Dari sebuah kantong terdapat 5 kelereng merah dan 3 kelereng putih,diambil 3 kelereng secara acak. Peluang terambil dua kelereng berwarna sama adalah...

5 kelereng merah dan 3 kelereng putih = 8 kelereng
Diambil 3 kelereng :
n(S) = 8C3 = 8!/(8-3)!3! = (8.7.6.5!)/(5!.3.2.1) = 56

Terambil 2 kelereng berwarna sama
2 merah 1 putih atau 2 putih 1 merah
n(A) = 5C2 . 3C1 + 3C2 . 5C1 = 10 . 3 + 3 . 5 = 30 + 15 = 45

Peluang = n(A)/n(S) = 45/56

[tex]\displaystyle \text{2 kelereng berwarna merah}\\P(A)=\frac{^5\text{C}_2\cdot{^3\text{C}_1}}{^8\text{C}_3}\\P(A)=\frac{\frac{5!}{2!3!}\cdot\frac{3!}{1!2!}}{\frac{8!}{3!5!}}\\P(A)=\frac{\frac{5\cdot4}{2}\cdot\frac{3}{1}}{\frac{8\cdot7\cdot6}{3\cdot2}}\\P(A)=\frac{5\cdot2\cdot3}{8\cdot7}\\P(A)=\frac{5\cdot3}{4\cdot7}\\P(A)=\frac{15}{28}[/tex]
[tex]\displaystyle \text{2 kelereng berwarna putih}\\P(B)=\frac{^5\text{C}_1\cdot{^3\text{C}_2}}{^8\text{C}_3}\\P(B)=\frac{\frac{5!}{1!4!}\cdot\frac{3!}{2!1!}}{\frac{8!}{3!5!}}\\P(B)=\frac{\frac{5}{1}\cdot\frac{3}{1}}{\frac{8\cdot7\cdot6}{3\cdot2}}\\P(B)=\frac{5\cdot3}{8\cdot7}\\P(B)=\frac{15}{56}\\\\P(A\cup B)=P(A)+P(B)\\P(A\cup B)=\frac{15}{28}+\frac{15}{56}\\\boxed{\boxed{P(A\cup B)=\frac{45}{56}}}[/tex]