Kunci Jawaban Matematika Kelas 8 Halaman 308-310 Uji Kompetensi 10 Bab 10, Bagian Esai Nomor 1-10 Semester 2

Alternatif:
f2 = 1/9
f2 = n(2)/M

n(2) = f2 x M
n(2) = 1/9 x (n(2) + 32)
9n(2) = n(2) + 32
9n(2) - n(2) = 32

8n(2) = 32
n(2) = 32/8
n(2) = 4

M = n(2) + 32
= 4 + 32
= 36

n(1) = 5
n(1') = 36 - 5 = 31

f1' = n(1')/M
= 31/36

5. Dalam percobaan melempar dadu sebanyak 450 kali, secara teoretik akan muncul mata dadu kurang dari 5 sebanyak ... kali.

Pembahasan:
n(S) = 6
Titik sampel kejadian A = {1, 2, 3, 4}
n(A) = 4

p(A) = n(A)/n(S)
= 4/6
= 2/3

N = 450
fh(A) = p(A) x N = 2/3 x 450 = 300