Jawaban:
n(s) = 9C3 = 9!/6!.3! = 7 x 8 x 9/6 = 84
bnyak kjadian trambil 2 merah =
5C2 = 5!/3!.2! = 4 x 5/2 = 10
bnyak kejaduan trambil 2 putih =
4C2 = 4!/2!.2! = 3x 4/2 = 6
saling lepas.
berarti peluang trambil 2 merah atau 2 putih =
10/84 + 6/84 =
16/84 =
4/21
Penjelasan dengan langkah-langkah:
Maaf kalau salah!!
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Jawaban:
n(s) = 9C3 = 9!/6!.3! = 7 x 8 x 9/6 = 84
bnyak kjadian trambil 2 merah =
5C2 = 5!/3!.2! = 4 x 5/2 = 10
bnyak kejaduan trambil 2 putih =
4C2 = 4!/2!.2! = 3x 4/2 = 6
saling lepas.
berarti peluang trambil 2 merah atau 2 putih =
10/84 + 6/84 =
16/84 =
4/21
Penjelasan dengan langkah-langkah:
Maaf kalau salah!!