Soal !! sebuah kotak berisi 8 bola merah, 7 bola putih dan 5 bola biru dan diambil secara acak. tentukan probabilitas terpilihnya : 1. bola merah 2. bola putih

jawab
n(M)= 8
n(P) = 7
n(B) = 5
n(S)= 20
diambil  1 bola

P(merah) = n(M)/ n(S) = 8/20 = 2/5
P(putih) = n(P) /n(S) = 7/20
P(biru)= n(B)/n(S) = 5/20 = 1/4
P(tidak merah) =P(putih) + P(biru) = 7/20 + 5/20 = 12/20= 3/5
P(merah atau putih) = P(merah ) + P( putih) = 8/20 + 7/20 = 15/20 = 3/4

n(Mrh) = 8
n(Pth) = 7
n(Br) = 5
n(S) = 8+7+5 = 20

1. P(Mrh) = 8/20 = 2/5
2. P(Pth) = 7/20
3. P(Br) = 5/20 = 1/4
4 P ( Pth & Br ) = (5+7)/20 = 12/20 = 3/5
5. P( Mrh / Pth ) = 8/20 + 7/20 = 15/20 = ¾