Page 10 - KEGIATAN PEMBELAJARAN 3 DISTRIBUSI PELUANG ACAK_DHINI MARLIYANTI
P. 10
b. Tentukan nilai peluang P ( X ≤ 1)
1 1
+ 1
≤ 1 = 0 ≤ ≤ 1 = න = න
4
0 0
1
1
= න( + 1)
4
0
1
1 1
2
= +
4 2
0
1 1 1
2
2
= 1 + 1 − 0 + 0
4 2 2
1 1 1
= 1 + 1 − 0 + 0
4 2 2
1 1
= + 1 − 0 + 0
4 2
1 3
= − 0
4 2
Jadi nilai peluang
1 3 3 3
= = ≤ 1 =
4 2 8 8
