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2 merah, 3 putih
n(m1)=2
n(p1)=3
n(S1)=5
Kotak 2:
4 merah, 2 putih
n(m2)=4
n(p2)=2
n(S2)=6
A. Peluang terambil merah
P(M)=P(m1) x P(m2)
=[n(m1)/n(S1)] x [n(m2)/n(S2)]
=2/5 x 4/6
=8/30=4/15
B. Peluang terambil putih
P(Put)=P(p1) x P(p2)
=[n(p1)/n(S1)] x [n(p2)/n(S2)]
=3/5 x 2/6
=6/30=3/15