Pada kantong terdapat 2 kelereng merah,2kelereng kuning,dan 6 kelereng biru. Seorang anak mengambil 3buah kelereng secara acak. Peluang terambil 3 kelereng biru

sampel keseluruhan
= 10C3
= 120

terambil kelereng biru
= 6C3
= 20

peluang terambil 3 kelereng biru
= 20/120
= 1/6

Langsung aja y

n(s) = 10C3
= 10! / 3!(10-3)!
= 10×9×8×7! / 3! 7!
= 10 × 9 × 8 / 3 × 2 × 1
= 10 × 3 × 4
= 120

n(a) = 6C3
= 6! / 3! (6-3)!
= 6 × 5 × 4 × 3! / 3 × 2 × 1 3!
= 5 × 4
= 20

P(a) = n(a) / n(s)
P(a) = 20 / 120
P(a) = 1/6

Semoga berguna +_+