题目内容
A、B两篮球队进行比赛,规定若一队胜4场则此队获胜且比赛结束(七局四胜制),A、B两队在每场比赛中获胜的概率均为
,ξ为比赛需要的场数,则Eξ=
.
1 |
2 |
93 |
16 |
93 |
16 |
分析:由题设知比赛需要的场数ξ为4,5,6,7.p(ξ=4)=(
)4+(
)4=
+
=
,p(ξ=5)=
(
)3(
)1•
+
(
)3(
)1•
=
,p(ξ=6)=
(
)3(
)2•
+
(
)3(
)2•
=
,p(ξ=7)=
(
)3(
)3•
+
(
)3(
)3•
=
,由此能求出Eξ.
1 |
2 |
1 |
2 |
1 |
16 |
1 |
16 |
1 |
8 |
C | 3 4 |
1 |
2 |
1 |
2 |
1 |
2 |
C | 3 4 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
4 |
C | 3 5 |
1 |
2 |
1 |
2 |
1 |
2 |
C | 3 5 |
1 |
2 |
1 |
2 |
1 |
2 |
5 |
16 |
C | 3 6 |
1 |
2 |
1 |
2 |
1 |
2 |
C | 3 6 |
1 |
2 |
1 |
2 |
1 |
2 |
5 |
16 |
解答:解:由题设知比赛需要的场数ξ为4,5,6,7.
p(ξ=4)=(
)4+(
)4=
+
=
,
p(ξ=5)=
(
)3(
)1•
+
(
)3(
)1•
=
,
p(ξ=6)=
(
)3(
)2•
+
(
)3(
)2•
=
,
p(ξ=7)=
(
)3(
)3•
+
(
)3(
)3•
=
,
∴Eξ=4×
+5×
+6×
+7×
=
.
故答案为:
.
p(ξ=4)=(
1 |
2 |
1 |
2 |
1 |
16 |
1 |
16 |
1 |
8 |
p(ξ=5)=
C | 3 4 |
1 |
2 |
1 |
2 |
1 |
2 |
C | 3 4 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
4 |
p(ξ=6)=
C | 3 5 |
1 |
2 |
1 |
2 |
1 |
2 |
C | 3 5 |
1 |
2 |
1 |
2 |
1 |
2 |
5 |
16 |
p(ξ=7)=
C | 3 6 |
1 |
2 |
1 |
2 |
1 |
2 |
C | 3 6 |
1 |
2 |
1 |
2 |
1 |
2 |
5 |
16 |
∴Eξ=4×
1 |
8 |
1 |
4 |
5 |
16 |
5 |
16 |
93 |
16 |
故答案为:
93 |
16 |
点评:本题考查离散型随机变量的分布列和数学期望,考查学生的运算能力,考查学生探究研究问题的能力,解题时要认真审题,注意n次独立重复试验恰有k次发生的试验概型的合理运用.
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