采用CSMA/CD进行介质访问，两个站点连续冲突3次后再次冲突的（ ）。
- A.1/2
- B.1/4
- C.1/8
- D.1/16
正确答案及解析
正确答案
解析
本题考查的是CSMACD中退避二进制指数算法的基本原理，每冲突一次，集合的范围就增大为原来的2倍，因此连续冲突3次之后，集合中的元素就是0～7，一共8个，再次冲突的概率为1/8。
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