Solution#3.1

For each frame chance of successful transmission = 0.8 So the complete message will arrive undamaged when all the frames are undamaged. This chance, p = (0.8)¹⁰ = 0.107 Now suppose that total k transmissions are required for the successful message transmission then, p = 1/k So k = 1/0.107 = 9.3 transmissions required.

Solution#3.2

The output string is 011110111110011111010.

Solution#3.6

For two adjacent codes either all the even bit or the odd bit block would change and hence the corresponding parity bit. So two adjacent differ in two bit positions. So Hamming distance = 2.

Solution#3.7

A single error cause both the horizontal and vertical parity checks to be wrong. Two errors will also be easily detected. Three errors will also be detected easily whether they all are in the same row (column) or in two or three different ones.

Solution#3.9

The remainder is 111 or x²+x+1.

Solution#3.10

CRC is computed during transmission and appended to the output stream as soon as the last bit goes out onto the network. If the CRC were in the header, it would be necessary to make a pass over a frame to compute CRC before transmitting. This would require each byte being handled twice-once for checksumming and once for transmitting. Using the trailer cuts the work in half.