2020-2022 – S2 – Mathematics – TEST 2 SOLUTIONS – page 1/3
IUT of Saint-Etienne – Sales and Marketing department
Mr Ferraris Prom 2020-2022 05/05/2021
MATHEMATICS – 2
ndsemester, Test 2 length: 2 hours – coefficient 1/2
SOLUTIONS
Exercice 1 : combinatorics (7 points)
The three questions of this exercise are independent
1) Ten coins are to be tossed, each of which can give the result "heads" or "tails". How many possible outcomes
are there to this experiment? 1.5 pt
Throws inevitably lead to p-lists, hence: 210 = 1024 possible outcomes.
2) a. In how many ways can four students be chosen at random from a group of 25 students? 1.5 pt Four students are chosen, without repetition and in no particular order.
There are a total of C425 =12650 ways to do this.
b. If this group consists of 10 men and 15 women, what is the probability of having chosen only one woman,
randomly selecting four students? 1.5 pt
The group of students is divided into two categories: men and women.
Let's draw a diagram of the set S of students and the desired draws D:
S D
men 10 3
women 15 1
total 25 4
The number of such draws D is: C310×C115=120 15× =1800. The probability of running into such a draw is 1800 %
p 14.23
12650
= ≈ .
3) We wish to determine the number of integers, between 0 and 9999, whose digits are different and which do not contain the digit 5. To do this, we will split up the problem according to the number of digits present:
we must count separately the numbers between 1000 and 9999, those between 100 and 999, those between 10 and 99, and the integers from 0 to 9. How many such numbers are there in all? 2.5 pts - numbers of one digit: the 9 numbers from 0 to 9 (except 5) are solutions;
- numbers of two digits: there are 8 possibilities for the first one (from 1 to 9 except 5), and then 8 for the second one (from 0 to 9 except 5 and except the first one), hence 8×8 = 64 solutions;
- numbers of three digits: there are 8 possibilities for the first one (from 1 to 9 except 5), and then the two others have to be taken among 8 figures (from 0 to 9 except 5 and except the first one), without repetition and in a precise order. There are A28=56 possible pairs for the two, and then: 8×A28=448 numbers of three digits that are different and not equal to 5;
- numbers of four digits: there are 8 possibilities for the first one (from 1 to 9 except 5), and then the three others have to be taken among 8 figures (from 0 to 9 except 5 and except the first one), without repetition and in a precise order. There are A38=336 ways to chose the three, hence: 8 A× 38=2688 numbers of four digits that are different and not equal to 5.
In all, we counted 9 + 64 + 448 + 2688 = 3209 numbers, between 0 and 9999, where all the digits are different and none of them is equal to 5.
2020-2022 – S2 – Mathematics – TEST 2 SOLUTIONS – page 2/3 Exercise 2 : probabilities (7 points)
A website offers a quizz with two series of questions one after the other. 60% of people answer the questions in the first series honourably (majority of correct answers). If this is the case, then the site chooses a second, more difficult series, in which only 30% of people have a majority of correct answers. On the contrary, if a person did not have an honourable result in the first round, then the site chooses a second, less difficult round, in which 50% of people have a majority of correct answers.
Event A : answer honourably to the first set of questions Event B : answer honourably to the second set of questions
1) Form either a tree or a table to represent the situation. 1.5 pt
B B B
B
2) If you take the quiz, how likely is it that...
a. you answer honourably to both series? 1 pt
( )
%p A∩B =0.6 0.3× =18 or p A
(
∩ =B)
10018 =18 %b. you answer honourably to the second series? 1.5 pt
( ) ( ) ( ) %
p B =p A∩ +B p A∩ =B 0.6 0.3 0.4 0.5× + × =38 or p B
( )
=10038 =38 %c. you answer honourably to the first series, given that you didn’t answer honourably to the second one?
1.5 pt
( ) ( )
( )
%B
p A B 0.6 0.7
p A 67.74
p B 1 0.38
∩ ×
= = ≈
− or B
( )
%p A 42 67.74
=62≈
3) Points are accumulated in case of success: passing series A awards 5 points and passing series B awards 10 points. The random variable X is the total score that can be reached for players who take part in both series in succession.
a. What are the possible values of X ? 0.5 pt
The event A∩B leads to 15 points, The event A∩B leads to 10 points, The event A∩B leads to 5 points and finally A∩B doesn’t give any point.
b. Give the probability distribution of this random variable. 1 pt
xi 15 10 5 0
pi 0.18 0.2 0.42 0.2
Exercise 3 : probability distributions (6 points)
Let's play spin a wheel. It is divided into 50 sectors of the same size. If you hit sector n°1, you win €80; if you hit one of the sectors n°2 to n°10, you win €3; otherwise, you win nothing. To be able to play (once), you have to pay 3 €. X is the random variable referring to the winnings (net: remove the €3 bet) at the end of a game.
1) What is the probability distribution of X? 2 pts
xi 77 0 –3
pi 1/50 9/50 40/50
2) a. Give the expected value and the standard deviation of X ; comment on the expectation. 2 pts E(X) = –0.86 € and σ(X) = 11.18 € If you play this game many times, you will expect to lose an average of about 86 cents per game.
A 0.6 A
0.4
0.3 0.7 0.5 0.5
2020-2022 – S2 – Mathematics – TEST 2 SOLUTIONS – page 3/3
b. How can the value of the standard deviation be used? (give the meaningof this parameter, and then the way it can be used to forecast the global or the average gain in a certain number of games, number to
be chosen) 2 pts
The €11.18 standard deviation is the average variation in gain between one game and the next.
Moreover, in n games, the expectation is multiplied by n while the standard deviation is to be multiplied by the square root of n.
For example:
* on 100 games, the expected overall gain is –86 €, with a standard deviation of 111.8 €;
the expected average gain is then –0.86 €, with a standard deviation of 1.118 €;
95% confidence interval of the overall gain:
[–86 – 1.96×111.8 ; –86 + 1.96×111.8] = [–305.13 ; 133.13]
95% confidence interval of the average gain:
[–0.86 – 1.96×1.118 ; –0.86 + 1.96×1.118] = [–3.0513 ; 1.3313]
* on 10,000 games, the expected overall gain is –8600 €, with a standard deviation of 1118 € (so the chances that players globally win money in 10,000 games are almost zero.);
the expected average gain is then –0,86 €, with a standard deviation of 0,1118 €;
95% confidence interval of the overall gain:
[–8600 – 1.96×1118 ; –8600 + 1.96×1118] = [–10791.28 ; –6408.72]
95% confidence interval of the average gain:
[–0.86 – 1.96×0.1118 ; –0.86 + 1.96×0.1118] = [–1.079128 ; –0.640872]
____________________ TEST END ____________________
