Cesar_Probabilidade

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Nome: Cesar Bassi Batista – SEGNA 3 
 
 
Ω = {Sul, Sudeste, Centro-oeste, Nordeste, Norte} 
 
 
• Ω = {1,2,3,4,5,6} 
• E1 = {3,6} 
• E2 = {1,4,6} 
 
 
 
 
• n(Ω) = 36 
• n(E1) = 9 
• n(E2) = 9 
 
 
n(Ω) = 52 
 
 
• n(Ω) = 6 
• E1 = {(co, ca, co); (ca,ca,co); (co,ca,ca); (ca,ca,ca)} 
• E2 = {(co,co,ca); (co,ca,co);(ca,co,co)} 
 
 
 
 
C 32,1 32!/1!(32-1)! = 32!/ 1! 31! = 32 * 31!/1!31! = 32 
C 32,2 32!/2!(32 - 2)! = 32!/2!30! = 32 * 31 * 30! /2! 30! = 992/2 = 496 
C 32,3 32!/3!(32 - 3)! = 32!/3!29! = 32 * 31 * 30 * 29! /3! 29! = 29760/6 = 4960 
 
 
15 * 17 = 255 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
E = {1,2,4,8,16,32,64} 
N(Ec) = 93 
 
 
 
 
E = {2,3,4,6,8,9,10,12,14,15,16,18,20,21,22,24,26,27,28,30,32,33,34,36,38,39,40} 
Ec = {5,7,11,13,17,19,23,25,29,31,35,37} 
 
 
 
 
• P(a) = 1/100 ou 1% 
• P(b) = 37/100 ou 37% 
• P( c) = 90/100 ou 90% 
 
 
S = {a,e,i,o,u,b,c,d,f,g} 
a) 
P(a) = 1/10 
b) 
P(b) = 1/10 
c) 
P(c) = 0/10 = 0 
d) 
 P(d) = 5/10 = 1/2 ou 50% 
e) 
P(e) = 4/10 = 2/5 ou 40% 
 
 
• P(a) = 1/52 
• P(b) = 13/52 
• P( c) = 48/52 = 12/13 
 
P(Carne) = 9/20 
 
a) 
P(a) = 6/36 
 
b) 
P(b) = 30/36 
 
c) 
S= {(1,6), (2,5), (3,4), (4,3) (5,2) (6,1)} 
P(c)= 6/36 
 
d) 
S={(1,1), (1,3), (2,2), (3,1), (1,5), (2,4), (3,3), (4,2),(5,1), (2,6), (3,5), (4,4), (5,3), (6,2),(4,6), (5,5), 
(6,4),(6,6)} 
P(d) = 18/36 = 3/6 = 1/ 2 ou 50% 
E ) 
S={(1,2),(1,4),(1,6),(2,1),(2,2)(2,3)(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2)(4,3),(4,4),(4,5),(4,6),(5,2),(5,4
),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} 
 
P(e) = 27/36 = 9/12 = 3/4 ou 75% 
 
 
 
a) 
a=22 ; b = 31; c =106 d = 48 ; e=71 
 
b) P(mulher não come carne) = 31 /200 
 
c ) P(carne frequentemente) = 81 / 200 
 
 
 
 
• P(a) = 6/10 
• P(b) = 3/10 
• P(c ) = 0/10 
 
 
 
a) 
1 1 2 2 3 3 
1 1 2 2 3 3 
P(a) = 4/36 
b) 
P(b) = 12/36 
2,2 2,2 2,2 2,2 1,3 1,3 1,3 1,3 3,1 3,1 3,1 3,1 
3 + 3 3 + 3 3 + 3 3 + 3 
 
 
C = {-1,0,1,2} 
P(raizes reais) = 4/5 ou 80% 
 
 
Somente A = 63 - 32 = 31% 
Somente B = 55 - 32 = 23% 
Ambas = 32% 
Que consomem A e ou B = 86% 
Não consomem nenhuma dessas marcas = 100% - 86% = 14% 
 
 
a) 
P(a) = 1/4 ou 25% 
b) 
P(b) = 3/4 ou 75%