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Assume that X and Y are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both X and Y attempt to transmit a frame, they wait to get the control of channel using binary exponential algorithm. The probability that both were successfully allowed to send the frame on fifth round of the algorithm (assuming every time both X and Y will collide in back-off race till 4th round) is ________.

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I think it will be 15/16.

After four collisions (Since it is given to assume four collision have happened I think we need not consider that probability), each host will choose a number between (0,$2^{4-1}$).

Probability of collision is 1/16 (Both choosing the same number).

Hence probability of success=15/16.
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i think 31/32 will be the ans
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Thanks for noticing, Updated it.

Here they are asking about the fifth iteration. That means we have to find the probability of success in 5th iteration not in 4th iteration.

For K=5, i.e. (0 to 25-1) = (0,31)PT.

So the probability of collision is 1/32 and hence the probability of success will be (1-1/32) = 31/32.

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Answer should be 15/16 not 31/32
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@Mayank

successfully allowed to send the frame on fifth round of the algorithm

Means the value of k will go up to 5 na, if value goes up to 5 i.e. (0 to (2^5)-1). Please suggest what I did wrong here.

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5th round successful means 4 times collision has occurred. K=4.
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Collision occurred 4 times but successful in only 5th time na
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When first collision occurred k=1 is used for second round

When  second collision occurred k=2 used for resolving the third round

When third collision occurred k=3 is used for resolving the 4th round

When 4th collision occurred k=4 is used for resolving the 5th round.

I think so.
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probability that both were successfully allowed to send the frame on fifth round of the algorithm

Isn't this means to find the success in 5th round. Please explain.