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→n(S) = 5C1 ×4C1 = 5×4 =20
→Peluang keduanya merah
n(A) = 3C1×2C1 = 3×2 = 6
=> P(A) = 6/20 =3/10
→Peluang keduanya biru
n(A) = 2C1×1C1 = 2
=> P(A) = 2/20 = 1/10
→3/10+1/10 = 4/10 = 0,4
Silahkan diteliti kembali