Combinatorics and probabilities MATHEMATICS

SALES AND MARKETING Department

MATHEMATICS

2 ^nd Semester

________ Combinatorics and probabilities ________

SOLUTIONS of tutorials and exercises

Online document : http://jff-dut-tc.weebly.com section DUT TC Maths S2

Exercise 1. (Tutorial for lesson page 8)

E is the set of the inhabitants of a city ; Card(E) = 2500.

A is the set of this city’s men ; Card(A) = 1220.

B is the set of this city’s retired people ; Card(B) = 670.

400 women are retired.

Create, then complete, a contingency table; tell how many men aren’t retired; tell how many people are women or retired people.

A A

B 270 400 670 = Card(B)

B 950 880 1830 = Card(B )

1220 = Card(A) 1280 = Card(A ) 2500 = Card(E) question 1 : {men AND non-retired} = A∩B . Card(A∩B ) = 950.

question 2 : {women OR retired} = A∪B . - using the union formula :

Card(A∪B ) = Card( A ) + Card(B) - Card(A∩B ) = 1280 + 670 – 400 = 1550 - using a Morgan law :

Card(A∪B ) = 2500 - Card(A∪B ) = 2500 – Card(A∩B ) = 2500 – 950 = 1550 Exercise 2.

Given A and B, two subsets of E, simplify the following expressions:

Exercise 3.

E = {2, 5, 8, 11, 14, 17, 20}. Let A be the subset of the even numbers of E and B the one of the multiples of 5.

1) Define the complement of A in E. Give its elements.

This is the set of every elements of E that are not in A, that is to say: the set of the odd numbers of E.

A = {5, 11, 17}.

2) Give the sets A∩B and A∩B. What is their union?

{ }

^;

{

^{; ;}

}

^;

( ) ( )

∩ = ∩ = ∩ ∪ ∩ =

A B 20 A B 2 8 14 A B A B A

( )

A∩ ∪ ∩A B B

(

^{A∪ ∩ ∩B}

) (

^{A B}

)

(

^{A∩ ∪ ∪B}

) ( )

^{A B}

(

^{A B}

) (

^{A B}

) (

^{A A}

)

^{B B B}

= ∪ ∩ ∪ = ∩ ∪ = ∅ ∪ =

( ) ( ) ^{( )}

A B A B A B A B

= ∩ ∩ ∪ = ∪ ∩ ∪ = ∅

(

^{A B}

) (

^{A B}

)

^A

( )

^{B B A E A}

= ∩ ∪ ∩ = ∩ ∪ = ∩ =

(

^A∩B

) (

^{∪ A∩B}

) (

^{= A∪A}

)

^{∩ = ∩ =B E B B}

(

^A∩B

)

^∪

(

^A∩B

)

^{= ∩ ∪A}

(

^{B B}

)

^{= ∩ =A E A A}∩ ∪ = ∩ ∩ = ∅ ∩ = ∅^{(A B) A}

(

^{A B}

)

^B

( ) ( ) ( ) ( )

A∪ ∩ =A B A∪ ∩ ∪ = ∩ ∪ =A A B A A B A

( ) ( ) ( ) ^{( )} ( ( ) ( ) )

( ) ( ( ) ) ^{( ) ( )}

( ) ( )

A B A B A B A B A B A B B

A B A B A A B A B A

A B B A A A B B

∪ ∩ ∩ ∩ ∪ = ∪ ∩ ∩ ∩ ∪ ∩

= ∪ ∩ ∩ ∩ ∪ ∅ = ∪ ∩ ∩ ∩

= ∪ ∩ ∩ ∩ = ∪ ∩ ∩ ∅ = ∅

Exercise 4.

There are three categories of water meters in a municipality:

A: Meters that are less than two years old and are therefore still under warranty;

B: Meters that are between 2 and 20 years old;

C: Meters that are more than 20 years old.

Let D be the set of defective water meters.

1) Which of these sets are mutually exclusive?

A, B et C sont disjoints (aucun compteur n’appartient à deux catégories à la fois).

2) What do the following sets represent?

( ) ( )

A ; D ; A∩D ; A∪D ; C∩D ; A∪D ; A∩ ∪D A∩D ; A∪B A : meters that are more than two years old ; D : non-defective meters;

A D : defective and less than two year old meters;

A D : defective or more than two year old meters;

C D : more than twenty year ol

∩

∪

∩

( ) ( )

d and non-defective meters;

A D A D : non-defective and more than two year old meters;

A D A D A : less than two year old meters ; A B : less than twenty year old meters

∪ = ∩

∩ ∪ ∩ = ∪

Exercise 5.

In a group of 25 students (including 17 women), 20 passed their exam (including 14 women).

1) Build a contingency table dispatching the information above.

2) How many men passed their exam? What is the name of the corresponding set?

B passed B didn’t pass

A men 6 2 8 6 men passed their exam

A women 14 3 17 This is the set A∩B.

20 5 25

Exercise 6.

Amongst 350 people interviewed for a survey, 244 own a computer with an Internet access, 287 own a smartphone, but inside this last group, 56 don’t have a computer with an Internet access.

1) Organize and fill a contingency table using these data.

B mob. phone B

A comp+inter 231 13 244

A 56 50 56

287 63 350

2) How many people own…

a. an Internet access but no smartphone ? 13 b. at least one of both? 350 – 50 = 300 c. only one of both? 13 + 56 = 69 Exercise 7.

After counting the answers from a survey conducted on a sample of 500 people, it appears that 154 of them go to the cinema at least once a month, 228 buy popcorn when they go to movies, and of those who go to the cinema less than once a month, 131 usually buy popcorn.

1) Let’s name A the set of people who go to the cinema at least once a month and B the set of people who buy popcorn when they go to movies; build the corresponding contingency table.

B popcorn B

A cine >= 1 97 57 154

A 131 215 346

228 272 500

2) Answer by naming the adequate subset and justifying the result if not already written in the table:

a. Among people who go to the cinema at least once a month, how many buy popcorn ? Card(A∩B) = 97 b. How many people go to the cinema less than once a month? Card(A) = 346

c. How many people are elements of A or B ?

Card(A∪B) = Card(A) + Card(B) - Card(A∩B) = 154 + 228 - 97 = 285

Exercise 8. (Tutorial for lesson page 9)

* by how many ways can we arrange two objects inside three drawers?

Two choices (p = 2) have to be made among three drawers (n = 3) with possible repetition; once a selection is done, the order is important. Each selection is then a p-list. Number of ways: 3² = 9.

* how many numbers composed with four figures only contain the figures 1, 2, 3?

Four figures have to be selected (p = 4) among three figures (n = 3) with possible repetition (if not, the experiment would be impossible!); once a selection is done, the order is important. Each number is then a p- list. Number of… numbers: 3⁴ = 81.

* how many words can be written by taking five letters chosen in the set {a, b, e, m, i, r, o}?

Five letters have to be selected (p = 5) among seven letters (n = 7) with possible repetition; once a selection is done, the order is important. Each word is then a p-list. Number of words: 7⁵ = 16807.

Exercise 9. (Tutorial for lesson page 10)

* how many pairs representative/assistant could have been elected from a group of 25 students?

Two students must be selected (p = 2) among the 25 (n = 25) without repetition; once a selection is done, the order is important. Each vote is then a permutation. Number of votes: P₂₅² = ×25 24=600.

* how many ways can 3 blocks be piled, taking them among 10 blocks of different colours?

Three blocks have to be selected (p = 3) among ten (n = 10) without repetition; once a selection is done, the order is important. Each pile is then a permutation. Number of piles: P₁₀³ =10 9 8× × =720.

* how many words can be written by taking five different letters chosen in {a, b, e, m, i, r, o}?

Five letters have to be selected (p = 5) among seven letters (n = 7) without repetition; once a selection is done, the order is important. Each word is then a permutation. Number of words: P₇⁵ = × × × × =7 6 5 4 3 2520. Exercise 10. (Tutorial for lesson page 11)

* How many couples of representatives could be elected from a group of 25 students?

Two students must be selected (p = 2) among the 25 (n = 25) without repetition; once a selection is done, the order has no importance. Each vote is then a combination. Number of votes:

!

2 25

25 24

C 300

2

= × = ^.

* How many different hands of 8 cards could be given from a deck of 32 playing cards?

Eight cards have to be selected (p = 8) among the 32 (n = 32) without repetition; once a selection is done, the order has no importance. Each hand is then a combination. Number of hands: C⁸₃₂=1 518 300.

* How many draws of 6 different integers are possible, taking them between 1 and 49?

Six cards have to be selected (p = 8) among the 32 (n = 32) without repetition; once a selection is done, the order has no importance. Each hand is then a combination. Number of hands: C⁶₄₉ =13 983 816.

Exercise 11. (Tutorial for lesson page 12)

This time, the initial set is parted into categories and our purpose is to get a certain number of elements of each category.

It is explained, in section 1.2.4, only for situations that lead to combinations, that combinations have to be calculated inside each subset (category) and then multiplied.

1) From a deck of 32 playing cards, how many 8-card hands own exactly 3 spades and 2 hearts?

initial set draw

spades 8 3 C³₈ =56

hearts 8 2 C²₈ =28

others 16 3 C³₁₆ =560

total 32 8

Number of such hands: C³₈× ×C²₈ C³₁₆=878 080

2) In a company, among 20 women and 20 men, 5 women and 3 men have to be chosen at random. How many possibilities are there?

initial set draw

women 20 5 C₂₀⁵ =15 504

men 20 3 C₂₀³ =1 140

total 40 8

Number of such groups: C⁵₂₀×C³₂₀ =17 674 560 Exercise 12.

1. dice

1.1 A die is being rolled three times. How many possible outcomes?

successive dice rolls : p-lists. 6³ = 216 possible outcomes.

1.2 Three dice are being rolled at the same time. How many possible outcomes?

simultaneous dice rolls : p-lists. 6³ = 216 possible outcomes.

2. numbers and letters

2.1 How many phone numbers of eight figures can theoretically exist ?

n = 10; p = 8. repetition (of the same figure in a phone number): possible; order (of the figures to create a phone number): essential. 10⁸ = 100 000 000 phone numbers

2.2 How many ways can six different integers be chosen among [1 ; 49]?

n = 49; p = 6. repetition (of the same integer): no, order (of the six integers once selected): not important. C⁶₄₉ =13 983 807

2.3 How many numbers are composed with three different figures (including 0)?

n = 10 available figures; p = 3. repetition (of the same figure in a number): forbidden, order (of the chosen figures to form a number): essential. P₁₀³ =720 numbers

2.4 How many different lists of 4 letters can be created for the vehicles’ number plates?

n = 26 available letters; p = 4. repetition (of the same letter in a list): allowed, order (of the four letters to define a plate): essential. 24⁴ = 456 976

2.5 How many anagrams of the word "MATHS" are there?

n = 5 available letters, p = 5; repetition (of the same letter into an anagram): no, order (of the letters to create a word): essential. 5 ! = 120 anagrams

2.6 How many words can be created, taking 4 letters from the word "BRACKET"?

n = 7 available letters; p = 4. repetition (of the same letter in a word): possible ,order (of the letters to create a word): essential;. 7⁴ = 2 401 words

2.7 How many 10 notes-long melodies can be written, taking notes among A,B,C,D,E,F,G?

n = 7 available musical notes ; p = 10. repetition (of the same musical note in a melody): authorized, order (of the chosen notes to create a melody): essential;. 7¹⁰ = 282 475 249 melodies

3. arrangements

3.1 How many ways can 5 objects be arranged into 8 boxes?

n = 8 boxes; p = 5 boxes to be selected. repetition (of the same box during the selection): not forbidden, order (of the selected boxes to arrange the different objects): yes. 8⁵ = 32 768 arrangements

3.2 Same question, but you can’t place more than one object per box.

n = 8 boxes; p = 5 boxes to be selected. repetition (of the same box during the selection): forbidden, order (of the selected boxes to arrange the different objects): yes. P₈⁵=6 720 arrangements

3.3 Paul drives a team of 5 people. Each month, he evaluates the work of one of them, chosen at random.

In a one year period, how many different lists of evaluated people could have been made?

n = 5 people ; p = 12 evaluations. repetition (of a person during 12 months): compelling, order (of the 12 names forming a list): yes. 5¹² = 244 140 625 lists

4. playing cards

4.1 How many 8 cards hands from a deck of 32 playing cards?

n = 32 available cards; p = 8. repetition (of the same card in a hand): impossible, order (of the 8 cards in a hand): no importance. C⁸₃₂=10 518 300 hands

4.2 How many 5 cards hands from a deck of 52 playing cards?

n = 52 available cards; p = 5. repetition (of the same card in a hand): impossible, order (of the 5 cards in a hand): no importance. C⁵₅₂ =2 598 960 hands

5. classifications, elections

5.1 How many possible different tiercés at the end of a 12 horses race?

n = 12 horses; p = 3. repetition (of the same horse in a tiercé): impossible, order (of the three horses):

essential. P₁₂³ =1 320 tiercés

5.2 How many possible different classifications at the end of a 12 horses race?

n = 12 horses; p = 12. repetition (of the same horse in a classification): impossible, order (of the twelve horses): essential. 12!=479 001 600 classifications

5.3 How many ways a delegation of 5 people can be chosen from a group of 40?

n = 40 people; p = 5. repetition (of the same person in the delegation): impossible, order (of the five selected people): unimportant. C⁵₄₀ =658 008 delegations

5.4 16 pilots are fighting at a formula 1 race. At the end, the only six first will score different numbers of points. How many possible distributions of points are there?

n = 16 pilots; p = 6. repetition (of the same pilot into the classification): impossible, order (of the six pilots, for the distribution of points): essential. P₁₆⁶ =5 765 760 distributions

5.5 How many possible podiums after a race in which 8 runners will compete?

n = 8 runners; p = 3. repetition (of the same runner on the podium): impossible, order (of the three runners on the podium): essential. P₈³=336 podiums

5.6 Fifteen people meet. Everyone gives a handshake to every other, once. How many handshakes?

n = 15 people ; p = 2 people to be selected for a handshake. repetition (of the same person for a handshake): of course not (nobody shakes her/his own hand), order (of both people to form a handshake): certainly not (the handshake A-B is the handshake B-A, it mustn’t be counted twice).

2

C15=105 handshakes

Exercise 13.

Let consider a group of 13 women and 8 men. 4 people have to be chosen.

1) How many possibilities are there?

n = 21 people; p = 4. repetition: no, order: no. C₂₁⁴ =5985 2) How many of them contain exactly one man?

1 man among 8 and 3 women among 13: C¹₈×C₁₃³ =2288 3) 2 men? 3 men? 4 men? no man?

2 man among 8 and 2 women among 13: C²₈×C²₁₃=2184 3 man among 8 and 1 women among 13: C³₈×C¹₁₃=728 4 man among 8 and 0 women among 13: C₈⁴×C₁₃⁰ =70 0 man among 8 and 4 women among 13: C₈⁰×C₁₃⁴ =715 Exercise 14.

Among the hands of 5 cards taken from a deck of 32 playing cards, how many contain:

a. the 4 aces?

4 aces among the 4 aces and 1 other card among the 28 others: C⁴₄×C¹₂₈ =28 b. a square?

square of aces, or square of kings, or…: 8 possible different squares, giving 8 separate groups of outcomes:

the numbers of hands must be added, and are the same for each group. 8 28× =224^. c. exactly 3 spades?

3 spades among 8 and 2 other cards among the 24 others: C³₈×C²₂₄ =15456 d. exactly 2 spades and one club?

2 spades among 8 and 1 club among 8 and 2 other cards among the 16 others (heart or diamond):

2 1 2

8 8 16

C × ×C C =26880 e. at least one king?

The contrary is easier to work on: 0 king among 4 and 5 other cards among 28: C⁰₄×C₂₈⁵ =98280. The whole number of hands (regardless the number of kings) is: C₃₂⁵ =201376.

"At least one king" is then realized in any other hand, whose number is: 201376 - 98280 = 103096 f. at least two jacks?

reversed: (0 jack among 8 and 5 others among 24) or (1 jack among 8 and 4 others among 24) So, for at least two jacks: ^{C325 −}

(

^C04×C528

) (

^{− C14×C428}

)

⁼²¹¹⁹⁶

g. exactly 3 diamonds and one king?

There might be the king of diamonds… or not ! These are two separated events, whose number of hands have to be added. (1 king of diamond among 1 and 2 diamonds among 7 and 2 other cards among 21 "nor king neither diamond") or (1 king among 3 (kings not diamond) and 3 diamonds among 7 (diamonds not king) and 1 other card among 21)

(

^{C11× ×C27 C221}

) (

^{+ C13× ×C37 C121}

)

^{=4410+2205=6615}

Exercise 15.

A jar contains balls: 2 white, 3 green, 5 red. 3 balls are to be taken simultaneously from the jar. Among the possible groups of 3 balls, how many contain…

a. one single colour? (3 red among 5) or (3 green among 3): C³₅+C³₃=10+ =1 11

b. the three colours? 1 red among 5 and 1 green among 3 and 1 white among 2: C¹₅× ×C¹₃ C¹₂ =30 c. two colours?

this is the contrary event of the union of both former events "one colour" and "three colours", hence we have to subtract their numbers of groups from the whole number (3 balls among 10): C³₁₀−11 30− =79 d. at least two colours? contrary event of "one colour", thus: C³₁₀−11=109

Exercise 16.

On loto game (former rules), a player has to select 6 different numbers from the set {1, 2, 3, …, 49}. Then, the official draw is performed, revealing the 6 winner numbers.

1) Calculate the whole number of possible selections.

n = 49 available numbers ; p = 6. repetition: no, order: no. C⁶₄₉=13 983 816 selections 2) Among them, how many would contain exactly…

a. the six winner numbers?

6 numbers selected among the 6 winners: only one possibility b. 5 winner numbers?

5 numbers selected among the 6 winners and 1 loser among the 43 losers: C⁵₆×C¹₄₃=258 c. 4 winner numbers?

4 numbers selected among the 6 winners and 2 loser among the 43 losers: C⁴₆×C²₄₃=13 545 d. 3 winner numbers?

3 numbers selected among the 6 winners and 3 loser among the 43 losers: C³₆×C³₄₃=246 820 e. no winner number? (find two ways )

1st way: 0 winner number among 6 and 6 losers among 43

0 6

6 43

C ×C =6 096 454 different possible selections

2nd way: let’s calculate the number of selections that own 2 winning numbers, the number of selections with only 1 winning number, then if we denote Tn the number of selections having n winning numbers (e.g. T5 = 258) and T the total number of possible selections (13983816), one can write:

T0 = T –

6

1

T_n

∑

n= up to you to perform the calculations and find out the result.

Exercise 17.

1) Independent questions

a. In order to build your team, you have to choose 5 people out of a group of 10 basket players. How many different teams could be made?

n = 10 available players; p = 5 players to select. Repetition: no ; order: no. C⁵₁₀=252

b. A company has 18 employees and its manager decides to give three awards : best employee, most punctual employee, and less bald employee. How many ways can these awards be given?

n = 18 employees; p = 3 to select. Repetition: possible (an employee can be awarded more than once);

order: yes (the awards are different and so are people): 18³ = 5832

c. On a chess board (8 × 8 tiles), how many ways can you put a king, a queen and a tower ?

n = 64 available tiles; p = 3 tiles to be chosen. Repetition: no (one tile for one object); order : yes (the objects are different). P₆₄³ =249 984

It would be better to notice that if we rotate the board 90°, then a permutation becomes another...

which is the same! Thus, each "storage" of three objects was counted four times in the previous count…

2) 20 chips have been put in a bag, numbered from 1 to 20. The chips from #1 to 10 are white; those from #11 to 16 are green; the last ones are red. You have to draw three chips, at random, simultaneously.

a. How many different possible draws are there?

A simultaneous draw is a combination. Here, n = 20 and p = 3. Number of combinations:. C³₂₀ =1140 b. How many draws can be made with three white chips?

3 white among 10 white (and 0 other among 10 others): C³₁₀ =120 c. How many draws would show three different colours?

1 white among 10 and 1 green among 6 and 1 red among 4: C¹₁₀× ×C¹₆ C¹₄=240 d. How many draws would show three chips of the same colour ?

3 white among 10 OR 3 green among 6 OR 3 red among 4: 120 + 20 + 4 = 144 combinations e. How many draws would show three even chips of the same colour?

3 even white among 5 even white OR 3 even green among 3 even green: 10 + 1 = 11 combinations

Exercise 18. (Tutorial for lesson page 15)

A random experiment consists in taking one integer, at random, among {1 ; … ; 20}.

1) Determine Card(Ω).

Ω, set of all possible outcomes, is the set of the integers from 1 to 20 itself. Card(Ω) = 20 2) Let’s name some events: A : "get at least 15" and B : "get an even number". Determine :

p(A) , p(A) , p(B) , p( A∩B) , p( A∪B).

p(A) = 6/20 = 0,3 ; p(A) = 1 – p(A) = 14/20 = 0,7 ; p(B) = 10/20 = 0,5 A∩B = {16 ; 18 ; 20}, hence p( A∩B) = 3/20 = 0,15

p( A∪B) = p(A) +p(B) - p( A∩B) = 0,3 + 0,5 – 0,15 = 0,65 Exercise 19. (Tutorial for lesson page 15)

A random experiment consists in a simultaneous drawing of 3 letters in our alphabet.

1) Determine Card(Ω)

Ω, set of the outcomes, is composed with every combination of three letters. Card(Ω) = C³₂₆ =2600. 2) b. We set the following events A: "get 3 consonants", B: " get 2 consonants", C: " get 1 consonant"

and D: "get 3 vowels"

a. Are they mutually exclusive?

They are indeed mutually exclusive. e.g. a draw showing 2 consonants (hence belonging to B) doesn’t own three consonants (hence does not belong to A), and reciprocally.

Same comments for the differences between A and C, A and D, B and C, B and D, C and D.

b. Do they represent a partition of Ω?

These four subsets do not only have their own elements without sharing a single one, but they also include all the elements of Ω. Hence, the set {A; B; C; D} is a partition of Ω.

c. Calculate their cardinal numbers and then their probabilities (writing four significant figures). Finally, check the sum of their cardinal numbers and the sum of their probabilities.

alphabet A B C D

consonants 20 3 2 1 0

vowels 6 0 1 2 3

total 26 3 3 3 3

Card(A) = C³₂₀× =C⁰₆ 1140 p(A) = 1140 / 2600 = 43,85 % Card(B) = C²₂₀× =C¹₆ 1140 p(B) = 1140 / 2600 = 43,85 % Card(C) = C¹₂₀× =C²₆ 300 p(C) = 300 / 2600 = 11,54 % Card(D) = C⁰₂₀× =C³₆ 20 p(D) = 20 / 2600 = 0,7692 % Exercise 20. (Tutorial for lesson page 15)

Random experiment: roll two dice and add both results.

1) What are the different possible sums?

{2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10 ; 11 ; 12}

2) Are they equally likely?

No. e.g.: there are four possibilities for our dice to reach a total of 5: (1, 4) ; (4, 1) ; (2, 3) ; (3, 2), but only one

possibility to make a sum of 2: (1, 1).

3) Build a sample space of equally likely outcomes.

If each side of a die has the same chances of being obtained than the other, then each pair of sides of two dice has the same chances than any other pair. We can therefore build the following sample space with equally likely pairs: Ω = {(1, 1) ; (1, 2) ; (2, 1) ; (2, 2) ; (1, 3) ; (3, 1) ; (2, 3) ; … ; (5, 6) ; (6, 5) ; (6, 6)}, whose cardinal number is 36.

4) We set the following events: A: "the sum equals 10" and B: "the sum is at least 10".

Determine p(A) ; p(A) ; p(B).

A = {(4, 6) ; (5, 5) ; (6, 4)}. Card(A) = 3. p(A) = 3/36 = 1/12. p(A) = 1 – p(A) = 33/36 = 11/12.

B = {(4, 6) ; (5, 5) ; (6, 4); (5, 6); (6, 5);(6, 6)}. Card(B) = 6. p(B) = 6/36 = 1/6.

Exercise 21. (Tutorial for lesson page 16)

E = {1, 2, 3, …, 10}, A: "even numbers of E", B : "multiples of 3 in E"

Venn diagram: probabilistic choice tree:

E 1/5

2 4 6

8 10 A 1/2 4/5

1 5 3 B

7 9 1/2 2/5

3/5 contingency table:

A A Inside, cardinal numbers of the

corresponding intersections must be placed, subtotals ("marginal

frequencies") and the overall total

B 1 2 3 = Card(B)

B ^{4 3} 7 = Card(B)

5 = Card(A) 5 = Card(A) 10 = Card(E) 1) Complete the contingency table.

2) Let’s choose a number between 1 and 10, at random, not looking at it.

a. What is the probability it would be even? p(A) = 5/10

b. What is the probability it would be a multiple of 3 ? p(B) = 3/10

3) Let’s choose a number between 1 and 10, at random, not looking at it, but we’re told it’s a multiple of 3.

a. What is the probability it would be even? 1/3

b. What is the result obtained by the corresponding formula?

( ) ( ) ( )

^//

B

p A B 1 10 1

p A

p B 3 10 3

= ∩ = =

4) Let’s choose a number between 1 and 10, at random, not looking at it, but we’re told it’s even.

a. What is the probability it would be a multiple of 3 ? 1/5

b. What is the result obtained by the corresponding formula?

( ) ( ) ( )

^//

A

p A B 1 10 1

p B

p A 5 10 5

= ∩ = =

Exercise 22. (Tutorial for lesson page 16)

A laboratory has developed a breathalyzer. A reliability test has been done on this product, with a test- population on which it's been stated that 2% exceed 0.5 g/L (event E) and so are out of law. Everyone exhales into the breathalyzer; the event P refers to a positive result given by this device.

The reliability test has given the following results:

- 95% of people who really exceed 0.5 g/L got a positive result by the breathalyzer;

- 96% of people who don't exceed 0.5 g/L got a negative result by the breathalyzer.

What is then your probability of really exceeding 0.5 g/L, given that your result is positive?

2;4;6;8;10

A

1;3;5;7;9

A

B

6

B

3;9 2;4;8;10

B

1;5;7

B

The information given above enable us to write: p(E) = 0.02 ; pE(P) = 0.95 ; ^pE

( )

^{P = 0.96.}

The exercise asks us to determine pP(E).

( ) ( )

( ) ( )

( ) ( )

^{. .. .. . .. . %}

P

p P E p P E 0 02 0 95 0 019

p E 32 65

p P p P E p P E 0 02 0 95 0 98 0 04 0 0582

∩ ∩ ×

= = = = =

∩ + ∩ × + ×

Exercise 23. (Tutorial for lesson page 16)

1) Taking back exercise 22: are E and P independent?

there are two ways to decide whether yes or not:

* ^{p P}

(

^{∩ =E}

) ( ) ( )

^{p P ×p E} ? they aren’t: 0.019 ≠ 0.0582 × 0.02 * p PE

( )

=pE

( )

P ? they aren’t: 0.95 ≠ 0.04 2) Taking back exercise 18: are A and B independent?

there are two ways to decide whether yes or not:

* ^{p A}

(

^∩B

) ( ) ( )

^{=p A ×p B} ? they are: 0.15 = 0.3 × 0.5 * pA

( )

B =pA

( )

B ? they are: 3/6 = 7/14 Exercise 24.

3 dice are rolled together.

1) How many different outcomes are possible?

n = 6 available figures; p = 3 obtained. The result of a throw is a p-list. Card(Ω) = 6³ = 216 triplets.

2) Calculate the probabilities of the following events:

a. A : "get a triple 6"

1 outcome of Ω corresponds: the list (6, 6, 6). Card(A) = 1. p(A) = 1/216.

b. B : "get a triple"

6 outcomes of Ω correspond. Card(B) = 6. p(B) = 6/216 = 1/36.

c. C : "get a 421"

One have to get at the same time a 4, a 2 and a 1, whatever the order. There are 3! = 6 permutations (orders) of three elements, so: 6 different outcomes in Ω that show 4 2, 1. p(C) = 6/216 = 1/36.

d. D : "get at least one 4"

The contrary is easier to treat: getting no 4.

This happens in case each of the three figures is taken among the figures 1,2,3,5,6.

The number of such p-lists is 5³ = 125. Card(D) = 216 – 125 = 91. p(D) = 91/216.

e. E : "get a sum of 10"

The list of the groups of three digits giving a total of 10 must be established; then, we have to tell how many different orders are possible between the three figures, for each group.

{1 ; 3 ; 6} 6 orders; {1 ; 4 ; 5} 6 orders; {2 ; 2 ; 6} 3 orders; {2 ; 3 ; 5} 6 orders; {2 ; 4 ; 4} 3 orders; {3 ; 3 ; 4}

3 orders. All in all, 27 triplets, out of 216, give a sum of 10. p(E) = 27/216 = 1/8.

3) a. Are B and D mutually exclusive?

No, they can be realized at the same time with the triplet (4, 4, 4).

b. Are C and E mutually exclusive?

Yes, a 4, a 2 and a 1 don’t give a sum of 10.

Exercise 25.

18 balls lay in a jar: 7 white, 9 red, 2 green. Three balls are simultaneously taken out.

1) What is the probability to see the three colours in your hand?

n = 18 available balls; p = 3. Repetition: no (simultaneous); order: no (simultaneous). Card

( )

Ω =C18³ =816. Setting A: "One ball for each colour". Card A

( )

=C¹7× ×C¹9 C¹2=126. p(A) = 126/816 ≈ 0.1544.

2) What is the probability to see only one colour in your hand?

The corresponding event, B, is the union of two mutually exclusive events: "3 white" OR "3 red", so, we must add their cardinal numbers and their probabilities. Card B

( )

= +C³7 C³9=119^. p B

( )

=119/816≈0.1458

3) a. Which third event makes a partition of Ω with both former ones?

That’s the event C: "two different colours among the three balls", that contain every other outcomes but the ones of A or B.

b. Deduce its probability. p(C) = 571/816

4) Calculate the probability to get no white ball or no green ball.

We must calculate the cardinal number of two events, not necessarily mutually exclusive (so we can’t only add both: we have to use the general formula). Setting D: "0 white" and E: "0 green".

Card(D∪E) = Card(D) + Card(E) – Card(D∩E) = C³11+C16³ −C³9=641^. p D

(

∪ =E

)

641/816≈0.7855^. Exercise 26.

A bag contains 20 coins: n are black and the 20-n others are white.

2 coins have to be picked up together.

1) Express (with n) the probabilities of the following events:

a. A: One black and one white.

20 available coins; p = 2 coins to be taken. Order: no, repetition: no. Card

( )

Ω =C²20=190. To get an element of A, we need: 1 black from n black and 1 white from 20-n white.

( ) ( ) ( ) ( )

1 1 .

20

Card A C C 20 p A 20 n

n n 190

n n

n n

−

= × = − = −

b. B: two black

To get an element of B, we need 2 black from n black.

( ) ( ) ( ) ( )

2 1 . 1

Card B C p B

2 380

n

n n− n n−

= = =

c. C: two white

Element of C: 2 white from 20-n white.

( ) ( )( ) ( ) ( )( )

2 .

20

20 19 20 19

Card C C p C

2 380

n

n n n n

−

− − − −

= = =

2) Check that the global probability equals 1.

The events A, B and C combined are a partition of Ω. Thus, the sum of their probabilities must be 1:

p(A) + p(B) + p(C) = ²

(

²⁰

) (

¹

) (

²⁰

)(

¹⁹

)

^{2 2 40 2 380 2 39 380} ₁

380 380 380

n − +n n n− + −n −n =− n + n+ − +n n +n − n= = 3) Determine, by solving an equation, the values of n such that p(C) > 0.5.

20-n and 19-n are two consecutive integers. Let’s test several values:

12×11 = 132 ; 13×12 = 156 ; 14×13 = 182 ; 15×14 = 210

As long as 20-n is more than or equal to 15 (and then 19-n ≥ 14),

(

²⁰⁻ⁿ

)(

¹⁹⁻ⁿ

)

is more than 190 (that’s what we need to get p(C) > 0.5). Therefore, n must be less than 6: n ≤ 5.

Exercise 27.

A taxi is involved in a pile-up at night. Two taxi companies, the Green and the Blue, operate in the city. We have the following data:

(a) 85 % of the taxis (in the city) are Green and 15 % are Blue.

(b) A witness identified the taxi responsible as a Blue.

The court tested the reliability of the testimony in such circumstances (accident at night) and concluded that the witnesses correctly identified the colours in 80% of cases and were wrong in 20% of cases.

What is the probability that the taxi involved in the accident was a Blue?

Make sure not to make a mistake: the 80% given probability is p(Identified Blue / Blue), whereas we’re asked about the probability p(Blue / Identified Blue)!

( ) ( )

( ) ( )

( ) ( ) ^{( ) ( ) ( )} ( ) ^{( ) ( )}

% % %

% % % % % %

b ib

b b

p b ib p b ib p b p ib

p b

p ib p b ib p b ib p b p ib p b p ib

15 80 12 12

41.38 %

15 80 85 20 12 17 29

∩ ∩ ×

= = =

∩ + ∩ × + ×

= × = = ≈

× + × +

This result may seem paradoxical (less than one chance in two that the taxi is actually blue!), but one must take into account the large number of green taxis (85% of taxis) for which the witness could be confused (in 20% of cases), which is far from being insignificant.

It is therefore necessary to learn to apply the Bayes formula automatically, without making mistakes, and leaving aside the (often false) result that one intuitively expects.

To support the above calculation, and this is often beneficial, a cross table can be built which will consider, for example, a total of 100 taxis:

Blue Green

identified Blue 12 17 29

identified Green 3 68 71

15 85 100

* To complete it, start with the last line respecting the 15% and 80% frequencies, then the four central boxes respecting the 80% and 20% frequencies within the "blue" column and then the "green" column.

Finally, the sub-totals of the rows can be calculated.

* We then see that, knowing that the taxi has been identified as blue, there are only 12 chances out of 29 that it is really blue!

Exercise 28.

A bank found that 2% of the checks issued by its clients aren't correctly worded (correctly worded : event W).

97% of correctly worded checks are correctly entered by the agent into the bank's data base (event E).

When it's not correctly worded, the agent is able to correct the mistakes 5 times out of 100.

A check has to be entered into data base. calculate the probabilities of the following events : 1) The agent doesn’t enter the check correctly.

Here, we're crossing two events and their contraries. We can represent it by a probabilistic choice tree.

Given W : "the check is correctly worded" and E : "the agent enters it correctly".

( ) ( ) ( ) ( )

^W

( ) ( )

^W

( )

^{. . . . . %}

p E =p E∩W +p E∩W =p W ×p E +p W ×p E =0 02 0 95 0 98 0 03× + × =4 84 0.05

0.02 0.95

0.98 0.97

0.03

2) The check has been correctly worded, given that the agent entered it correctly

( ) ( )

( ) ( ) ( )

( )

^{. . . . %}

- . .

W E

p E W p W p E 0 98 0 97 0 9506

p W 99 89

p E 1 p E 1 0 0484 0 9516

∩ × ×

= = = = =

−

3) The check has not been correctly worded, given that the agent didn’t enter it correctly

( ) ( )

( ) ( ) ( )

( )

^{W . . . .. . %}

E

p E W p W p E 0 02 0 95 0 019

p W 39 26

p E p E 0 0484 0 0484

∩ × ×

= = = = =

One can notice that a little bit less than 60% of the checks for which the agent made a mistake were actually correctly worded by the client.

Contingency table to answer the questions (total example of 10000 cheques):

not correctly worded correctly worded

entered correctly 10 9506 9516

not entered correctly 190 294 484

200 9800 10000

p(not entered correctly) = 484/10000 p(correctly worded / entered correctly) = 9506/9516 p(not correctly worded / not entered correctly) = 190/484

W

W

E E E E

Exercise 29.

35 % of people in a city are employees. Among them, 8 people on 10 use their car every day, whereas 30 % of the unemployed do (employee: event E; use car every day: event D). We are to select one person in this city, at random.

1) Display the different categories of people in a probabilistic choice tree.

0.8

0.35 0.2

0.65 0.3

0.7

2) a. What is the probability that this random individual be unemployed?

( ) ( )

p E = −1 p E = 0.65

b. What is the probability that this person be an employee who uses his/her car every day?

( ) ( )

^E

( )

^{. . %}

p E∩D =p E ×p D =0 35 0 8× =28

c. What is the probability that he/she uses his/her car every day?

( ) ( ) ( )

^.

( )

^E

^{( )}

^{. . . . %}

p D =p D∩ +E p D∩E =0 28 p E+ ×p D =0 28+0 65 0 3× =47 5

d. Given that this person drives his/her car every day, probability he/she is an employee?

( ) ( )

( )

^{. . %}

D

p E D 0 28

p E 58 95

p D 0.475

= ∩ = =

e. Are the events E and D independent?

two ways to answer it:

* ^{p E}

(

^∩D

) ( ) ( )

^{=p E ×p D} ? no: 0.28 ≠ 0.35 × 0.475 * ^{p DE}

( )

^=pE

( )

^D ? no: 0.8 ≠ 0.3 Contingency table to answer the questions (total example of 1000 people):

employee unemployed

every day 280 195 475

not every day 70 455 525

350 650 1000

p(unemployed) = 650/1000 p(employee and car every day) = 280/1000 p(car every day) = 475/1000 p(employee / car every day) = 280/475 Exercise 30.

Following the discovery in one country of the first cases of a contagious disease, a major vaccination campaign was carried out: 70% of the inhabitants were vaccinated. A study revealed that 5% of the vaccinated had been affected to varying degrees by the disease, while 60% of the unvaccinated had been affected to varying degrees. Calculate:

1) The probability that an individual randomly selected from the population has been affected by this disease.

Let's represent a tree (which is not necessary for the exercise); V: vaccinated, M: ill

0.05 0.035

0.7 0.95 0.665

0.3 0.6 0.18

0.4 0.12

( ) ( ) ( )

p M =p M∩V +p M∩V =0.035 0.18+ =0.215=21.5 % E

E

D D D D

V

V

M M

M M

2) The probability that an individual has been vaccinated, given that he or she has been affected.

( ) ( )

( )

M

p V M 0.035

p V 0.1628 16.28 %

p M 0.215

= ∩ = ≈ =

Contingency table to answer the questions (total example of 1000 people):

vaccinated not vaccinated

ill 35 180 215

not ill 665 120 785

700 300 1000

p(ill) = 215/1000 p(vaccinated / ill) = 35/215 Exercise 31.

Four girls and three boys are required to take an oral examination. The examiner decides to draw up a random list of candidates establishing the order in which they are to take the test. To do this, he puts the names (all supposedly different) of the seven candidates in an envelope and then draws the seven names one after the other.

Let’s name F1 the event: "The first candidate interviewed was a girl", and F2 the event: "The second candidate interviewed was a girl".

1) What is the probability that the first two candidates interviewed are girls?

Let's represent a tree of the first two choices; G : girl, B : boy

3/6 2/7

4/7 3/6 2/7

3/7 4/6 2/7

2/6 1/7

(

1 2

) ( )

1 G1

( )

2

4 3 2

p G G p G p G 28.57 %

7 6 7

∩ = × = × = ≈

2) What is the probability that the first candidate interviewed is a girl given that the second candidate interviewed is a girl?

( ) ( )

( )

2

1 2

G 1

2

p G G 27 1

p G 50 %

2 2

p G 2

7 7

= ∩ = = =

+

Contingency table to answer the questions (total example of 700 trials):

1st = boy 1st = girl

2nd = boy 100 200 300

2nd = girl 200 200 400

300 400 700

p(1st = girl AND 2nd = girl) = 200/700 p(1st = girl / 2nd = girl) = 200/400 Exercise 32.

During a TV game show, a contestant stands in front of three boxes, only one of which contains the jackpot, and chooses one box at random. Before opening it, the presenter points out one of the other two boxes that does not contain the jackpot.

The presenter then suggests two strategies to the candidate:

STRATEGIE 1 : the candidate maintains his or her first choice;

STRATEGIE 2 : the candidate does not maintain his or her initial choice and chooses the door not designated by the presenter.

Which one is the best strategy for the player?

G1

B1

G2

B2

G2

B2

Note that if the player's first choice is not the right one, then the player is sure to win if he changes his mind afterwards (but he doesn't know if his first choice is the right one!).

What happens to a player who has decided not to change his mind, no matter what?

Only his first choice counts, so he has a one in three chance of winning.

What happens to a player who has decided to change his mind, no matter what?

If he had made the right choice at the start, then he will lose (prob = one chance out of three); but if he had made the wrong one, then a change of mind will win him for sure! (prob = two chances out of three).

Obviously, deciding to change our mind doubles our chances of winning.

Let's represent a probability tree of the possible sequences of the game, assuming that a player is

undecided and has a one in two chance of changing his or her mind (or, equivalently, a tree of proportions representing the games played by a very large number of people, half of whom decide to change their mind).

Let's define events : G = good choice of the player, C = the player changes his mind afterwards.

1er choix 1/2 loser, prob = 1/6

1/3 1/2 winner, prob = 1/6

2/3 1/2 winner, prob = 2/6

1/2 loser, prob = 2/6

The question is: does changing one's mind increase the probability of winning?

( ) ( )

( ) ( )

( )

C

2 1

p G C

p C win 3 2 2

p win

p C p C 1 3

2

∩ ×

= ∩ = = = : two thirds of the players who changed their mind won.

( ) ( )

( ) ( )

( )

C

1 1

p C win p G C 3 2 1

p win

1 3

p C p C

2

∩ ∩ ×

= = = = : one third of those who did not change their mind won.

One last note: according to this tree, the probability of winning is equal to 1/2. Indeed, this tree takes into account people who have decided to change their mind (and of whom two thirds have won) as well as those who have not changed their mind (and of whom one third have won). By combining these two categories, half of the people who played won (1/2 is the average of 1/3 and 2/3).

Exercise 33. (Tutorial for lesson page 16)

A lottery is held. 100 tickets are to be sold, €1 each. One ticket is a €30 winner, two are €15 winners, and seven would make the buyer win €1. Considering we want to purchase one ticket, X is the random variable of the net gain (the €1 expense taken into account).

1) Give the probability distribution of X.

xi 29 14 0 -1

p(X = xi) or pi 0.01 0.02 0.07 0.90

2) If we’re playing this lottery the same way many times, can we expect to be a long-term winner?

(begin by an estimate of what would be likely to occur after a thousand attempts) On 1000 attempts, our expected gain is negative: 10×29 + 20×14 + 70×0 - 900×1 = -330.

This result means that we can expect an average loss of 33 cents per attempt.

G

G

C C C C

Even though this accurate result is not likely to occur, the law of large numbers tells us that the more we’ll play this lottery, the closer our actual average gain will be to this average expected gain. One cannot expect to be a long-term winner.

Exercise 34. (Tutorial for lesson page 16)

1) Calculate the expected value and the standard deviation, with the data of exercise 33. Comment.

Using the calculator, on stat mode: let’s enter the gains on list 1 and their probabilities on list 2.

Casio: CALC SET: 1Var X: List 1 and 1Var F: List 2, then EXIT, 1VAR.

TI: CALC: Stat1Var L1,L2.

Results: E(X) = -0.33 and σ(X) = 3.622.

This means that in n attempts, with n big enough, the actual average gain is a random variable, normally distributed, centred on -0.33, with a standard deviation of 3 622.

n . e.g. in 10 000 attempts, the standard deviation of the possible average gain (around -0.33) is approximately 0.036 €. The fact that its distribution is normal implies for instance that:

* there are 68.3 % chances that the actual average gain would be between -0.33 – 0.036 and -0.33 + 0.036, so between -0.366 and -0.294, and then that we would lose between €2940 and €3660;

* there are 95.4 % chances that the actual average gain would be between -0.33 – 2×0.036 and -0.33 + 2×0.036, so between -0.402 and -0.258, and then that we would lose between €2580 and €4020.

Globally, on this number of attempts, the standard deviation of the average gain is quite low and makes it virtually impossible to make money. The bigger n is, the lower the standard deviation is, the less likely it is that our loss approaches zero and the greater the chances that our average gain is close to € -0.33. The calculation of this standard deviation is a numerical translation of the law of large numbers.

2) If the possible gains and loss were doubled (in the initial array), what would these parameters become?

Logically, they would be doubled. You may check this thanks to your calculator.

3) If the values of X were increased by € 0.5, what would these parameters become?

The expected value would too, and then would become non negative: € +0.17.

However, the standard deviation wouldn’t be affected: increase a list of values by the same number only shifts them, but does not change the distances between them. Just check!

Exercise 35.

There are two hospitals in one city. In the larger one, about 45 babies are born every day, while in the smaller one, about 15 babies are born every day. As you know, about 50% of all babies are boys. However, the exact percentage on any given day varies: sometimes it is more than 50%, sometimes less.

Over a period of one year, each hospital recorded the days on which more than 60% of babies born were boys.

In your opinion, which hospital recorded the most days of this type?

Here, let's call the event B "birth of a boy", with a probability of 50%. The law of large numbers shows that the greater the number of experiences, the more likely the actual frequency of an event tends towards its probability. In a hospital where a lot of babies are born, the frequency of B is therefore less likely to deviate from 50%, compared to a hospital where fewer babies are born.

The variability of the frequency of B is therefore greater in the small hospital than in the large one.

Exercise 36.

A game consists of two identical boxes each having 10 chips numbered 1 through 10. The experiment is to pick a chip in each box.

1) a. Describe one of the possible outcomes.

An outcome is an ordered list of two chips. e.g.: (2, 5) ; (10, 8) ; (5, 2), etc.

b. Explain why the sample space’s cardinal number is 100.

These lists consist of two components (p = 2) chosen from 10 (n = 10), with a possible repetition and order to take into account: (5, 2) and (2, 5) are two different results. The sample space is thus a set of p- lists, and Card(Ω) = 10² = 100.

c. What is the probability of choosing two even numbers?

These p-lists consist of two elements (p = 2) taken from 5 even elements (n = 5). Number of lists: 5² = 25.

probability: 25/100 = 0.25.

d. Prove that the probability of two different even numbers is 0.2.

These p-lists consist of two elements (p = 2) chosen from 10 (n = 10), without repetition of the same element. Thus, we’re here looking for a number of permutations: P₅²=20.

One can give another explanation, noticing that on the 25 lists of even numbers, 5 are a double, and then 20 aren’t… Finally, the requested probability is 20/100 = 0.2.

2) For one game, you have to spend € 1. If you get two different even numbers, you win € 1; if you get two identical numbers except 1 and 1, you earn € 6; 1 and if you get the double one, you win € 50; in all other cases, no gain. The random variable X gives the gain at the end of the game, regardless of the initial € 1 bet.

a. Give the probability distribution of X.

xi 50 6 1 0

p(X = xi) or pi 0.01 0.09 0.20 0.70

b. Give the expected value of X.

E(X) = 0.01×50 + 0.09×6 + 0.2×1 = 1.24 €

c. Can we expect to win money on playing this game a lot?

This expectation is bigger than the bet, with a € 0.24 deviation in favour of the player. One can then expect an actual average gain close to € 0.24 per game, playing this game a lot (e.g.: € 240 € in 1000 games).

Exercise 37.

A bag contains 5 white and 10 black balls. You bet € 2 for a 3 balls draw together. Get 3 white makes you earn

€ 100; 2 white: € 10; 1 white: € 2; 3 black: nothing. The random variable X is your gain at the end of a test, once deduced the bet.

1) Give the probability distribution of X.

Total number of different possible draws: ^Card

( )

^{Ω =C153 =455}

Number of draws that show 3 white: ^{Card A}

( )

^{= ×C35 C010=10}

Number of draws that show 2 white: Card B

( )

= ×C²5 C¹10 =100 Number of draws that show 1 white: ^{Card C}

( )

^{= ×C15 C210=225}

Number of draws that show 0 white: ^{Card D}

( )

^{= ×C05 C310 =120}

Probability distribution of X, net gain:

xi 98 8 0 -2

p(X = xi) or pi 10/455 100/455 225/455 120/455 2) Give the expected value and the standard deviation of X.

E(X) = 1540/455 = € 3.385. σ(X) = € 14.65

3) If you play a hundred times, what gain is the most likely? 338 €

Exercise 38.

A game consists of a random draw of a letter from the alphabet (which contains 20 consonants and 6 vowels A, E, I, O, U, Y). Each letter is assigned a number according to the following table:

A B C D E F G H I J K L M

1 2 3 4 5 6 7 8 9 10 11 12 13

N O P Q R S T U V W X Y Z

14 15 16 17 18 19 20 21 22 23 24 25 26

We'll set the events C: "the letter is a consonant" and M: "its number is at least 17".

Part 1

1) Build a probabilistic choice tree (1st level: C and its contrary; 2nd level: M and its contrary) into which the simple, conditional and intersection probabilities will be placed.

8/20 8/26

20/26 12/20 12/26

6/26 2/6 2/26

4/6 4/26

2) Given that a vowel has been drawn, what’s the probability its number is more than 16?

( )

C

2 1

p M

6 3

= =

3) Given that its number is more than 16, what’s the probability it’s a vowel?

( ) ( )

( )

M

p C M 226 2 1

p C

8 2

p M 10 5

26 26

= ∩ = = =

+

4) Are the events M and C independent?

( )

^{8 .}

( ) ( )

^{20 10 .}

p C M 0 3077 and p C p M 0 2959

26 26 26

∩ = ≈ × = × ≈

These events are not completely independent (slightly dependent, or related).

Part 2

The event C∩M, is awarded a € 10 gain and the event C ∩M would make you lose € 5; as for the other possibilities: they don’t lead to either gain or loss. X is the random variable “gain after one draw”.

1) Give the probability distribution of X.

xi 10 0 -5

p(X = xi) or pi 2/26 16/26 8/26 2) Give the expected value of X and its meaning.

E(X) = -20/26 = € -0.7692. This is the average gain expected on a long series of attempts.

3) Give the standard deviation of X and its meaning.

σ(X) ≈ € 3.846: average gain fluctuation around E(X) at each attempt.

C

C

M M M M

Exercise 39. (Tutorial for lesson page 19)

From a jar that contains 7 white balls and 3 black balls, let’s draw two balls, one after the other and without putting the first one back. We name X1 the random variable corresponding to 1 point in case the first ball is black and 0 point in case it’s white ; we name X2 the random variable corresponding to 1 point in case the first ball is black and 0 point in case it’s white.

a. Give p(X1 = 0) and p(X1 = 1). p(X1 = 0) = 7/10 and p(X1 = 1) = 3/10 b. Give pX1 = 0(X2 = 0) and pX1 = 0(X2 = 1), then pX1 = 1(X2 = 0) and pX1 = 1(X2 = 1).

pX1 = 0(X2 = 0) = 6/9 and pX1 = 0(X2 = 1) = 3/9, pX1 = 1(X2 = 0) = 7/9 and pX1 = 1(X2 = 1) = 2/9 c. Complete the probabilistic choice tree and then the associated probability table.

X1

X2

0 1

PX2

0 42/90 21/90 7/10

1 21/90 6/90 3/10

PX1 7/10 3/10 1

d. Is the knowledge of both marginal distributions sufficient for the knowledge of the joint distribution?

Knowing the subtotals of this table doesn’t allow us to attribute values to the four empty cells that represent the intersections.

e. Compare p(X1 = 0)×p(X2 = 0) to p((X1 = 0)∩(X2 = 0)). Are the variables X1 and X2 independent?

7/10 × 7/10 ≠ 42/90. These variables are not independent.

Exercise 40.

Two sales agents A and B of a cooperative work in team for two weeks to obtain orders from potential customers. A is responsible for placing new contracts to existing customers and B is responsible for prospecting new customers.

Let’s name: XA the random variable measuring the number of contracts obtained by A and XB the random variable measuring the number of contracts obtained by B.

It’s assumed that XA can only take its values in {0 ; 1 ; 2 ; 3} and XB in {0 ; 1}.

The joint distribution of XA and XB is given through the following table:

XA

XB

0 1 2 3

0 0.05 0.15 0.20 0.10

1 0.1 0.2 0.15 0.05

1) a. Determine the margin distributions of XA and of XB.

By adding the matching crossed probabilities, these distributions are visible:

margin distribution of XA :

k 0 1 2 3

p(XA = k) 0.15 0.35 0.35 0.15 margin distribution of XB :

k 0 1

p(XB = k) 0.50 0.50 3/10 7/10

6/9 3/9

7/9 2/9 X1 = 0

X1 = 1

X2 = 0 X2 = 1 X2 = 0 X2 = 1

b. Are these variables independent?

Let’s consider one of the possible crossings: XA = 0 and XB = 0.

p((XA = 0) and (XB = 0)) = 0.05 (as seen in the first table)

p(XA = 0) × p(XB = 0) = 0.15 × 0.5 = 0.075 (both intermediate values coming from the 2nd and 3rd tables) These results are not equal, thus XA and XB are not independent.

2) Let’s set a new variable, X “total number of obtained contracts”, by X = XA + XB. a. Give the probability distribution of X.

We have to consider every possible crossing between one value of XA and one value of XB, and then give the corresponding value of X:

XA 0 1 2 3 0 1 2 3

XB 0 0 0 0 1 1 1 1

prob 0.05 0.15 0.20 0.10 0.10 0.20 0.15 0.05

XA + XB 0 1 2 3 1 2 3 4

Finally, the probability distribution of X is given by consolidating identical values:

k 0 1 2 3 4

p(X = k) 0.05 0.25 0.40 0.25 0.05

b. Calculate E(X) and V(X).

E(X) = 0×0.05 + 1×0.25 + 2×0.4 + 3×0.25 + 4×0.05 = 2

V(X) = E(X²) – E(X)² = 0²×0.05 + 1²×0.25 + 2²×0.4 + 3²×0.25 + 4²×0.05 – 2² = 4.9 – 4 = 0.9