lebrongu 发表于 2015-2-27 13:07
美女你好。
First throw, we get a number. Then the probability of getting a “new” number would be ...

帅哥你好。
我就是发个问题确认一下。
帅哥再见。:time:

daenerysi402 发表于 2015-2-27 08:36
对了，请问一下楼主，最后一题是几何分布吧？只不过X的 support从6 开始？support(x)=(6,7,.....)?

...

美女你好。
First throw, we get a number. Then the probability of getting a “new” number would be 5/6. It follows a geometric distribution pattern. Then the expectation would be 6/5. After hitting that “new ” number, the probability of getting the next “new” number would be 4/6. Again, the expectation of that is 6/4. We keep adding up until 6/1.
Note that we assume each event is the occurrence of a “new” number. They are not independent. But the formula E(X+Y) = E(X) + E(Y) does not require independency.

本帖最后由 daenerysi402 于 2015-2-27 23:29 编辑

对了，请问一下楼主，最后一题是几何分布吧？只不过X的 support从6 开始？support(x)=(6,7,.....)?
搞混了，那个生小孩儿一男一女那个才是几何分布从2开始。。。

本帖最后由 guoao1992 于 2015-2-26 15:33 编辑

谢谢分享