Dalam suatu kantong berisi 15 kelereng ber- nomer 1,2,3,4,...,15. Dua kelereng di ambil secara acak, nilai kemungkinan terambil kelereng keduanya bernomer ganj

ganjil; 1,3,5,7,9,11,13,15 ada 8
banyak cara terambil 2 nomer ganjil= 8C2=8!/6!2!=28
kejadian seluruhnya= 15C2=15!/13!.2!=105
Peluang terambil keduanya ganjil= 28/105

Cara biasa:
A = {(1, 3), (1, 5), (1, 7), (1, 9), (1, 11), (1, 13), (1, 15), (3, 5), (3, 7), (3, 9), (3, 11), (3, 13), (3, 15), (5, 7), (5, 9), (5, 11), (5, 13), (5, 15), (7, 9), (7, 11), (7, 13), (7, 15), (9, 11), (9, 13), (9, 15), (11, 13), (11, 15), (13, 15)}

n(A) = 28

S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (1, 12), (1, 13), (1, 14), (1, 15), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (2, 11), (2, 12), (2, 13), (2, 14), (2, 15), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (3, 10), (3, 11), (3, 12), (3, 13), (3, 14), (3, 15), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (4, 10), (4, 11), (4, 12), (4, 13), (4, 14), (4, 15), (5, 6), (5, 7), (5, 8), (5, 9), (5, 10), (5, 11), (5, 12), (5, 13), (5, 14), (5, 15), (6, 7), (6, 8), (6, 9), (6, 10), (6, 11), (6, 12), (6, 13), (6, 14), (6, 15), (7, 8), (7, 9), (7, 10), (7, 11), (7, 12), (7, 13), (7, 14), (7, 15), (8, 9), (8, 10), (8, 11), (8, 12), (8, 13), (8, 14), (8, 15), (9, 10), (9, 11), (9, 12), (9, 13), (9, 14), (9, 15), (10, 11), (10, 12), (10, 13), (10, 14), (10, 15), (11, 12), (11, 13), (11, 14), (11, 14), (11, 15), (12, 13), (12, 14), (12, 15), (13, 14), (13, 15), (14, 15)}

n(S) = 105

P(A) = n(A)/n(S)
P(A) = 28/105
P(A) = 4/15

Cara cepat:
P(A) = n(A)/n(S)
= 8C2 ÷ 15C2
= 8!/(2!(8 - 2))! ÷ 15!/(2!(15 - 2)!)
= 8!/(2!6!) ÷ 15!/(2!13!)
= (8 × 7 × 6!)/(2!6!) ÷ (15 × 14 × 13!)/(2!13!)
= (8 × 7)/(2!) ÷ (15 × 14)/(2!)
= (8 × 7)/(2 × 1) ÷ (15 × 14)/(2 × 1)
= 56/2 ÷ 210/2
= 28 ÷ 105
= 28/105