Question

Sa se calculeze probabilitatea ca alegand unul dintre nr permutari de 3, aranjamente de 6 luate cate 2 si combinari de 4 luate cate 7 acesta sa fie divizibil cu 5

Answer 1

$P_{3}=3! = 1 \times 2 \times 3 = 6 \: \not \vdots \:\:5$

$A_{6}^{2} = \frac{6!}{(6 - 2)!} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6}{4!} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6}{1 \times 2 \times 3 \times 4} = 5 \times 6 = 30 \: \: \vdots \: \: 5$

$C_{7}^{4} = \frac{7!}{4!(7 - 4)!} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}{1 \times 2 \times 3 \times 4 \times 3!} = \frac{5 \times 6 \times 7}{1 \times 2 \times 3} = 5 \times 7 = 35 \: \: \vdots \: \: 5$

$= > nr. \: cazuri \: favorabile = 2$

$= > nr. \: cazuri \: posibile = 3$

$P = \frac{nr. \: cazuri \: favorabile}{nr. \: cazuri \: posibile} = \frac{2}{3}$