问题 解答题

一只袋子装有大小相同的2个红球和8个黄球,从中随机连取三个球,每次取一个.记“恰有一红球”为事件A,“第三个球是红球”为事件B,求在下列情况下A、B的概率.

(1)取后不放回;              

(2)取后放回.

答案

(1)根据题意,袋子中共有2+8=10个球,

若不放回的从中取出3个,有A103=720种取法,

事件A即恰有1个红球的取法有C21C82A33=336种取法,则P(A)=

336
720
=
7
15

事件B即第三个球是红球,其取法有C21A92=144种,则P(B)=

144
720
=
1
5

(2)根据题意,袋子中共有2+8=10个球,

若有放回抽取,每次抽取时,袋中球的数目不变,则每次取到红球的概率都是

2
10
=
1
5
,则取到白球的概率为
4
5

事件B即第三个球是红球,易得其概率P(B)=

1
5

事件A即恰有1个红球,即3次试验中恰有1次发生,其概率为P(A)=C31

1
5
)(
4
5
2=
48
125

阅读理解

阅读理解。

     Mr Green is a teacher. His only son, Bill, is a student. He is twelve. The child likes to fly kites and play some

other games. One Monday afternoon, Bill flies kites with his friends after school. But Bill's kite is in the tree. and

he can not get it. Bill and his friends do not know what to do. They worry very much and Bill goes to his father

for help. His father wants to get the kite down from the tree, but he can't. So Bill asks him to make a new one for

him.The man makes a new kite. When the boy gets the new kite, he is very glad. He thanks his father for the kite

and goes out to play with his friends.

1 .How many children are there in Mr Green's family?    

A. Two            

B. One            

C. Three          

D. We don't know

2 .Bill likes to ________.  

A. play other games                

B. fly kites only

C. make kites                        

D. play games

3.One day_______.  

A. Bill finds a kite in a tree            

B. Bill's kite is broken and it is in the tree

C. Bill's friends mend his kite        

D. Bill makes a kite for his friends

4 .Bill is glad because________.     

A. he helps his friends get the kite down from the tree

B. his friends do not know what to do

C. his father helps him get him get the kite down

D. Mr Green makes a new kite for him

5.From the text we can see_____.

A .Bill can not get the kite down but his friends don't' worry very much

B. Bill's father can not get the kite down, so he makes a new kite

C. Mr Green can not get the kite down and wants to make a new one for Bill

D. Bill's friends break the kite and help Bill make a new one.

单项选择题

大爆炸理论的最直接的证据来自于对遥远星系光线特征的研究。在20世纪20年代美国天文学家埃德温哈勃测量了18颗恒星(它们距离是已知的)发出来的光,发现它们都全部存在着红移。哈勃得出结论,这些恒星一定相对于我们(观察者)在后退。因为根据多普勒效应,恒星一边后退一边发光而且光速(相对于观测者)是不变的话,我们收到光的波长就会长于原来的值。就如远去的声音将变向低音,光将偏向红光。哈勃认为,遥远星系的光波变长(红化)是由于宇宙正在膨胀的结果。在一个正在膨胀着的宇宙中任何一点来观测,其周围的任何星系都是离它退行的,离它越远的星系退行速度越高。 

如果宇宙正在膨胀,它在过去必定比较小(这也是蕴涵在膨胀观念中的逻辑结论)。如果能倒放这部“宇宙影片”,我们会发现,所有的星系在遥远的过去是聚合在一起的。并且根据现在的膨胀速度,我们还可以推断这种聚合状态必定出现在好几十亿年前。科学家们现在将之定在150亿年左右。 

对大爆炸宇宙学有力支持的第二个观测证据出现在1965年。两位美国科学家彭齐亚斯和威尔逊偶然的检测到弥漫在全天空的微波背景辐射。这种辐射以相同的强度从空间的各个方向射向地球。它的光谱线与达到某种热动平衡态的熔炉内的发光情况<>相符,由于符合程度非常之好,因而不可能是一种巧合。这个发现被称为“宇宙微波背景辐射”。对它的测量表明,它的有效温度大约比绝对零度(约等于-273℃)高3度,可写为3K,只能将微波背景辐射解释为这是宇宙原初阶段(大爆炸阶段)的直接遗迹,把它看作为宇宙诞生时灼热火焰的余辉,是从大爆炸散落的残余辐射由于宇宙膨胀而冷却所具有的。实际上最早提出微波宇宙背景辐射假说的科学家就意识到,如果宇宙起始于遥远过去的某种既热而密的状态,那就应当留下某种从这个爆发式开端洒落的辐射。

文中划线处“红移”的意思是()

A.恒星一边后退一边发光,观测者收到光的光波变长

B.恒星发出的光呈现红色,观测者可以看到它缓缓后移

C.恒星如果用不变的光速发出光线,那么光线将偏向红色

D.恒星与地球的距离是已知的,它们向地球发出红色的光