sebuah kantong berisi 6 kelereng kuning dan 4 kelereng biru diambil 2 kelereng sekaligus,tentukan terambilnya A.2 kelereng biru B.2 kelereng kuning C.1 kelereng

a. 2 kelereng biru
smoga bnar

6 kuning 4 biru = 10 kelereng
Diambil 2 kelereng sekaligus :
n(S) = 10C2 = 10!/(10 - 2)!2! = (10.9.8!)/(8!.2.1) = 45

A) terambil 2 kereng biru
n(A) = 4C2 = 4!/(4 - 2)!2! = (4.3.2!)/(2!2.1) = 6
Peluang : P(A) = 6/45 = 2/15

B) terambil 2 kelereng kuning
n(B) = 6C2 = 6!/(6 - 2)!2! = (6.5.4!)/(4!2.1) = 15
Peluang : P(B) = 15/45 = 1/3

C) terambil 1 kelereng biru dan 1 kelereng kuning
n(C) = 4C1 . 6C1 = 4 . 6 = 24
Peluang : P(C) = 24/45 = 8/15