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Lampu yang bagus = 12-2 = 10
n(S) = 12
Terdapat 3 kemungkinan diantaranya
1. kemungkinan (bagus,bagus, rusak)
P(A) = 10/12 x 9/11 x 2/10
= 18/132
= 9/66
2. Kemungkinan (rusak , bagus , rusak)
P(A) = 2/12 x 10/11 x 1/10
= 1/6 x 1/11
= 1/66
3. kemungkinan (bagus , rusak , rusak)
P(A) = 10/12 x 2/11 x 1/10
= 1/12 x 2/11
= 1/66
peluang pembeli ke 3 mendapat lampu rusak
P(A) = 9/66 + 1/66 + 1/66
= 11/66
= 1/6 D