For each of the five questions in this exercise, only one of the four proposed answers is correct. No justification is required. A wrong answer, a multiple answer or the absence of an answer to a question neither awards nor deducts points.
We consider L a list of numbers consisting of consecutive terms of an arithmetic sequence with first term 7 and common difference 3, the last number in the list is 2023, namely: $$\mathrm{L} = [7, 10, \ldots, 2023].$$
Question 1: The number of terms in this list is:

Answer A	Answer B	Answer C	Answer D
2023	673	672	2016

Question 2: We choose a number at random from this list. The probability of drawing an even number is:

Answer A	Answer B	Answer C	Answer D
$\frac{1}{2}$	$\frac{34}{673}$	$\frac{336}{673}$	$\frac{337}{673}$

We choose a number at random from this list. We are interested in the following events:

• Event $A$: ``obtain a multiple of 4''
• Event $B$: ``obtain a number whose units digit is 6''

We are given $p(A \cap B) = \frac{34}{673}$.
Question 3: The probability of obtaining a multiple of 4 having 6 as the units digit is:

Answer A	Answer B	Answer C	Answer D
$\frac{168}{673} \times \frac{34}{673}$	$\frac{34}{673}$	$\frac{17}{84}$	$\frac{168}{34}$

Question 4: $P_B(A)$ is equal to:

Answer A	Answer B	Answer C	Answer D
$\frac{36}{168}$	$\frac{1}{2}$	$\frac{33}{168}$	$\frac{34}{67}$

Question 5: We choose, at random, successively, 10 elements from this list. An element can be chosen multiple times. The probability that none of these 10 numbers is a multiple of 4 is:

\begin{tabular}{ c } Answer A

$\left(\frac{505}{673}\right)^{10}$

&

Answer B

$1 - \left(\frac{505}{673}\right)^{10}$

&

Answer C

$\left(\frac{168}{673}\right)^{10}$

&

Answer D

$1 - \left(\frac{168}{673}\right)^{10}$

\hline \end{tabular}

This exercise is a multiple choice questionnaire. For each of the following questions, only one of the four proposed answers is correct. To answer, indicate on your paper the number of the question and the letter of the chosen answer. No justification is required.
A wrong answer, an absence of answer, or a multiple answer, neither gives nor removes points.
The 200 members of a club are girls or boys. These members practice rowing or basketball according to the distribution shown in the table below.

\cline { 2 - 4 } \multicolumn{1}{c|}{}	Rowing	Basketball	Total
Girls	25	80	105
Boys	50	45	95
Total	75	125	200

We choose a member at random and consider the following events: $F$ : the member is a girl; $A$ : the member practices rowing.

1. The probability of $F$ given $A$ is equal to : a. $\frac { 25 } { 100 }$ b. $\frac { 25 } { 75 }$ c. $\frac { 25 } { 105 }$ d. $\frac { 75 } { 105 }$
2. The probability of the event $A \cup F$ is equal to : a. $\frac { 9 } { 10 }$ b. $\frac { 1 } { 8 }$ c. $\frac { 31 } { 40 }$ d. $\frac { 5 } { 36 }$

To get to work, Albert can use either the bus or the train. The probability that the bus breaks down is equal to $b$. The probability that the train breaks down is equal to $t$. Bus and train breakdowns occur independently.
3. The probability $p _ { 1 }$ that the bus or the train breaks down is equal to : a. $p _ { 1 } = b t$ b. $p _ { 1 } = 1 - b t$ c. $p _ { 1 } = b + t$ d. $p _ { 1 } = b + t - b t$
4. The probability $p _ { 2 }$ that Albert can get to work is equal to : a. $p _ { 2 } = b t$ b. $p _ { 2 } = 1 - b t$ c. $p _ { 2 } = b + t$ d. $p _ { 2 } = b + t - b t$
5. We consider a coin for which the probability of obtaining HEADS is equal to $x$. We flip the coin $n$ times. The flips are independent. The probability $p$ of obtaining at least one HEADS in the $n$ flips is equal to a. $p = x ^ { n }$ b. $p = ( 1 - x ) ^ { n }$ c. $p = 1 - x ^ { n }$ d. $p = 1 - ( 1 - x ) ^ { n }$

There are four distinct balls labelled $1, 2, 3, 4$ and four distinct bins labelled A, B, C, D. The balls are picked up in order and placed into one of the four bins at random. Let $E_i$ denote the event that the first $i$ balls go into distinct bins. Calculate the following probabilities.
(i) $\Pr[E_4]$
(ii) $\Pr[E_4 \mid E_3]$
(iii) $\Pr[E_4 \mid E_2]$
(iv) $\Pr[E_3 \mid E_4]$.
Notation: $\Pr[X] =$ the probability of event $X$ taking place. $\Pr[X \mid Y] =$ the probability of event $X$ taking place, given that event $Y$ has taken place.

Bag A contains 5 cards with the numbers $1,2,3,4,5$ written on them, one each, and Bag B contains 5 cards with the numbers $6,7,8,9,10$ written on them, one each. One card is randomly drawn from each of the two bags A and B. When the sum of the two numbers on the drawn cards is odd, what is the probability that the number on the card drawn from Bag A is even? [3 points]
(1) $\frac { 5 } { 13 }$
(2) $\frac { 4 } { 13 }$
(3) $\frac { 3 } { 13 }$
(4) $\frac { 2 } { 13 }$
(5) $\frac { 1 } { 13 }$

A die is rolled twice. Given that the number 6 does not appear at all, what is the probability that the sum of the two numbers is a multiple of 4? [3 points]
(1) $\frac { 4 } { 25 }$
(2) $\frac { 1 } { 5 }$
(3) $\frac { 6 } { 25 }$
(4) $\frac { 7 } { 25 }$
(5) $\frac { 8 } { 25 }$

A bag contains 1 white ball marked with 1, 1 white ball marked with 2, 1 black ball marked with 1, and 3 black balls marked with 2. We perform a trial of simultaneously drawing 3 balls from the bag. Let $A$ be the event that among the 3 balls drawn, 1 is white and 2 are black, and let $B$ be the event that the product of the numbers on the 3 balls is 8. What is the value of $\mathrm { P } ( A \cup B )$? [3 points]
(1) $\frac { 11 } { 20 }$
(2) $\frac { 3 } { 5 }$
(3) $\frac { 13 } { 20 }$
(4) $\frac { 7 } { 10 }$
(5) $\frac { 3 } { 4 }$

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{2}{5}$
(D) $\frac{1}{5}$

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. A fair die is thrown three times. If $r _ { 1 } , r _ { 2 }$ and $r _ { 3 }$ are the numbers obtained on the die, then the probability that $\omega ^ { r _ { 1 } } + \omega ^ { r _ { 2 } } + \omega ^ { r _ { 3 } } = 0$ is
A) $\frac { 1 } { 18 }$
B) $\frac { 1 } { 9 }$
C) $\frac { 2 } { 9 }$
D) $\frac { 1 } { 36 }$

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is
(1) $\frac { 1 } { 4 }$
(2) $\frac { 1 } { 3 }$
(3) $\frac { 1 } { 8 }$
(4) $\frac { 1 } { 9 }$

There are two fair six-sided dice A and B: The numbers on A are $1, 2, 5, 6, 7, 9$, The numbers on B are $1, 3, 4, 5, 6, 9$. The relationship between the numbers on A and B is recorded in the table below. For example: if the numbers on A and B are 5 and 3 respectively, it is recorded as ``A wins''; if both A and B show 5, it is recorded as ``tie''.

\multirow{2}{*}{}		\multicolumn{6}{|c|}{A}
	Number	1	2	5	6	7	9
\multirow{6}{*}{B}	1	Tie	A wins	A wins	A wins	A wins	A wins
	3	B wins	B wins	A wins	A wins	A wins	A wins
	4	B wins	B wins	A wins	A wins	A wins	A wins
	5	B wins	B wins	Tie	A wins	A wins	A wins
	6	B wins	B wins	B wins	Tie	A wins	A wins
	9	B wins	B wins	B wins	B wins	B wins	Tie

If a person rolls both dice A and B simultaneously, what is the probability that B shows 6 given that A's number is greater than B's number?
(1) $\frac { 1 } { 6 }$
(2) $\frac { 1 } { 9 }$
(3) $\frac { 1 } { 16 }$
(4) $\frac { 1 } { 18 }$
(5) $\frac { 1 } { 32 }$

A bag contains 10 balls numbered from 1 to 10.
Given that the sum of the numbers on two balls randomly drawn from the bag is 15, what is the probability that ball number 7 was drawn?
A) $\frac { 2 } { 3 }$
B) $\frac { 2 } { 5 }$
C) $\frac { 2 } { 7 }$
D) $\frac { 1 } { 2 }$
E) $\frac { 1 } { 3 }$