1 kantong berisi 5 kelereng merah 7 kelereng hijau, dari kantong tersebut di ambil 2 kelereng sekaligus A. Peluang terambil keduanya merah B. Peluang terambil k

Merah = 5
Hijau = 7
Kelereng = 12

n(S) = 12C2
12C2 = 12!/2!10!
12C2 = 66

A) Keduanya merah
n(A) = 5C2
» 5C2 = 5!/2!3!
» 5C2 = (5×4×3×2×1)/(2×1×3×2×1)
» 5C2 = 10

P(A) = n(A)/n(S)
= 10/66
= 5/33

B) Keduanya hijau
n(B) = 7C2
» 7C2 = 7!/2!5!
» 7C2 = (7×6×5×4×3×2×1)/(2×1×5×4×3×2×1)
» 7C2 = 21

P(B) = n(B)/n(S)
= 21/66
= 7/22

C) Kelereng merah dan hijau
n(C) = 5C1 × 7C1
» n(C) = (5!/1!4!) × (7!/1!6!)
» n(C) = 5×7
» n(C) = 35

P(C) = n(C)/n(S)
= 35/66