题目内容
桌面上有三颗均匀的骰子(6个面上分别标有数字1,2,3,4,5,6).重复下面的操作,直到桌面上没有骰子:将骰子全部抛掷,然后去掉哪些朝上点数为奇数的骰子.记操作三次之内(含三次)去掉的骰子的颗数为X.(Ⅰ)求P(X=1); (Ⅱ)求X的分布列及期望EX.
分析:(1)P(X=1)=
(
)3(
)2(
)2+(
)3
(
)3(
)2+(
)3(
)3
(
)3=
.
(2)由题设知,X的取值为0,1,2,3,P(X=0)=
,P(X=1)=
,P(X=2)=
,P(X=3)=
,由此能求出X的分布列和期望.
| C | 1 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| C | 1 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| C | 1 3 |
| 1 |
| 2 |
| 21 |
| 521 |
(2)由题设知,X的取值为0,1,2,3,P(X=0)=
| 1 |
| 512 |
| 21 |
| 251 |
| 147 |
| 512 |
| 343 |
| 512 |
解答:解:(1)P(X=1)=
(
)3(
)2(
)2+(
)3
(
)3(
)2+(
)3(
)3
(
)3=
.
(2)由题设知,X的取值为0,1,2,3,
P(X=0)=(
)3(
)3(
)3=
,
P(X=1)=
(
)3(
)2(
)2+(
)3
(
)3(
)2+(
)3(
)3
(
)3=
,
P(X=2)=
,P(X=3)=
,
∴X的分布列是
Ex=0×
+1×
+2×
+3×
=
.
| C | 1 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| C | 1 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| C | 1 3 |
| 1 |
| 2 |
| 21 |
| 521 |
(2)由题设知,X的取值为0,1,2,3,
P(X=0)=(
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 512 |
P(X=1)=
| C | 1 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| C | 1 3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| C | 1 3 |
| 1 |
| 2 |
| 21 |
| 521 |
P(X=2)=
| 147 |
| 512 |
| 343 |
| 512 |
∴X的分布列是
| X | 0 | 1 | 2 | 3 | ||||||||
| P |
|
|
|
|
| 1 |
| 512 |
| 21 |
| 251 |
| 147 |
| 512 |
| 343 |
| 512 |
| 21 |
| 8 |
点评:本题考查离散型随机变量的分布列和期望,解题时要认真审题,注意挖掘题设中的隐含条件仔细解答,避免不必要的错误.
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