(C) $1 / 8$
(D) $1 / 49$

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DIRECTIONS : Question numbers $26 - 35$ carry 3 marks each and may have more than one correct answers. All correct answers must be marked to get any credit in these questions.

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) $\frac{1}{2}$

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] $\frac { 5 } { 11 }$

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 $\_\_\_\_$.

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) $\frac { 7 } { 55 }$

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) $\dfrac{7}{55}$

Let $A$ be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5 is:
(1) $\frac { 1 } { 5 }$
(2) $\frac { 122 } { 297 }$
(3) $\frac { 97 } { 297 }$
(4) $\frac { 2 } { 9 }$

Let $S$ be the sample space of all five digit numbers. If $p$ is the probability that a randomly selected number from $S$, is a multiple of 7 but not divisible by 5 , then $9 p$ is equal to
(1) 1.0146
(2) 1.2085
(3) 1.0285
(4) 1.1521

Let M be the maximum value of the product of two positive integers when their sum is 66 . Let the sample space $S = \left\{ x \in Z : x ( 66 - x ) \geq \frac { 5 } { 9 } M \right\}$ and the event $\mathrm { A } = \{ \mathrm { x } \in \mathrm { S } : \mathrm { x }$ is a multiple of $3 \}$. Then $\mathrm { P } ( \mathrm { A } )$ is equal to
(1) $\frac { 15 } { 44 }$
(2) $\frac { 1 } { 3 }$
(3) $\frac { 1 } { 5 }$
(4) $\frac { 7 } { 22 }$

Let the sum of two positive integers be 24. If the probability, that their product is not less than $\frac { 3 } { 4 }$ times their greatest possible product, is $\frac { m } { n }$, where $\operatorname { gcd } ( m , n ) = 1$, then $n - m$ equals
(1) 10
(2) 9
(3) 11
(4) 8

Two numbers $\mathrm { k } _ { 1 }$ and $\mathrm { k } _ { 2 }$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $\mathrm { i } ^ { \mathrm { k } _ { 1 } } + \mathrm { i } ^ { \mathrm { k } _ { 2 } } , ( \mathrm { i } = \sqrt { - 1 } )$ is non-zero, equals
(1) $\frac { 1 } { 2 }$
(2) $\frac { 3 } { 4 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 2 } { 3 }$

Q80. Let the sum of two positive integers be 24 . If the probability, that their product is not less than $\frac { 3 } { 4 }$ times their greatest possible product, is $\frac { m } { n }$, where $\operatorname { gcd } ( m , n ) = 1$, then $n - m$ equals
(1) 10
(2) 9
(3) 11
(4) 8