| ||
|
1 |
4 |
则有放回连续取3次,其中2次取得次品的概率:
P=
C | 23 |
1 |
4 |
1 |
4 |
9 |
64 |
(2)依题知X的可能取值为0、1、2、3(6分)
且P(X=0)=
9 |
12 |
3 |
4 |
P(X=1)=
3 |
12 |
9 |
11 |
9 |
44 |
P(X=2)=
3 |
12 |
2 |
11 |
9 |
10 |
9 |
220 |
P(X=3)=
3 |
12 |
2 |
11 |
1 |
10 |
9 |
9 |
1 |
220 |
则X的分布列如下表:
X | 0 | 1 | 2 | 3 | ||||||||
P |
|
|
|
|
EX=0×
3 |
4 |
9 |
44 |
| ||
|
1 |
4 |
C | 23 |
1 |
4 |
1 |
4 |
9 |
64 |
9 |
12 |
3 |
4 |
3 |
12 |
9 |
11 |
9 |
44 |
3 |
12 |
2 |
11 |
9 |
10 |
9 |
220 |
3 |
12 |
2 |
11 |
1 |
10 |
9 |
9 |
1 |
220 |
X | 0 | 1 | 2 | 3 | ||||||||
P |
|
|
|
|
3 |
4 |
9 |
44 |