With three identical valves $V_1, V_2$ and $V_3$, we manufacture a hydraulic circuit. The circuit is operational if $V_1$ is operational or if $V_2$ and $V_3$ are simultaneously operational.
We treat as a random experiment the fact that each valve is or is not operational after 6000 hours. We denote:

• $F_1$ the event: ``valve $V_1$ is operational after 6000 hours''.
• $F_2$ the event: ``valve $V_2$ is operational after 6000 hours''.
• $F_3$ the event: ``valve $V_3$ is operational after 6000 hours''.
• $E$: the event: ``the circuit is operational after 6000 hours''.

We assume that the events $F_1, F_2$ and $F_3$ are pairwise independent and each have probability equal to 0.3.

1. The probability tree shown represents part of the situation. Reproduce this tree and place the probabilities on the branches.
2. Prove that $P(E) = 0.363$.
3. Given that the circuit is operational after 6000 hours, calculate the probability that valve $V_1$ is operational at that time. Round to the nearest thousandth.

Sofia wishes to go to the cinema. She can go by bike or by bus.

Part A: Using the bus

We assume in this part that Sofia uses the bus to go to the cinema. The duration of the journey between her home and the cinema (expressed in minutes) is modelled by the random variable $T _ { B }$ which follows the uniform distribution on [12; 15].

1. Prove that the probability that Sofia takes between 12 and 14 minutes is $\frac { 2 } { 3 }$.
2. Give the average duration of the journey.

Part B: Using her bike

We now assume that Sofia chooses to use her bike. The duration of the journey (expressed in minutes) is modelled by the random variable $T _ { v }$ which follows the normal distribution with mean $\mu = 14$ and standard deviation $\sigma = 1,5$.

1. What is the probability that Sofia takes less than 14 minutes to go to the cinema? What is the probability that Sofia takes between 12 and 14 minutes to go to the cinema? Round the result to $10 ^ { - 3 }$.

Part C: Playing with dice

Sofia is hesitating between the bus and the bike. She decides to roll a fair 6-sided die. If she gets 1 or 2, she takes the bus, otherwise she takes her bike. We denote:

• $B$ the event ``Sofia takes the bus'';
• $V$ the event ``Sofia takes her bike'';
• C the event ``Sofia takes between 12 and 14 minutes to go to the cinema''.

1. Prove that the probability, rounded to $10 ^ { - 2 }$, that Sofia takes between 12 and 14 minutes is 0.49.
2. Given that Sofia took between 12 and 14 minutes to go to the cinema, what is the probability, rounded to $10 ^ { - 2 }$, that she used the bus?

During a fair, a game organizer has, on one hand, a wheel with four white squares and eight red squares and, on the other hand, a bag containing five tokens bearing the numbers $1, 2, 3, 4$ and 5. The game consists of spinning the wheel, each square having equal probability of being obtained, then extracting one or two tokens from the bag according to the following rule:

• if the square obtained by the wheel is white, then the player extracts one token from the bag;
• if the square obtained by the wheel is red, then the player extracts successively and without replacement two tokens from the bag.

The player wins if the token(s) drawn all bear an odd number.

1. A player plays one game and we denote by $B$ the event ``the square obtained is white'', $R$ the event ``the square obtained is red'' and $G$ the event ``the player wins the game''. a. Give the value of the conditional probability $P _ { B } ( G )$. b. It is admitted that the probability of drawing successively and without replacement two odd tokens is equal to 0.3. Copy and complete the following probability tree.
2. a. Show that $P ( G ) = 0.4$. b. A player wins the game. What is the probability that he obtained a white square by spinning the wheel?
3. Are the events $B$ and $G$ independent? Justify.
4. The same player plays ten games. The tokens drawn are returned to the bag after each game. We denote by $X$ the random variable equal to the number of games won. a. Explain why $X$ follows a binomial distribution and specify its parameters. b. Calculate the probability, rounded to $10 ^ { - 3 }$, that the player wins exactly three games out of the ten games played. c. Calculate $P ( X \geqslant 4 )$ rounded to $10 ^ { - 3 }$. Give an interpretation of the result obtained.
5. A player plays $n$ games and we denote by $p _ { n }$ the probability of the event ``the player wins at least one game''. a. Show that $p _ { n } = 1 - 0.6 ^ { n }$. b. Determine the smallest value of the integer $n$ for which the probability of winning at least one game is greater than or equal to 0.99.

A company calls people by telephone to sell them a product.

• The company calls each person a first time:
• the probability that the person does not answer is equal to 0.6;
• if the person answers, the probability that they buy the product is equal to 0.3.
• If the person did not answer on the first call, a second call is made:
• the probability that the person does not answer is equal to 0.3;
• if the person answers, the probability that they buy the product is equal to 0.2.
• If a person does not answer on the second call, we stop contacting them.

We choose a person at random and consider the following events: $D _ { 1 }$: ``the person answers on the first call''; $D _ { 2 }$: ``the person answers on the second call''; $A$: ``the person buys the product''.

Part A

1. Copy and complete the weighted tree opposite.
2. Using the weighted tree, show that the probability of event $A$ is $P ( A ) = 0.204$.
3. We know that the person bought the product. What is the probability that they answered on the first call?

Part B

We recall that, for a given person, the probability that they buy the product is equal to 0.204.

1. We consider a random sample of 30 people. Let $X$ be the random variable that gives the number of people in the sample who buy the product. a. We admit that $X$ follows a binomial distribution. Give, without justification, its parameters. b. Determine the probability that exactly 6 people in the sample buy the product. Round the result to the nearest thousandth. c. Calculate the expected value of the random variable $X$. Interpret the result.
2. Let $n$ be a non-zero natural number. We now consider a sample of $n$ people. Determine the smallest value of $n$ such that the probability that at least one person in the sample buys the product is greater than or equal to 0.99.

Exercise 2

We have two opaque urns $\mathrm { U } _ { 1 }$ and $\mathrm { U } _ { 2 }$. Urn $\mathrm { U } _ { 1 }$ contains 4 black balls and 6 white balls. Urn $\mathrm { U } _ { 2 }$ contains 1 black ball and 3 white balls. Consider the following random experiment: We randomly draw a ball from $U _ { 1 }$ which we place in $U _ { 2 }$, then we randomly draw a ball from $\mathrm { U } _ { 2 }$. We denote:

• $N _ { 1 }$ the event ``Drawing a black ball from urn $\mathrm { U } _ { 1 }$''.
• $N _ { 2 }$ the event ``Drawing a black ball from urn $\mathrm { U } _ { 2 }$''.

For any event $A$, we denote $\bar { A }$ its complementary event.
PART A

1. Consider the probability tree opposite. a. Justify that the probability of drawing a black ball from urn $\mathrm { U } _ { 2 }$ given that a white ball was drawn from urn $\mathrm { U } _ { 1 }$ is 0.2. b. Copy and complete the probability tree opposite, showing on each branch the probabilities of the events concerned, in decimal form.
2. Calculate the probability of drawing a black ball from urn $U _ { 1 }$ and a black ball from urn $\mathrm { U } _ { 2 }$.
3. Justify that the probability of drawing a black ball from urn $\mathrm { U } _ { 2 }$ is equal to 0.28.
4. A black ball was drawn from urn $\mathrm { U } _ { 2 }$. Calculate the probability of having drawn a white ball from urn $\mathrm { U } _ { 1 }$. The result will be given in decimal form rounded to $10 ^ { - 2 }$.

PART B $n$ denotes a non-zero natural number. The previous random experiment is repeated $n$ times in an identical and independent manner, that is, urns $\mathrm { U } _ { 1 }$ and $\mathrm { U } _ { 2 }$ are returned to their initial configuration, with respectively 4 black balls and 6 white balls in urn $U _ { 1 }$ and 1 black ball and 3 white balls in urn $\mathrm { U } _ { 2 }$, between each experiment. We denote $X$ the random variable that counts the number of times a black ball is drawn from urn $\mathrm { U } _ { 2 }$. We recall that the probability of drawing a black ball from urn $\mathrm { U } _ { 2 }$ is equal to 0.28 and that of drawing a white ball from urn $\mathrm { U } _ { 2 }$ is equal to 0.72.

1. Determine the probability distribution followed by $X$. Justify your answer.
2. Determine by calculation the smallest natural number $n$ such that: $$1 - 0{,}72 ^ { n } \geqslant 0{,}9$$
3. Interpret the previous result in the context of the experiment.

PART C In this part urns $\mathrm { U } _ { 1 }$ and $\mathrm { U } _ { 2 }$ are returned to their initial configuration, with respectively 4 black balls and 6 white balls in urn $U _ { 1 }$ and 1 black ball and 3 white balls in urn $\mathrm { U } _ { 2 }$.
Consider the following new random experiment: We simultaneously draw two balls from urn $\mathrm { U } _ { 1 }$ which we place in urn $\mathrm { U } _ { 2 }$, then we randomly draw a ball from urn $\mathrm { U } _ { 2 }$.

1. How many possible draws of two balls simultaneously from urn $\mathrm { U } _ { 1 }$ are there?
2. How many possible draws of two balls simultaneously from urn $\mathrm { U } _ { 1 }$ containing exactly one white ball and one black ball are there?
3. Is the probability of drawing a black ball from urn $\mathrm { U } _ { 2 }$ with this new experiment greater than the probability of drawing a black ball from urn $\mathrm { U } _ { 2 }$ with the experiment in part A? Justify your answer. You may use a weighted tree diagram modeling this experiment.

A company that manufactures toys must perform conformity checks before their commercialization. In this exercise, we are interested in two tests performed by the toy company: a manufacturing test and a safety test. Following a large number of verifications, the company claims that:

• $95 \%$ of toys pass the manufacturing test;
• Among toys that pass the manufacturing test, $98 \%$ pass the safety test;
• $1 \%$ of toys pass neither of the two tests.

A toy is chosen at random from the toys produced. We denote:

• F the event: ``the toy passes the manufacturing test'';
• S the event: ``the toy passes the safety test''.

Part A

1. From the data in the statement, give the probabilities $P ( F )$ and $P _ { F } ( S )$.
2. a. Construct a probability tree that illustrates the situation with the data available in the statement. b. Show that $P _ { \bar { F } } ( \bar { S } ) = 0.2$.
3. Calculate the probability that the chosen toy passes both tests.
4. Show that the probability that the toy passes the safety test is 0.97 rounded to the nearest hundredth.
5. When the toy has passed the safety test, what is the probability that it passes the manufacturing test? Give an approximate value of the result to the nearest hundredth.

Part B
A batch of $n$ toys is randomly selected from the company's production, where $n$ is a strictly positive integer. We assume that this selection is made from a sufficiently large quantity of toys to be assimilated to a succession of $n$ independent draws with replacement. Recall that the probability that a toy passes the manufacturing test is equal to 0.95. Let $S _ { n }$ be the random variable that counts the number of toys that have passed the manufacturing test. We admit that $S _ { n }$ follows the binomial distribution with parameters $n$ and $p = 0.95$.

1. Express the expectation and variance of the random variable $S _ { n }$ as a function of $n$.
2. In this question, we set $n = 150$. a. Determine an approximate value to $10 ^ { - 3 }$ of $P \left( S _ { 150 } = 145 \right)$. Interpret this result in the context of the exercise. b. Determine the probability that at least $94 \%$ of the toys in this batch pass the manufacturing test. Give an approximate value of the result to $10 ^ { - 3 }$.
3. In this question, the non-zero natural integer $n$ is no longer fixed.

Let $F _ { n }$ be the random variable defined by: $F _ { n } = \frac { S _ { n } } { n }$. The random variable $F _ { n }$ represents the proportion of toys that pass the manufacturing test in a batch of $n$ toys selected. We denote $E \left( F _ { n } \right)$ the expectation and $V \left( F _ { n } \right)$ the variance of the random variable $F _ { n }$. a. Show that $E \left( F _ { n } \right) = 0.95$ and that $V \left( F _ { n } \right) = \frac { 0.0475 } { n }$. b. We are interested in the following event $I$: ``the proportion of toys that pass the manufacturing test in a batch of $n$ toys is strictly between $93 \%$ and $97 \%$''. Using the Bienaymé-Chebyshev inequality, determine a value $n$ of the size of the batch of toys to select, from which the probability of event $I$ is greater than or equal to 0.96.

Solve the following two independent problems.
(i) A mother and her two daughters participate in a game show. At first, the mother tosses a fair coin.
Case 1: If the result is heads, then all three win individual prizes and the game ends. Case 2: If the result is tails, then each daughter separately throws a fair die and wins a prize if the result of her die is 5 or 6. (Note that in case 2 there are two independent throws involved and whether each daughter gets a prize or not is unaffected by the other daughter's throw.)
(a) Suppose the first daughter did not win a prize. What is the probability that the second daughter also did not win a prize?
(b) Suppose the first daughter won a prize. What is the probability that the second daughter also won a prize?
(ii) Prove or disprove each of the following statements.
(a) $2 ^ { 40 } > 20!$
(b) $1 - \frac { 1 } { x } \leq \ln x \leq x - 1$ for all $x > 0$.

There is a basket containing at least 10 white balls and at least 10 black balls, and an empty bag. Using one die, the following trial is performed.
Roll the die once. If the result is 5 or more, put 2 white balls from the basket into the bag. If the result is 4 or less, put 1 black ball from the basket into the bag.
When the above trial is repeated 5 times, let $a _ { n }$ and $b _ { n }$ denote the number of white balls and black balls in the bag after the $n$-th trial ($1 \leq n \leq 5$) respectively. Given that $a _ { 5 } + b _ { 5 } \geq 7$, what is the probability that there exists a natural number $k$ ($1 \leq k \leq 5$) such that $a _ { k } = b _ { k }$? If this probability is $\frac { q } { p }$, find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

There is one bag and two boxes A and B. The bag contains 4 cards with the numbers $1, 2, 3, 4$ written on them, one number per card. Box A contains more than 8 white balls and more than 8 black balls, and box B is empty. Using this bag and the two boxes A and B, the following trial is performed.
A card is randomly drawn from the bag, the number on the card is confirmed, and the card is returned to the bag.
If the confirmed number is 1, 1 white ball from box A is placed into box B. If the confirmed number is 2 or 3, 1 white ball and 1 black ball from box A are placed into box B. If the confirmed number is 4, 2 white balls and 1 black ball from box A are placed into box B.
After repeating this trial 4 times, given that the total number of balls in box B is 8, find the probability that the number of black balls in box B is 2. [4 points]
(1) $\frac{3}{70}$
(2) $\frac{2}{35}$
(3) $\frac{1}{14}$
(4) $\frac{3}{35}$
(5) $\frac{1}{10}$

There are 16 balls and six empty boxes with the natural numbers 1 through 6 written on them. A trial is performed using one die.
When the die is rolled and the result is $k$: If $k$ is odd, place 1 ball each in the boxes labeled $1, 3, 5$, and if $k$ is even, place 1 ball each in the boxes labeled with the divisors of $k$.
After repeating this trial 4 times, given that the sum of all balls in the six boxes is odd, what is the probability that the number of balls in the box labeled 3 is 1 more than the number of balls in the box labeled 2? [4 points]
(1) $\frac { 1 } { 8 }$
(2) $\frac { 3 } { 16 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 5 } { 16 }$
(5) $\frac { 3 } { 8 }$

A signal which can be green or red with probability $\frac { 4 } { 5 }$ and $\frac { 1 } { 5 }$ respectively, is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is $\frac { 3 } { 4 }$. If the signal received at station $B$ is green, then the probability that the original signal was green is
A) $\frac { 3 } { 5 }$
B) $\frac { 6 } { 7 }$
C) $\frac { 20 } { 23 }$
D) $\frac { 9 } { 20 }$

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) $\frac{2}{7}$

Suppose that
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
(A) $\frac { 15 } { 256 }$
(B) $\frac { 3 } { 16 }$
(C) $\frac { 5 } { 52 }$
(D) $\frac { 1 } { 8 }$

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 $\_\_\_\_$ .

When a missile is fired from a ship, the probability that it is intercepted is $\frac { 1 } { 3 }$ and the probability that the missile hits the target, given that it is not intercepted, is $\frac { 3 } { 4 }$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:
(1) $\frac { 3 } { 8 }$
(2) $\frac { 1 } { 27 }$
(3) $\frac { 1 } { 8 }$
(4) $\frac { 3 } { 4 }$

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac { m } { n }$, where $\operatorname { gcd } ( m , n ) = 1$, then $m + n$ is equal to:
(1) 4
(2) 14
(3) 13
(4) 11

$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of 5 before $B$ throws a sum of 8, and $B$ wins if he throws a sum of 8 before $A$ throws a sum of 5. The probability that $A$ wins if $A$ makes the first throw, is
(1) $\frac{8}{17}$
(2) $\frac{9}{19}$
(3) $\frac{9}{17}$
(4) $\frac{8}{19}$