考虑具有四个节点S1，S2，S3和S4的LAN。时间被划分为固定大小的时隙，并且节点只能在时隙的开始处开始其传输。如果有多个节点在同一时隙中传输，则据说发生了冲突。由S1，S2，S3和S4在时隙中生成帧的概率分别为0.1、0.2、0.3和0.4。这四个站中的任何一个在第一个时隙中发送帧而没有任何冲突的概率为_________。
(A) 0.462
(B) 0.711
(C) 0.5
(丁)0.652答案： (A)
解释：
The probability of sending a frame in the first slot
without any collision by any of these four stations is
sum of following 4 probabilities
Probability that S1 sends a frame and no one else does +
Probability thatS2 sends a frame and no one else does +
Probability thatS3 sends a frame and no one else does +
Probability thatS4 sends a frame and no one else does
= 0.1 * (1 - 0.2) * (1 - 0.3) *(1 - 0.4) +
(1 -0.1) * 0.2 * (1 - 0.3) *(1 - 0.4) +
(1 -0.1) * (1 - 0.2) * 0.3 *(1 - 0.4) +
(1 -0.1) * (1 - 0.2) * (1 - 0.3) * 0.4
= 0.4404 这个问题的测验