Hello, the answer has to be typed here, not in paper to avoid bad handwritting.
ID: 3756912 • Letter: H
Question
Hello, the answer has to be typed here, not in paper to avoid bad handwritting. Thanks
5. (10 points) Assume that and are two arbitrarily chosen 16-bit integers with the UDP checksum of . and are sent along with their checksum by host A through an unreliable network toward host B. In the destination, B receives three 16- bit integers , , , and . If B cannot detect any error using checksum, what is the probability that the integers are transferred with error? Assume that the bits are received by B in the same order that they are sent by A. Also, assume that every bit can be altered over the network with the probability of T, independent of other bits and0: i.e. it is almost impossible that more than two bits get flipped.Explanation / Answer
Probability of error= probability of error when only 1 bit is flipped +
probability of error when 2 bits are flipped
No. of 16-bit numbers when 1 bit is flipped= 1 (correct one) + 16 (incorrect)
= 17
and error numbers = 16 (incorrect)
probability of error when only 1 bit is flipped = 16/17
No. of 16-bit numbers when 1 bit is flipped= 1 (correct one) + 16C2 (incorrect)
= 121
and error numbers = 16C2 (incorrect) = 120
probability of error when 2 bits are flipped = 120/121
Total Probability = 16/17 + 120/121
And in the form of Pi; probability of error = pi/16 + pi/(16*15)
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