Determine the probability that the 1000-bit frame would be received without an error. MY AWNSER The probability that a frame arrives with one or more errors, P2 = 1-P1 = 1-(1-10-^6) ^1000 = 1- 0.999 = 0.001 1 out of 1000 frames has an error The number of frames transmission one day is 64000 bits/ sec * 24 hours/ day* 3600 second /hr * frame / 1000bits = 5.529 * 10^6 frames/day (5.529 * 10^6) * (0.001)= 5529.6 frames per day

SOLVE THIS PART>>>> Assuming that the Stop-and-wait flow-control is implemented in data-link layer communication, determine the probability that the sender transmits 3 of same frames to receive a positive acknowledgement. Assume that no packet loss occurs during these transmissions.