概率论，求解答，谢谢

Question

想要详细些的过程，通俗易懂，谢谢

Answer 1

X和Y取值分别为0,1,2和0,1,2
P｛X=x｝= C『2,x』(0.2^x × 0.8^(2-x))
P｛Y=y｝= C『2,y』(0.5^y × 0.5^(2-y))
由于X，Y相互独立，P｛X|Y｝=P｛X｝
当Y=0时,X=0
P｛X≤0,Y=0｝= P｛X=0｝P｛Y=0｝
= C『2,0』0.8^2 C『2,0』0.5^2
=16/25 × 1/4
=4/25
当Y=1时，X=0或1
P｛X≤1,Y=1｝= (P｛X=0｝+P｛X=1｝)P｛Y=1｝
= (C『2,0』0.8^2 + C『2,1』0.2×0.8)C『2,1』0.5^2
=(16/25+2×4/25) × 2×1/4
=24/25×1/2
=12/25
当Y=2时，X=0或1或2
P｛X≤2,Y=2｝= P｛X≤2｝P｛Y=2｝
= 1×C『2,2』0.5^2
= 1×1/4
= 1/4
综上，P｛X≤Y｝
=P｛X≤0,Y=0｝+P｛X≤1,Y=1｝+P｛X≤2,Y=2｝
=4/25 + 12/25 + 1/4
=16/100 + 48/100 + 25/100
=89/100

Answer 2

供参考，请笑纳。