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S = {(A,1), (A,2), (A,3), (A,4), (A,5), (A,6), (G,1), (G, 2), (G, 3), (G,4), (G,5), (G,6)}
n(S) = 12
A = muncul angka dadu genap
B = muncul gambar
n(A) = 6
n(B) = 6
n(AnB) = 3
n(AUB) = n(A) + n(B) - n(AnB) = 6 + 6 - 3 = 9
Jadi peluang munculnya angka dadu genap dan gambar pada koin = n(AnB) / n(S) = 9/12 = 3/4