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A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is
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- 0.5
- 0.625
- 0.75
- 1.0
- 0.5
Correct Option: B
In the first back off race, there are two conditions 0 and 1 In the second back off race there are four conditions 0, 1, 2 and 3 Hence, winning probability of A can be calculated as
1/2 × 3/4 + 1/2 × 1/2
= 3/8 + 1/4
= 5/8 = 0.625
