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Consider a LAN with four nodes $S_1, S_2, S_3,$ and $S_4$. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmits in the same slot. The probabilities of generation of a frame in a time slot by $S_1, S_2, S_3,$ and $S_4$ are $0.1, 0.2, 0.3$ and $0.4$ respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is__________________.
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82 votes
$P= P(S1) P(\neg S2) P(\neg S3) P(\neg S4)$
$\qquad + P(\neg S1) P(S2) P(\neg S3) P(\neg S4)$
$\qquad + P(\neg S1) P(\neg S2) P(S3) P(\neg S4) $
$\qquad + P(\neg S1) P(\neg S2) P(\neg S3) P(S4) $

$ \quad = $ $ 0.1 * 0.8 * 0.7 * 0.6 \\ + 0.9 * 0.2 * 0.7 * 0.6 \\ + 0.9 * 0.8 * 0.3 * 0.6 \\ + 0.9 * 0.8 * 0.7 * 0.4$

$ \quad = \textbf{0.4404}$
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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
9 votes
9 votes

Required probability

=P(only one frame can transmit in a given time slot)

=P(only S1 sends)+ P(only S2 sends)+ P(only S3 sends)+ P(only S4 sends)

=(0.1*0.8*0.7*0.6)+(0.9*0.2*0.7*0.6)+(0.9*0.8*0.3*0.6)+(0.9*0.8*0.7*0.4)

=0.0336+ 0.0756 + 0.1296 + 0.2016

=0.4404

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Pi is the probability of frame being generated by station si

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