2016-2018 – S2 – Mathematics – TEST 2 – SOLUTIONS
IUT of Saint-Etienne – Sales and Marketing department
Mr Ferraris Prom 2016-2018 05/2017
MATHEMATICS – 2
ndsemester, Test 2 length : 2 hours – coefficient 1/2
SOLUTIONS
Exercise 1 : MCQ (2 points) – tick the right boxes below
One correct answer only per question - 0 point in case of wrong/missing/multiple answer at a question 1)
(
A∩B) (
∪ A∩B)
...⊂A ⊃B ⊃ ∩A B ⊂ ∪A B
2) If Card
(
A∩ =B)
Card( )
A −Card( )
B , then:A = B Card
(
A∪ = ×B)
2 Card( )
B Card( )
A =Card( )
B A and B intersect 3) If two events A and B are independent, then:p(A) + p(B) = 1 pA(B) = pB(A) p(A∩B) = p(B) p (B) = p (B) A A 4) Which one is the correct inequality?
P C
p p p
n n
n ≤ ≤ np ≤Cnp ≤Pnp Cnp ≤Pnp ≤np Cnp ≤np ≤Pnp
Exercise 2 : Cardinal numbers (4 points)
A survey consists in analysing the sales quantities of two products a and b in a shop. Within 200 clients, 57 bought the objet a, 103 bought the objet b, 38 bought both objects. We name A the set of clients who bought the object a and B the set of clients who bought the object b.
1) Calculate Card
(
A∪B)
and then give a concrete meaning of this number. 1 pt(
A B)
57 103 38 122Card ∪ = + − = . 122 clients bought at least one of both objects.
2) Build a contingency table for A, B and their contraries. 1 pt
A A
B 38 65 103
B 19 78 97
57 143 200
3) Thanks to this table, say (justify your answers and name the corresponding sets):
a. How many people who bought neither a nor b. 0.5 pt
(
A B)
78 200(
A B)
Card ∩ = = −Card ∪
b. How many people bought exactly one of both objects. 0.5 pt
( ) ( )
(
A B A B)
65 19 84(
A B) (
A B)
Card ∩ ∪ ∩ = + = =Card ∪ −Card ∩
4) What is the probability that a client bought the object b, given that this person bought the object a.
1 pt
( ) ( )
( )
. %A
A B 38 2
B 66 67
A 57 3
p Card
Card
= ∩ = = ≈
Exercise 3 : Counting (2 points)
Seven candidates (A, B, C, D, E, F and G) stand in municipal elections. You will answer the following questions by justifying the counting tools used.
1) At the count of the first round, these candidates are ranked in descending order of the number of votes obtained. How many different rankings are possible for these seven candidates? 1 pt This is the number of permutations in the whole set of candidates: 7! = 7×6×5×4×3×2×1 = 5040.
2) Both candidates who scored the best are qualified for the second round. How many different second
rounds are possible? 1 pt
Number of ways to choose two people among seven: C27 =21.
2016-2018 – S2 – Mathematics – TEST 2 – SOLUTIONS Exercise 4 : Counting and probabilities (5 points)
A magician shows a set of 32 playing cards (4 colours, 8 cards per colour: hearts, spades, clubs, diamonds) to the audience who only sees the reverse side.
1) First game: the magician asks person A to pick a card at random, memorise it, and then put it back in the set he immediately mixes. He does the same with three other people B, C and D.
a. How many different draws, in the order A, B, C, D, are possible? 0.5 pt Draws with putting back and taking the order into account: 324 = 1,048,576.
b. What is the probability that A would pick a heart? 0.5 pt
8/32 = 0.25.
c. What is the probability that A would pick a heart and B would pick a spade? 0.5 pt As the successive draws are performed with putting back, both events are independent: the probability of their intersection is the product of both probabilities: 0.25×0.25 = 1/16 = 0.0625.
2) Second game: the magician asks person A to make a simultaneous draw of four cards.
a. Calculate the number of different possible draws. 0.5 pt
Simultaneous draws lead to combinations: C432=35960.
b. What is the probability that A would pick four clubs? 1 pt
This is a "combination and partition" case.
4 0
8 24
C ×C =70.
prob = 70/35960 = 0.001947
set draw
clubs 8 4
others 24 0
c. What is the probability that A would pick exactly two kings? 1 pt This is a "combination and partition" case.
2 2
28 4
C × =C 2268.
prob = 2268/35960 = 0.06307
set draw
clubs 4 2
others 28 2
d. How many possible draws would own exactly two kings and two clubs? 1 pt This is a "combination
and partition" case, but this event corresponds to two different kinds of draws.
set draw set draw
king of clubs 1 1 king of clubs 1 0
other kings 3 1 other kings 3 2
other clubs 7 1 other clubs 7 2
others 21 1 others 21 0
1 1 1 1 0 2 2 0
1 3 7 21 1 3 7 21
C × × ×C C C + × × ×C C C C =441 63+ =504. Exercise 5 : Conditional probabilities (3 points)
A territorial organization conducted a study in the arts and crafts (fr: artisanat) sector. Its intention is to allocate a subsidy (fr: subvention) to craftspeople (fr: artisans) who request it, and preferably to those who have a professional development project and who wish to carry it out.
* 30 % of craftspeople really have such a project and, among them, 92% ask for a subsidy (and will get it);
* Among the other 70%, one fifth ask for a subsidy, will get it, but won't carry some project out.
If at least two thirds of the allocated subsidies will contribute to real and accomplished projects, the organization will consider its "subsidy operation" as a success.
1) According to the results of the study, build, as you wish: a probabilistic choice tree or a contingency table (you may then choose a total number of 1000 craftspeople). 1.5 pt
0.92 S
0.3 P
0.7 P
P P
S 276 140 416
S 24 560 584 300 700 1000 S
S S 0.08 0,2 0.8
2016-2018 – S2 – Mathematics – TEST 2 – SOLUTIONS
2) Calculate the probability that a craftsperson would carry his (or her) project out, given that the
organization allocated a subsidy to him (or her). Can the "subsidy operation" be regarded as a success?
1.5 pt Using the tree:
( ) ( )
( )
S
P S 0.3 0.92 0.276
P 0.6635
S 0.3 0.92 0.2 0.7 0.416 p p
p
∩ ×
= = = ≈
× + × Using the table: S
( )
P 276 0.6635
p =416≈
This probability is slightly less than two thirds. The operation is "almost" a success.
Exercise 6 : Random variable (4 points)
A company is used to manufacture three products A, B and C, and is forecasting its commercialisation costs.
A represents 20 % of the production, B represents 45%, and C represents 35 %. Typical commercialisation costs, for each produced unit of A, B, or C, are in the same order: €30, €36, €42, except in 30 % cases (same for A, B, or C), where this cost rises by €6 (extra costs for exports).
1) Build a probabilistic choice tree that displays the six different possible situations of costs. 0.5 pt
0.20 A 0.45 B 0.35 C
2) Give the probability distribution of the variable X : "commercialisation cost of a unit". 1 pt
xi 30 36 42 48
prob(X = xi) 0.14 0.375 0.38 0.105
3) What is the probability that X may exceed €40? 0.5 pt
0.38 + 0.105 = 0.485
4) Calculate the expectation of X. Interpret its value. 1 pt
E(X) = 0.14×30 + 0.375×36 + 0.38×42 + 0.105×48 = 38.7 : average commercialisation cost per unit.
5) Give an estimate of the commercialisation cost of 5,000 produced units, following the information given
by the directions of this exercise. 1 pt
38.7×5,000 = €193,500
cost
30 0.14
36 0.06
36 0.315
42 0.135 42 0.245
48 0.105
prob 0.7
0.3 0.7 0.3 0.7 0.3
