设随机变量X的概率颁布为P｛x=k｝＝ak/18,(k=1,2,…,9)求常数a求概率P｛x=1或x=4｝求概率P｛-1≤x＜7/2｝

Question

设随机变量X的概率颁布为P｛x=k｝＝ak/18,(k=1,2,…,9)
求常数a
求概率P｛x=1或x=4｝
求概率P｛-1≤x＜7/2｝

Answer 1

(a/18)(1+2+3+...+8+9)=1
所以 a=18/45
=2/5
概率P｛x=1或x=4｝=P{x=1}+P{x=4}
=2/(18*5) + 2*8/(18*5)
=1/5
求概率P｛-1≤x＜7/2｝=P{x=1}+P{x=2}+P{x=3}
=(2+2*2+2*3)/(18*5)
=2/15