Matematici speciale

Probleme clasice in teoria

probabilitatilor

Aprilie 2018

ii

“Regula 50 − 50 − 90: de fiecare data cand ai o sansa de 50 − 50 saintelegi ceva corect exista o probabilitate de 90% sa intelegi gresit.”

Andy Rooney

8Probabilitati. Probleme clasice

Problema intalnirii

Un barbat si o femeie decid sa se intalneasca intr-un restaurant dupa ora 21.Restaurantul se inchide la ora 24. Din cauza programului incarcat al fiecaruiaei decid ca in cazul in care unul dintre ei va intarzia fiecare sa astepte dupacelalalt un anumit timp. Barbatul este dispus sa astepte o ora iar femeia doar15 minute!

Care este probabilitatea ca cei doi sa se intalneasca la acel restaurant?

1

Solutie: Vom modela matematic problema in felul urmator: notam cu 𝑥timpul la care soseste femeia la restaurant si cu 𝑦 timpul la care soseste barbatul.Putem sa consideram ora 21 ca fiind timpul 0 si atunci 24 va fi reprezentat de3. Asadar 𝑥, 𝑦 ∈ [0, 3]. Toate posibilitatile sunt reprezentate de punctele (𝑥, 𝑦)din interiorul patratului [0, 3] × [0, 3] de mai jos.

Daca barbatul soseste primul, adica 𝑦 ≤ 𝑥 atunci cei doi se vor intalni daca𝑥 − 𝑦 ≤ 1 (adica timpul la care soseste femeia este cu cel mult o ora pestecel al sosirii barbatului). Toti timpii de sosire care satisfac aceste restrictiisunt continuti in regiunea gri, din interiorul patratului, cuprinsa intre primabisectoare 𝑦 = 𝑥 si dreapta 𝑥− 𝑦 = 1

Daca femeia soseste prima, adica 𝑥 ≤ 𝑦 atunci cei doi se intalnesc daca𝑦 − 𝑥 ≤ 1

4 . Toti timpii de sosire care satisfac aceste restrictii sunt continuti inregiunea gri, din interiorul patratului, cuprinsa intre prima bisectoare 𝑦 = 𝑥 sidreapta 𝑦 − 𝑥 = 1

4Probabilitatea ca cei doi sa se intalneasca va fi:

𝑃 =numar cazuri favorabile

numar cazuri posibile.

Evident sunt o infinitate de cazuri favorabile si o infinitatea de cazuri posi-bile. Putem totusi sa estimam probabilitatea fara sa numaram toate punctele(𝑥, 𝑦) care corespund cazurilor favorabile sau posibile si anume utilizand ari-ile regiunilor care corespund multimii cazurilor favorabile, respectiv multimiicazurilor posibile.

Probabilitatea ca cei doi sa se intalneasca :

𝑃 =aria regiunii gri

aria patratului=

10332

32≈ 35%.

2

Problema Monty Hall

Sa presupunem ca esti la un concurs TV in fata a trei usi si trebuie saalegi una pentru a castiga o masina. In spatele unei usi se afla masina iar inspatele celorlalte doua se afla cate o capra. Ai ales o usa, sa zicem usa nr. 1.Prezentarorul emisiunii, care stie unde se afla masina, deschide apoi una dintreusi, in spatele careia se va afla evident o capra, sa presupunem ca e vorba deusa nr. 3.

Apoi prezentatorul se joaca cu mintea ta si spune:“Iti voi da posibilitatea sa iti schimbi optiunea si sa alegi usa 2, daca doresti.”Vei alege usa nr.2 sau vei ramane cu alegerea initial facuta ( usa nr.1) ?

3

Tehnici de numarare:

∙ multiplication rule of counting: if a task consists of a sequence of 𝑛 choicesin which there are 𝑝1 ways to make the first choice, 𝑝2 ways to make the second,etc., then the task can be done in:

𝑝1 · 𝑝2 · . . . · 𝑝𝑛

different ways.∙ permutations of 𝑛 distinct objects taken 𝑘 at a time: the number of

arrangements of 𝑘 objects chosen from 𝑛 objects in which:

· the 𝑛 objects are distinct

· repeats are not allowed

· order matters

is given by the formula 𝐴𝑘𝑛 =

𝑛!

(𝑛− 𝑘)!∙ combination of 𝑛 distinct objects taken 𝑘 at a time: the number of ar-

rangements of 𝑛 objects using 𝑘 ≤ 𝑛 of them, in which:

· the 𝑛 objects are distinct

· repeats are not allowed

· order does not matter

is given by the formula 𝐶𝑘𝑛 = 𝑛!

(𝑛−𝑘)!𝑘!

∙ arrangements with non-distinct items (repetitions): the number of permu-tations of 𝑛 objects, where there are 𝑛1 of the 1st type, 𝑛2 of the 2nd type,. . . ,𝑛𝑘 of 𝑘-th type is:

𝑛!

𝑛1!𝑛2! · . . . · 𝑛𝑘!, 𝑛1 + 𝑛2 + . . . + 𝑛𝑘 = 𝑛.

In continuare vom nota prin |𝐴| numarul de elemente ale multimii 𝐴.Teorema: ( Principiul incluziunii si excluziunii)For any finite sequence 𝐴1, 𝐴2, . . . , 𝐴𝑛 of subsets of a finite set 𝑋 we have:

𝑛⋃

𝑘=1

𝐴𝑘

=

𝑛∑𝑖=1

|𝐴𝑖|−∑

1≤𝑖<𝑗≤𝑛

|𝐴𝑖∩𝐴𝑗 |+. . .+(−1)𝑝−1∑

1≤𝑖1<𝑖2<...<𝑖𝑝≤𝑛

|𝐴𝑖1∩𝐴𝑖2∩. . .∩𝐴𝑖𝑝 |

+ . . . + (−1)𝑛−1|𝐴1 ∩𝐴2 ∩ . . . 𝐴𝑛|.The principle expressed in its complementary form:

𝑛⋂

𝑘=1

𝐴𝑘

=

𝑋 ∖

𝑛⋃𝑘=1

𝐴𝑘

= |𝑋| −

𝑛∑𝑖=1

|𝐴𝑖| +∑

1≤𝑖<𝑗≤𝑛

|𝐴𝑖 ∩𝐴𝑗 | + . . .

+(−1)𝑝∑

1≤𝑖1<𝑖2<...<𝑖𝑝≤𝑛

|𝐴𝑖1 ∩𝐴𝑖2 ∩ . . . ∩𝐴𝑖𝑝 | + . . . + (−1)𝑛|𝐴1 ∩𝐴2 ∩ . . . 𝐴𝑛|.

4

Scheme clasice de probabilitate:

∙ Poincare’s theorem:

𝑃

(𝑛⋃

𝑘=1

𝐸𝑘

)=

𝑛∑𝑘=1

𝑃 (𝐸𝑘) −𝑛−1∑𝑘=1

𝑛∑𝑗=𝑘+1

𝑃 (𝐸𝑘 ∩ 𝐸𝑗)+

+

𝑛−2∑𝑘=1

𝑛−1∑𝑗=𝑘+1

𝑛∑𝑖=𝑗+1

𝑃 (𝐸𝑘 ∩ 𝐸𝑗 ∩ 𝐸𝑖) − . . . + (−1)𝑛−1𝑃 (𝐸1 ∩ 𝐸2 . . . ∩ 𝐸𝑛)

e.g. for 𝑛 = 3 :

𝑃 (𝐴 ∪𝐵 ∪ 𝐶) = 𝑃 (𝐴) + 𝑃 (𝐵) + 𝑃 (𝐶) − 𝑃 (𝐴 ∩𝐵) − 𝑃 (𝐴 ∩ 𝐶) − 𝑃 (𝐵 ∩ 𝐶)

+𝑃 (𝐴 ∩𝐵 ∩ 𝐶)

∙ the multiplication rule:

𝑃

(𝑛⋂

𝑘=1

𝐸𝑘

)= 𝑃 (𝐸1) · 𝑃 (𝐸2|𝐸1) · 𝑃 (𝐸3|𝐸1 ∩ 𝐸2) · . . . · 𝑃

(𝐸𝑛|

𝑛−1⋂𝑘=1

𝐸𝑘

)

if the events are independent:

𝑃

(𝑛⋂

𝑘=1

𝐸𝑘

)= 𝑃 (𝐸1) · 𝑃 (𝐸2) · . . . · 𝑃 (𝐸𝑛).

∙ the probability that an event 𝐴 will occur simultaneously with one of theevents 𝐻1, 𝐻2, . . . ,𝐻𝑛 forming a system of mutually exclusive events (hypothe-ses) is given by the total probability formula:

𝑃 (𝐴) =𝑛∑

𝑘=1

𝑃 (𝐻𝑘)𝑃 (𝐴|𝐻𝑘)

where∑𝑛

𝑘=1 𝑃 (𝐻𝑘) = 1∙ the probability 𝑃 (𝐻𝑗 |𝐴) of the hypothesis 𝐻𝑗 after the event 𝐴 occurred

is given by the Bayes’ formula:

𝑃 (𝐻𝑗 |𝐴) =𝑃 (𝐻𝑗)𝑃 (𝐴|𝐻𝑗)

𝑛∑𝑘=1

𝑃 (𝐻𝑘)𝑃 (𝐴|𝐻𝑘)

Experimentul binomial

∙ is a statistical experiment that has the following properties:· the experiment consists of 𝑛 repeated trials· each trial can result in just two possible outcomes: we call one of these

outcomes a success and the other, a failure· the probability of success, denoted by 𝑝, is the same on every trial.· the probability of failure, denoted 𝑞 = 1 − 𝑝, is the same on every trial

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· the trials are independent: the outcome on one trial does not affect theoutcome on other trials.

∙ the binomial probability refers to the probability that a binomial experi-ment results in exactly 𝑘 successes out of 𝑛 trials:

𝑃 = 𝐶𝑘𝑛 · 𝑝𝑘 · 𝑞𝑛−𝑘

e.g.: flip a coin 6 times, the probability to get 4 heads is:

𝑃 = 𝐶46 ·(

1

2

)4

·(

1

2

)6−4

∙ the probability that a binomial experiment results in at least 𝑘 successesis:

𝑃 = 1 −𝑘−1∑𝑖=0

𝐶𝑖𝑛 · 𝑝𝑖 · 𝑞𝑛−𝑖

∙ the probability that the 𝑘-th success is obtained after exactly 𝑟 trials is:

𝑃 = 𝐶𝑘−1𝑟−1 𝑝

𝑘(1 − 𝑝)𝑟−𝑘, 𝑟 ≥ 𝑘.

Experimentul multinomial:

∙ generalizes the binomial experiment:· now each trial can have 𝑘 outcomes: 𝐸1, 𝐸2, . . . , 𝐸𝑘

· each of these outcomes have the probabilities 𝑝1, 𝑝2, . . . 𝑝𝑘· the 𝑛 trials are again independent.∙ the multinomial probability is the probability that 𝐸1 occurs 𝑛1 times, 𝐸2

occurs 𝑛2 times,. . .𝐸𝑘 occurs 𝑛𝑘 times:

𝑃 =𝑛!

𝑛1!𝑛2! · . . . · 𝑛𝑘!𝑝𝑛11 𝑝𝑛2

2 · . . . · 𝑝𝑛𝑘

𝑘

where 𝑛 = 𝑛1 + 𝑛2 + . . . + 𝑛𝑘

Schema Poisson:

∙ let 𝐴1, 𝐴2, ..., 𝐴𝑛, be 𝑛 independent events of an experiment.∙ denote by 𝑝𝑖 the probability of 𝐴𝑖 to occur and by 𝑞𝑖 = 1 − 𝑝𝑖, 𝑖 = 1, 𝑛

the probability of the complementary event.∙ the probability that 𝑘 events, out of those 𝑛, occur is given by the coefficient

of 𝑋𝑘 in the expression:

(𝑝1𝑋 + 𝑞1) · (𝑝2𝑋 + 𝑞2) · ... · (𝑝𝑛𝑋 + 𝑞𝑛)

6

Probleme rezolvate

Problema 1. Un muncitor a lucrat 𝑛 piese. Sa notam cu 𝐴𝑖, 𝑖 = 1, 𝑛evenimentul care consta ın faptul ca cea de a ”𝑖 -a” piesa lucrata estedefecta. Sa se descrie matematic folosind limbajul teoriei multimilor ur-matoarele evenimente:a) Nici una dintre piesele lucrate nu este defecta,b) Cel putin una dintre piesele lucrate este defecta,c) Numai una dintre piesele lucrate este defecta,d) Exact doua piese sunt defecte,e) Cel putin doua piese nu sunt defecte,f) Cel mult doua piese sunt defecte.

Solutie: Fie 𝐴𝑖 evenimentul ca cea de a ”𝑖 -a” piesa este defecta si 𝐴𝑖 estebuna.

a) 𝐴1 ∩𝐴2 ∩ ...𝐴𝑛

b) 𝐴1 ∪𝐴2 ∪ ... ∪𝐴𝑛

c)

𝑛⋃𝑖=1

(𝐴1 ∩𝐴2 ∩ ... ∩𝐴𝑖−1 ∩𝐴𝑖 ∩𝐴𝑖+1 ∩ ... ∩𝐴𝑛

)d)

𝑛⋃𝑖=1𝑖<𝑗

(𝐴1 ∩𝐴2 ∩ ... ∩𝐴𝑖−1 ∩𝐴𝑖 ∩𝐴𝑖+1 ∩ ... ∩𝐴𝑗−1 ∩𝐴𝑗 ∩𝐴𝑗+1 ∩ ... ∩𝐴𝑛

)e) Evenimentul ”cel putin doua piese nu sunt defecte” este complementar

evenimentului ”cel mult o piesa nu este defecta”

(𝑛⋂

𝑖=1

𝐴𝑖

)∪

[𝑛⋃

𝑖=1

𝐴1 ∩𝐴2 ∩ ...𝐴𝑖−1 ∩𝐴𝑖 ∩𝐴𝑖+1 ∩ ... ∩𝐴𝑛

]

f)

(𝑛⋂

𝑖=1

𝐴𝑖

)∪

[𝑛⋃

𝑖=1

(𝐴1 ∩𝐴2 ∩ ... ∩𝐴𝑖−1 ∩𝐴𝑖 ∩𝐴𝑖+1 ∩ ... ∩𝐴𝑛

)]∪

∪

⎡⎢⎢⎣ 𝑛⋃𝑖=1𝑖<𝑗

(𝐴1 ∩𝐴2 ∩ ... ∩𝐴𝑖−1 ∩𝐴𝑖 ∩𝐴𝑖+1 ∩ ... ∩𝐴𝑗−1 ∩𝐴𝑗 ∩𝐴𝑗+1 ∩ ... ∩𝐴𝑛

)⎤⎥⎥⎦

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Problema 2. Dintre studentii prezenti la un curs de MS se alege laintamplare unul. Sa notam cu:𝐴 - evenimentul ca studentul ales este baiat,𝐵 - evenimentul ca studentul ales este nefumator,𝐶 - evenimentul ca studentul ales locuieste ın camin.Se cer urmatoarele:a) Sa se descrie evenimentul 𝐴 ∩𝐵 ∩ 𝐶,

b) In ce conditii are loc identitatea 𝐴 ∩𝐵 ∩ 𝐶 = 𝐴?c) Cand este adevarata relatia 𝐶 ⊆ 𝐵?d) Cand va putea avea loc egalitatea 𝐴 = 𝐵?

Solutie: a) A fost ales un baiat care nu fumeaza si care nu locuieste ın camin.b) Cand toti baietii locuiesc ın camin si nici unul nu fumeaza.c) Cand toti studentii care nu stau ın camin sunt nefumatori.d) Are loc daca nicio fata nu fumeaza si ın acelasi timp toti baietii fumeaza.

Problema 3. Intr-o societate compusa din 𝑛 perechi (sot si sotie) sedanseaza (se presupune ca formarea perechilor la dans este egal probabila).Care este probabilitatea ca la un moment dat fiecare barbat sa nu dansezecu sotia sa ? Sa se calculeze limita acestei probabilitati cand 𝑛 → ∞.

Solutie: Fie 𝐴 evenimentul ca fiecare barbat nu danseaza cu sotia sa si fie 𝐴𝑘

evenimentul ca la un moment dat sa se formeze perechea ”𝑘”daca am numerotatperechile ”sot - sotie” de la 1 la 𝑛.

Se vede ca

𝑃 (𝐴𝑖1 ∩𝐴𝑖2 ∩ ... ∩𝐴𝑖𝑘) =(𝑛− 𝑘)!

𝑛!

deoarece daca s-au format 𝑘 perechi sot-sotie, atunci celelalte (𝑛 − 𝑘) perechi”barbat-femeie” se pot forma ın (𝑛− 𝑘)! moduri.

Fie 𝐴 evenimentul ca la un moment dar s-a format cel putin o pereche.Evident 𝑃

(𝐴)

= 1 − 𝑃 (𝐴). Evenimentele 𝐴1, 𝐴2, ..., 𝐴𝑛 sunt compatibile si

dependente, iar 𝐴 = 𝐴1 ∪𝐴2∪ ...∪𝐴𝑛.Conform formulei lui Poincare, avem:

𝑃(𝐴)

=

𝑛∑𝑖=1

𝑃 (𝐴𝑖) −𝑛∑

𝑖,𝑗=1𝑖<𝑗

𝑃 (𝐴𝑖 ∩𝐴𝑗) + ... + (−1)𝑛−1

𝑃

(𝑛⋂

𝑖=1

𝐴𝑖

)

𝑃(𝐴)

= 𝐶1𝑛𝑃 (𝐴𝑖) − 𝐶2

𝑛𝑃 (𝐴𝑖 ∩𝐴𝑗) + ... + (−1)𝑛−1

𝐶𝑛𝑛𝑃

(𝑛⋂

𝑖=1

𝐴𝑖

)

=𝑛!

1! (𝑛− 1)!

(𝑛− 1)!

𝑛!− 𝑛!

2! (𝑛− 2)!

(𝑛− 2)!

𝑛!+ ... + (−1)

𝑛−1 1

𝑛!

𝑃(𝐴)

=1

1!− 1

2!+

1

3!− 1

4!+ ... + (−1)

𝑛−1 1

𝑛!

𝑃 (𝐴) = 1 − 𝑃(𝐴)

= 1 −(

1

1!− 1

2!+

1

3!− 1

4!+ ... + (−1)

𝑛−1 1

𝑛!

)8

de unde rezulta

lim𝑛→∞

𝑃 (𝐴) =1

𝑒.

Problema 4. In doua grupe, asezate ın aceeasi sala, formate din 22respectiv 24 studenti avem: 14 studenti buni, 6 mediocri, 2 slabi ın primagrupa si 12 buni, 4 mediocrisi 8 slabi, respectiv ın a doua grupa.a) Fiind ascultat un student, care este probabilitatea ca studentul sa fieunul mediocru?b) Ascultand un student bun, care este probabilitatea ca sa fie din a douagrupa?

Solutie: Fie 𝐵1 evenimentul de a alege studentul din prima grupa si 𝐵2

evenimentul de a alege studentul din a doua grupa.a) Fie 𝐴 evenimentul ca studentul ascultat sa fie mediocru. Atunci avem:

𝑃 (𝐵1) =22

46=

11

23, 𝑃 (𝐵2) =

24

46=

12

23,

𝑃𝐵1 (𝐴) =6

22=

3

11, 𝑃𝐵2 (𝐴) =

4

24=

1

6.

Formula probablilitatii totale implica

𝑃 (𝐴) = 𝑃 (𝐵1) · 𝑃𝐵1(𝐴) + 𝑃 (𝐵2) · 𝑃𝐵2

(𝐴) =11

23

3

11+

12

23

1

6=

5

23.

sau direct

𝑃 (𝐴) =6 + 4

22 + 24=

5

23.

b) Fie 𝐴 evenimentul ca studentul ascultat sa fie bun. Folosim formula luiBayes si avem

𝑃𝐴 (𝐵2) =𝑃 (𝐵2) · 𝑃𝐵2

(𝐴)

𝑃 (𝐵1) · 𝑃𝐵1(𝐴) + 𝑃 (𝐵2) · 𝑃𝐵2

(𝐴)

unde

𝑃𝐵1(𝐴) =

14

22=

7

11; 𝑃𝐵2

(𝐴) =12

24=

1

2

si rezulta

𝑃𝐴 (𝐵2) =1223

12

1223

12 + 11

23711

=6

13.

sau direct

𝑃𝐴 (𝐵2) =

12

23

1

214 + 12

22 + 24

=6

13.

Problema 5. Problema Monty Hall

Solutie: Conform informatiilor de la inceputul fisierului concurentul a alesusa 1 iar gazda emisiunii a deschis usa 3. Ce ne propunem sa calculam esteprobabilitatea ca masina sa se afle in spatele usii 2 daca gazda a deschis usa 3

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Pentru a aborda problema folosind formule de tip Bayes va trebui sa definimipotezele:

𝐻1: masina se afla in spatele usii 1𝐻2: masina se afla in spatele usii 2𝐻3: masina se afla in spatele usii 3Definim evenimentul:𝐸 : moderatorul emisiunii deschide usa 3Acum evenimentul a carui probabilitate dorim sa o calculam este: 𝐻2 con-

ditionat de aparitia lui 𝐸 !! =⇒ 𝑝(𝐻2|𝐸) asadar probabilitatea unei ipoteze inconditiile in care evenimentul a avut loc.

Prin definitie:

𝑝(𝐻2|𝐸) =𝑝(𝐻2)𝑝(𝐸|𝐻2)

𝑝(𝐻1)𝑝(𝐸|𝐻1) + 𝑝(𝐻2)𝑝(𝐸|𝐻2) + 𝑝(𝐻3)𝑝(𝐸|𝐻3)

Va trebui sa discutam semnificatia fiecarei probabilitati din formula de maisus:

𝑃 (𝐻1) : probabilitatea ca masina sa fie in spatele usii 1 =⇒ 𝑃 (𝐻1) = 13

𝑃 (𝐻2) : probabilitatea ca masina sa fie in spatele usii 2 =⇒ 𝑃 (𝐻2) = 13

𝑃 (𝐻3) : probabilitatea ca masina sa fie in spatele usii 3 =⇒ 𝑃 (𝐻3) = 13

Apoi probabilitatile conditionate:𝑝(𝐸|𝐻1): probabilitatea ca gazda sa deschida usa 3 daca masina este in

spatele usii 1 =⇒ atunci gazda stie ca in spatele usii 2 si 3 este o capra decipoate deschide pe oricare dintre acestea, din moment ce concurentul a ales usa1 =⇒ 𝑝(𝐸|𝐻1) = 1

2𝑝(𝐸|𝐻2): probabilitatea ca gazda sa deschida usa 2 daca masina este in

spatele usii 2 =⇒ gazda poate acum sa deschida doar usa 3 din moment cemasina este in spatele usii 2 si concurentul a ales usa 1 =⇒ 𝑝(𝐸|𝐻1) = 1

𝑝(𝐸|𝐻3): probabilitatea ca gazda sa deschida usa 3 daca masina este inspatele usii 3 este evdident nula =⇒ 𝑝(𝐸|𝐻1) = 0

Inlocuind aceste informatii in formula se obtine:

𝑝(𝐻2|𝐸) =2

3=≈ 66%

Sansa de castig prin alegerea usii 2 este surprinzator de mare, insusi marelematematician Pal Erdos nu a fost convins de exactitatea acestui rezultat panain momentul in care a vazut o simulare pe calculator a problemei.

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Probleme propuse

Problema 1. Fiind date numerele 1, 2, 3,..., 𝑛 scrise ıntr-o anumita ordine,care este probabilitatea ca numerele 1 si 2 sa fie consecutive?

Problema 2. Intr-un lot pus ın vanzare la un magazin se afla produsele atrei fabrici 𝐹𝑖, 𝑖 = 1, 2, 3, ın cantitatile 300, 420, 540 produse si se stie dinverificarile statistice ca fiecare dintre fabrici livreaza produse defecte ın proportiede 1% , 2% si 2, 5% respectiv. O cantitate de produse ın valoare de 6000 lei afost restituita magazinului de catre cumparatori ın cadrul conventiei de garantieca fiind inutilizabile.

Sa se determine sumele ce trebuie imputate fabricilor, daca nu se stie de lacare dintre fabrici au provenit produsele defecte.

Problema 3. De pe un submarin se lanseaza asupra unui distrugator 4 torpile.Probabilitatea ca o torpila sa loveasca distrugatorul este 0, 3. Pentru scufundareadistrugatorului sunt suficiente doua torpile, iar daca o singura torpila lovestedistrigatorul el se scufunda cu probabilitatea 0, 6. Sa se gaseasca probabilitateaca distrugatorul sa se scufunde.

Problema 4. Gasiti probabilitatea ca printre 7 persoane:

a) Sa nu existe doua nascute in aceeasi zi a saptamanii

b) Cel putin doua sa fie nascute in aceeasi zi

c) Doua persoane sa fie nascute duminica si doua martea

Problema 5. Popescu stie raspunsurile la una dintre cele 10 intrebari cu raspun-suri multile ale examenului de MS. El a absentat la multe dintre cursuri si vatrebui sa ghiceasca raspunsurile la celelalte 9 intrebari. Presupunand ca fiecareintrebare are patru rapsunsuri care este probabilitatea ca el sa nimereasca 7raspunsuri corecte ? Fiecare raspuns valoreaza un punct si este nevoie de 5puncte pentru a promova examenul. Care este probabilitatea ca Popescu sa pro-moveze examenul ?

Problema 6. Suppose you’re a volleyball tournament organizer. There are10 teams signed up for the tournament, and it seems like a good idea for eachteam to play every other team in a ”round robin” setting, before advancing tothe playoffs. How many games are possible if each team plays every other teamonce?

Problema 7. Eight students are to be put up in a student residence in threerooms, two of which have three beds and one has two beds. In how many wayscan the students be distributed over the three rooms ?

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Problema 8. How many full houses are possible in poker?

Problema 9. A straight consists of five cards with values forming a string offive consecutive values (with no wrap around). For example, 45678, 𝐴2345 and10𝐽𝑄𝐾𝐴 are considered straights, but 𝐾𝑄𝐴23 is not (suits are immaterial forstraights.) How many different straights are there in poker?

Problema 10. How many of the integers 1, 2, 3, . . . , 2017 are not divisible byany of the numbers 4, 5, 6 ?

Problema 11. How many rectangles are contained in an 𝑚×𝑛 checkerboard?Start with the normal chess board.

Problema 12. a) An experiment of drawing a random card from an ordinaryplaying cards deck is done with replacing it back. This was done ten times. Findthe probability of getting 2 spades ♠, 3 diamonds ♢, 3 clubs ♣ and 2 hearts ♡.

b) The numbers 1, 2, 3, . . . 𝑛 are written in random order. What is the prob-ability of having 1 and 2 on consecutive positions ?

Problema 13. At some moment in a backgammon game you have to roll a 6or the sum of the two numbers to be 6 to put your opponent’s rear checker onthe bar. What is the probability of hitting your opponent’s checker ? What isthe probability of hitting at least twice in 4 attempts?

Problema 14. Ten shots are fired at a target consisting of an inner circle andtwo concentric annuli. The probabilities of hitting these regions in one shot are0.15, 0.20 and 0.25, respectively. Find the probability that there will be three hitsin the circle, two in the first annulus and two in the second annulus.

Problema 15. A question posed in the mid-17th century to Blaise Pascal by aFrench nobleman and inveterate gambler, the Chevalier de Mere, which markedthe birth of probability theory. One of de Mere’s favorite bets was that at leastone six would appear during a total of four rolls of a die. From past experience,he knew that this gamble paid off more often than not. Then, for a change, hestarted betting that he would get a double-six on 24 rolls of two dice. However,he soon realized that his old approach to the game was more profitable. He askedhis friend Pascal why. Can you give him the answer ?

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Problema 16. a) A single card is chosen at random from a standard deck of52 playing cards. What is the probability of choosing a king (𝐾) or a club (♣)?

b) A professor gives only two types of exams, ”easy” and ”hard”. You will geta hard exam with probability 0.80. The probability that the first question on theexam will be marked as difficult is 0.90 if the exam is hard and is 0.15 otherwise.What is the probability that the first question on your exam is marked as difficult? What is the probability that your exam is hard given that the first question onthe exam is marked as difficult?

Problema 17. Two friends decide to meet at 21 : 00 pm in a restaurant. Theydecided that who ever reaches the restaurant earlier will wait for the other personfor 20 minutes. The restaurant closes at 23 : 00 pm. What’s the probability thatthose friends do meet?

Problema 18. A person wrote 5 letters, sealed them in envelopes and wrotethe different addresses randomly on each of them. Find the probability that atleast one of the envelopes has the correct address.

Problema 19. Find the probability of drawing a king, a queen, a king and aknave in this order, from a deck of 52 cards, in four consecutive draws. Thecards drawn are not replaced.

Problema 20. In USA 40% of the registered voters are Republicans, 45% areDemocrats and 15% are Independent. When the voters were asked about increas-ing military spending 20% of the Republicans opposed it, 65% of the Democratsopposed it and 55% of the Independents opposed it. What is the probability thata randomly selected voter opposes increased military spending ?

Problema 21. A telegraphic communications system transmits the signals dot∙ and dash ¯. Assume that the statistical properties of the obstacles are suchthat an average of 0.4 of the dots and 0.25 of the dashes are changed. Supposethat the ratio between the transmitted dots and the transmitted dashes is 5 : 3.What is the probability that a received signal will be the same as the transmittedsignal if:

a) the received signal is a dot.

b) the received signal is a dash.

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