Răspuns :
A={1,2,3,4,....,2013}
card A=2013⇒2013 cazuri posibile
k³ ∈ {1,8,27,64,125,216,343,512,729,1000,1331,1728}
⇒ 12 cazuri favorabile
P= nr. cazuri favorabile/ nr. cazuri posibile
P=12/2013
P=4/671
card A=2013⇒2013 cazuri posibile
k³ ∈ {1,8,27,64,125,216,343,512,729,1000,1331,1728}
⇒ 12 cazuri favorabile
P= nr. cazuri favorabile/ nr. cazuri posibile
P=12/2013
P=4/671
cub =ceva la puterea a 3
1-1
2-8
3-27
4-64
5-125
6-216
7-343
8-512
9-729
10-1000
11-1331
12-1728
13-2197
deci 12
probabilitate inseamna nr bune pe toate
12supra2013
4supra671