mohon bantuannya?????

[tex]\displaystyle P(A)=\frac{n(A)}{n(S)}\\P(A)=\frac{^2\text{C}_2}{^8\text{C}_2}\\P(A)=\frac{\frac{2!}{2!0!}}{\frac{8!}{2!6!}}\\P(A)=\frac{\frac{2}{2}}{\frac{8\cdot7}{2}}\\P(A)=\frac{1}{4\cdot7}\\\boxed{\boxed{P(A)=\frac{1}{28}}}[/tex]

8 lampu = 2 rusak dan 6 baik

Dibeli 2 lampu :
n(S) = 8C2 = 8!/(8 - 2)!.2! = (8 . 7 . 6!)/(6!. 2 . 1) = 28

Yang terbeli 2 lampu rusak :
n(A) = 2C2 = 1

Peluang : P(A) = n(A)/n(S) = 1/28