9
(1,1) (1,2)(1,3)(1,4)(1,5)(1,6) 3 cazuri (1,1)(1,2)1,3)
(2,1) (2,2)(2,3)(2,4)(2,5)(2,6) 2 cazuri (2,1)(2,2)
(3,1) (3,2)(3,3)(3,4)(3,5)(3,6) 1 caz (3,1)
(4,1) (4,2)(4,3)(4,4)(4,5)(4,6)
(5,1) (5,2)(5,3)(5,4)(5,5)(5,6)
(6,1) (6,2)(6,3)(6,4)(6,5)(6,6)
cazuri favorabile 3+2+1=6
cazuri posibile 6x6=36
probabilitatea=6/36=1/6
10.
a=8/x
x∈{1,2,3,4,5,6,7,8,9}
ca sa obtinem a∈N x trebuie sa fie divizorul lui 8
cazuri favorabile 4 (adica 1,2,4,si 8)
cazuri posibile 9
probabilitatea =4/9
11.
p₁=multiplu de 12
intre 1 si 100 avem 8 multipli de 12 (12,24,36,48,60,72,84,90)
cazuri favorabile 8
cazuri posibile 100
probabilitatea =8/100=2/25
p₂=nr care sa se divida cu 5 (adica multiplu de 5)
intre 1 si 100 avem 20 multipli de 5 (5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95)
cazuri favorabile 20
cazuri posibile 100
probabilitatea =20/100=1/5
