IUT of Saint-Etienne – Sales and Marketing department Mr Ferraris Prom 2018-2020 05/2019 MATHEMATICS – 2

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2018-2020 – S2 – Mathematics – TEST 2 « partiel » SOLUTIONS – page 1 / 3

IUT of Saint-Etienne – Sales and Marketing department

Mr Ferraris Prom 2018-2020 05/2019

MATHEMATICS – 2

^nd

semester, Test 2 length: 2 hours – coefficient 1/2

SOLUTIONS

Exercise 1 - sets (3 points)

Here, A and B are two subsets of a set E.

1) To what does the set A refer? 0.5 pt

Complement of A inside E (set of the elements of E that are do not belong to A).

2) What does the sentence "A and B are mutually exclusive" mean? 0.5 pt A and B don’t share any element. Their intersection is empty.

3) What does the sentence "A and B form a partition of E" mean? 1 pt A and B are mutually exclusive and their union is E.

4) Simplify, on giving details, the expression

(

^A^{∩ ∪}^B

) (

^A^∩^{B .}

)

^{1 pt}

(

^A^{∩ ∪}^B

) (

^A^∩^B

)

^{= ∩ ∪}^A

(

^B ^B

)

^{= ∩ =}^A ^E ^{A .}

Exercise 2 - Cardinal numbers (3 points)

A store offers two relatively complementary objects A and B. To study the purchasing behaviour on these objects, the management recruited a trainee, who gave him/her the following information: on a number of the last receipts, it appeared that 127 people bought both objects, 64 bought only one of them (those who bought only the object A represent a quarter of them) and 109 bought neither A nor B

1) These indications are incomplete. Organize a contingency table crossing the purchase (and non-purchase) quantities of the objects A and B, then complete this table. 1.5 pt

B B

A 127 16 143

A 48 109 157

175 125 300

The total number of receipts is 127+109+64 = 300 2) Calculate:

a. the rate of the clients who bought the object A. 0.5 pt

143/300 ≈ 47.67 %.

b. the rate of the clients who only bought the object A. 0,5 pt

16/300 ≈ 5.33 %.

c. the rate of the clients who bought the object B, among those who bought the object A. 0.5 pt 127/143 ≈ 88.81 %.

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2018-2020 – S2 – Mathematics – TEST 2 « partiel » SOLUTIONS – page 2 / 3 Exercise 3 - combinatorics (4.5 points) – the three questions are independent

1) A bag contains eight coins numbered from 1 to 8. We have to draw three of them, successively and without putting back, so as to create a number (like 327 for instance). How many numbers are possible?

1 pt We must choose three coins (p = 3) among eight (n = 8) without putting back; the order of the three matters (if changed, the result isn’t the same). Each number is then a permutation. There are in total:

= × × =

3

A8 8 7 6 336 numbers.

2) While walking the streets of a big city, one has the choice at each crossroads to turn to the left, to turn to the right, or to go straight. Our route must go through 6 intersections. How many different routes could

we do? 1 pt

We must choose a direction six times (p = 6) among three possibilities (n = 3) with possible repetition; the order of the decisions matters (if changed, the route isn’t the same). Each route is then a p-list. There are in total: 3⁶ =729 possible routes.

3) 80 candidates are competing in a first round of a game. Only 5 of them will be qualified for the second round.

a. How many possible qualified groups are there? 1 pt

We have to choose five people (p = 5) among 80 (n = 80) without possible repetition; the order of the five doesn’t matter (if changed, the group of five is the same). Each group is then a combination. There are in total: C⁵80 =24 040 016 possible groups.

b. If the candidates consist of 30 men and 50 women, how many groups of five people can be composed

with two men and three women? 1.5 pt

Considering combinations again, we have to choose two men among 30 and three women among 50. The number of possibilities is: C²₃₀×C³₅₀ =8 526 000 .

Exercise 4 - probabilities (4.5 points)

We are in the middle of winter. The chances that it will snow tomorrow (event N) are estimated at 2 out of 3. If it snows tomorrow, 120 students out of 160 of the promotion will be present in class. If it does not snow, 150 students will be present. The next day, in the list of 160 students, a name will be drawn at random. Let’s name A the event "the student is absent".

1) Built a probabilistic choice tree with the events A and N. 1 pt 1/4 A 8/48

2/3 N

1/3 N

( )

⁼

( )

⁼

N N

1 1

p A , p A

4 16 : a quarter of the students will be absent if it snows, but only one student out of 16 if it doesn’t.

2) What is the probability that a name drawn at random corresponds to a present student? 1 pt

( ) ( ) ( )

^/ ^/ ^/ ^/

p A =p A∩ +N p A∩ =N 24 48 15 48+ =39 48=13 16=0.8125

3) Given that a student is present, what is the probability that it’s snowing? 1.5 pt

( ) ( )

( )

/

A /

p A N 24 48 24 8

p N 0.6154

p A 39 48 39 13

= ∩ = = = ≈

4) Are the events N and A independent? 1 pt

We saw that N

( )

N

( ) ( )

1 1 3

p A , p A and p A

4 16 16

= = = . These probabilities being different, we can conclude that N and A are not independent (here, the fact it snows increases the number of absent students).

Instead, we could check that ^{p A}

(

^∩^N

) ( ) ( )

^≠^{p A} ^×^{p N :}

( )

^/

( ) ( )

^/ ^/ ^/

p A∩N =8 48≈0.1667 and p A ×p N =3 16 2 3× =6 48=0.125 A

A 3/4 1/16 15/16

24/48 1/48 15/48 A

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2018-2020 – S2 – Mathematics – TEST 2 « partiel » SOLUTIONS – page 3 / 3 Exercise 5 - random variable and probability distribution (5 points)

A new scratch game consists in a ticket on which are located 30 golden discs. Four of these discs are "winners":

if we scratched them, we would discover a star; the other 26 are "losers": no picture under the gilding (fr.:

“dorure”). To play, we must scratch three discs at random, not more, not less.

1) Four events are possible in the end: to get three stars, or two, or one, or none. Explain why their probabilities are in the same order: 0.0009852; 0.03842; 0.3202; 0.6404. 1.5 pt We must choose three discs and scratch them (p = 3) among 30 (n = 30) without possible repetition; the order of the three doesn’t matter (if changed, your selection remains the same). Each group is then a combination. We have: C³₃₀=4060 possibilities.

To get three stars, we must have chosen 3 winner discs among the 4 (and none among the 26 losers). There are: C³₄×C⁰₂₆=4 possibilities. The probability to get 3 stars is then: 4/4060 = 0.0009852.

To get two stars, we must have chosen 2 winner discs among the 4 and 1 loser among the 26. There are:

2 1

4 26

C ×C =156 possibilities. The probability to get 2 stars is then: 156/4060 = 0.03842.

To get one star, we must have chosen 1 winner discs among the 4 and 2 losers among the 26. There are:

1 2

4 26

C ×C =1300 possibilities. The probability to get 1 star is then: 1300/4060 = 0.3202.

To get no star, we don’t have any winner disc among the 4 and 3 losers among 26. There are:

0 3

4 26

C ×C =2600 possibilities. The probability to get no star is then: 2600/4060 = 0.6404.

2) You have to spend €2 to buy a ticket. The fixed gains are the following: €500 with 3 stars, €10 with 2 stars,

€2 with one star and €0 if no star is discovered. We set X as the variable “gain after a game, minus the ticket’s cost”.

a. Thanks to the data given by the question 1, give the probability distribution of X. 0.5 pt

Gain X 498 8 0 -2

probability 0.0009852 0.03842 0.3202 0.6404

b. Give its expected value E(X) and its standard deviation σ(X). Explain these values in details. 2 pts E(X) ≈ – 0.4828 € and σ^(X)≈ 15.78 €.

The expected value is the value towards which the real average gain tends if many games are played (law of large numbers); for example, after 10000 games, the real cumulative loss for players will be close to 4828 €.

The standard deviation makes it possible to refine this forecast; for example, on a basis of 10000 games:

* There are 68.3% chances that the average gain would be located between 15.78

0.4828 0.6406 €

10000

− − ≈ − and 15,78

0,4828 0,3250 €

10000

− + ≈ − ,

* There are 99% chances that the average gain would be located between 15.78

0.4828 2.58 0.8900 €

10000

− − × ≈ − and 15.78

0.4828 2.58 0.0756 €

10000

− + × ≈ − , that’s to say: the

cumulative gain, for the owners, would be between 756 € and 8900 €.

c. On buying 500 tickets and scratching all of them, what average global gain can be expected? 0.5 pt 0.4828 500 241.4

− × = − . I can forecast to lose around 241.4 €.

d. What percentage of the players’ expenses is actually redistributed in gains? 0.5 pt For each trial, the €2 bet and the average player’s loss is €0.4828. Thus, the average player’s gain (not considering the bet) is €1.5172.

The company redistributes 1.5172 / 2 = 0.7586 = 75.86 % of the bets in gains.

____________________ TEST END ____________________