A system uses odd parity. Two bytes arrive: 1101 0110 and 0110 0110. Which one contains an error?
- Count the 1s in 1101 0110: 1 + 1 + 0 + 1 + 0 + 1 + 1 + 0 = 5. Five is odd, so this byte passes.
- Count the 1s in 0110 0110: 0 + 1 + 1 + 0 + 0 + 1 + 1 + 0 = 4. Four is even.
- With odd parity an even count means a bit has been changed.
Answer: The byte 0110 0110 contains an error.