题目内容
从一批含有13只正品,2只次品的产品中,不放回地抽取3次,每次抽取1只,设抽得次品数为X,则E(5X+1)=
3
3
.分析:确定X的取值,求出相应的概率,可得期望,进而可求E(5X+1).
解答:解:由题意,X的取值为0,1,2,则
P(X=0)=
×
×
=
;P(X=1)=
×
×
+
×
×
+
×
×
=
P(X=2)=
×
×
+
×
×
+
×
×
=
所以期望E(X)=0×
+1×
+2×
=
,
所以E(5X+1)=
×5+1=3
故答案为3.
P(X=0)=
| 13 |
| 15 |
| 12 |
| 14 |
| 11 |
| 13 |
| 22 |
| 35 |
| 2 |
| 15 |
| 13 |
| 14 |
| 12 |
| 13 |
| 13 |
| 15 |
| 2 |
| 14 |
| 12 |
| 13 |
| 13 |
| 15 |
| 12 |
| 14 |
| 2 |
| 13 |
| 12 |
| 35 |
P(X=2)=
| 13 |
| 15 |
| 2 |
| 14 |
| 1 |
| 13 |
| 2 |
| 15 |
| 13 |
| 14 |
| 1 |
| 13 |
| 2 |
| 15 |
| 1 |
| 14 |
| 13 |
| 13 |
| 1 |
| 35 |
所以期望E(X)=0×
| 22 |
| 35 |
| 12 |
| 35 |
| 1 |
| 35 |
| 14 |
| 35 |
所以E(5X+1)=
| 14 |
| 35 |
故答案为3.
点评:本题考查数学期望的计算,考查概率的求解,确定变量的取值,正确求概率是关键.
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