（2）Y?(X?12)，当y?0时，FY(y)?P(Y?y)?P(X?12?y)?0
当1?y?0时，FY(y)?P(Y?y)?P(X?12?y)?P(X?2y?1)
?FX(2y?1)当y?1时，FY(y)?1
Y?X?12的概率概率密度函数为故
?1?,1?y?0fY(y)??2?0,其他?
（3）Y?X2，当y?0时，FY(y)?P(Y?y)?P(X
当y?0时，FY(y)?P(Y?y)?P(X?FX(Y?X的概率概率密度函数为y??1e2,y?0故 ?fY(y)??2?y?y?0?0,22?y)?0
y?X?y)2?y)?P(?y)
y)?FX(?
?1??3x?1?27、?fX?x???8?0?20?x?2其它
?x??0,2??y??x??0,4?? 当y?0时,FY?y??0 0?y?4? FY?y??P??Xy2??y?P????y?18?X?y?? ??y ???y?fx?x?dx???0?3x?1?dx
? 16
y?4? FY?y??P????y??X?y??????2180?3x?1?dx?1
?当y?0,y?4?时 ?1?3???'? fY?y??FY?y???8???0?1?1???2?y??y0?y?4?y?0,y?4?
1?3???fY?y???16?16?y??00?y?4?其它
28、因为X 与 Y相互独立，且服从正态分布N(0,?2)
12??2?x?y2?222 f(x,y)?fX(x)fY(y)? 由Z?X22e
?Y知， 当z?0时，fZ(z)?0
xz?x222 当z?0时，FZ(z)????x12?2x?y2?222?z?x2??edydx=?z0?2?12???r220e22?rd?dr?1?e?z222?
?z?z2?2e fZ(z)????0?(2?2),z?0
,其他29、
1??2,?1?x?1fX(x)????0,其他fZ(z)?
????fX(z?y)fY(y)dy??z?112?(1?y)2z?1dy?12?(arctan(z?1)?arctan(z?1))
30、当z?0时，fZ(z)?0
17
当z?0时
?zfZ(z)????fX(z?y)fY(y)dy??0?e??(z?y)?ye2??ydy??e3??zz22
?1,31、fX(x)???0,0?x?1其他，
?1,fY(y)??0,0?y?1其他，
fZ(z)??????zdy?z,??0?1fX(z?y)fY(y)dy???dy?2?z,z?1??0,?0?z?11?z?2其他
32 解
??（1）f?x???X???23?3xedy?f?x,y?dy???02?0?x?0?3e?3x??0x?0??1???2?0?x?0x?0
fY?y??????????3?3xedx?f?x,y?dx???02?0?0?y?2其它0?y?2其它
（2）FX?x???x???0?fX?t?dt??x?3t3edt???0?0?y?1fY?t?dt???dt02???1x?0?0???3xx?0?1?ex?0x?0y?0
FY?y???y???0??10?y?2??y?2y?2??1y?00?y?2 y?2z?00?z?2 z?2?Fmax?Z??P?max?X,Y??Z??FX?Z??FY?Z??0??1?3z??1?ez?2?3Z??1?e?? 18
（3）P??1??1??Z?1??Fmax?1??Fmax?? ?2??2?12 ??1?e??33?1???1?e22???1?? ?2??14?12e?3?14e?32
1??l,率密度为fX(x)????0,0?x?l其他1）上服从均匀分布，概33、（1）X在（0，
1??l, （2）两个小段均服从(0,l)上的均匀分布,fX1(x)????0,y2 Y?min(X1,X2), FY(y)?1?(1?)
l?2(l?y),0?y?l?2l fY(y)??
?其他?0,0?x?1其他
34、（1）U的可能取值是0，1，2，3
P{U?0}?P{X?0,Y?0}?1123P{U?1}?P{X?0,y?1}?P{X?1,Y?0}?P{X?1,Y?1}?2 P{U?2}?P{X?2,Y?0}?P{X?2,Y?1}?P{X?2,Y?2}?P{X?0,Y?2}
?P{X?1,Y?2}?29120120P{U?3}?P{X?3,Y?0}?P{X?3,Y?1}?P{X?3,Y?2}?1U P 0 112 1 23 2 29120 3 1120 (2) V的可能取值为0,1,2 19
P{V?0}?P{X?0,Y?0}?P{X?0,Y?1}?P{X?0,Y?2}?P{X?1,Y?0}?P{X?2,Y?0}?P{X?3,Y?0}?2740 P{V?1}?P{X?1,Y?1}?P{X?1,Y?2}?P{X?2,Y?1}?P{X?3,Y?1}?13P{V?2}?040
V P 0 2740 1 1340 2 0 (3) W的可能取值是0，1，2，3，4，5 P{W?0}?P{X?0,Y?0}?1121212P{W?1}?P{X?1,Y?0}?P{X?0,Y?1}?5P{W?2}?P{X?2,Y?0}?P{X?1,Y?1}?P{X?0,Y?2}?5P{W?3}?P{X?3,Y?0}?P{X?2,Y?1}?P{X?1,Y?2}?1P{W?4}?P{W?5}?0
12W P 0 112 1 512 2 512 3 112
概率统计第三章习题解答
1、P{X?4}?P{X?5}?P{X?6}? E(X)?29515,P{X?7}?25
429,P{Y?5}?529,P{Y?6}?629,P{Y?7}?14292、P{Y?4}? E(Y)?
17529
3、设X为取到的电视机中包含的次品数， P{X?k}?C2C10C312k3?k,k?0,1,2
20