题目
题型:不详难度:来源:
3 |
4 |
2 |
3 |
(Ⅰ)求该射手恰好命中一次得的概率;
(Ⅱ)求该射手的总得分X的分布列及数学期望EX.
答案
由题意知P(B)=
3 |
4 |
2 |
3 |
由于A=B
. |
C |
. |
D |
. |
B |
. |
D |
. |
B |
. |
C |
根据事件的独立性和互斥性得
P(A)=P(B
. |
C |
. |
D |
. |
B |
. |
D |
. |
B |
. |
C |
. |
C |
. |
D |
. |
B |
. |
D |
. |
B |
. |
C |
=
3 |
4 |
2 |
3 |
2 |
3 |
3 |
4 |
2 |
3 |
2 |
3 |
3 |
4 |
2 |
3 |
2 |
3 |
=
7 |
36 |
(II)根据题意,X的所有可能取值为0,1,2,3,4,5
根据事件的对立性和互斥性得
P(X=0)=P(
. |
B |
. |
C |
. |
D |
3 |
4 |
2 |
3 |
2 |
3 |
1 |
36 |
P(X=1)=P(B
. |
C |
. |
D |
3 |
4 |
2 |
3 |
2 |
3 |
1 |
12 |
P(X=2)=P(
. |
B |
. |
D |
. |
B |
. |
C |
. |
B |
. |
D |
. |
B |
. |
C |
3 |
4 |
2 |
3 |
2 |
3 |
3 |
4 |
2 |
3 |
2 |
3 |
1 |
9 |
P(X=3)=P(BC
. |
D |
. |
C |
3 |
4 |
2 |
3 |
2 |
3 |
3 |
4 |
2 |
3 |
2 |
3 |
1 |
3 |
P(X=4)=P(
. |
B |
3 |
4 |
2 |
3 |
2 |
3 |
1 |
9 |
P(X=5)=P(BCD)=
3 |
4 |
2 |
3 |
2 |
3 |
1 |
3 |
故X的分布列为