Use even and odd parity checks
Calculate a parity bit and test received data using even or odd parity.
Section 2 · Data transmission
This sequence follows Cambridge IGCSE Computer Science 0478, 2026–2028 syllabus Version 6. It covers subject content sections 2.1–2.3 only. The separately published 2029–2031 syllabus is not mapped here. Practice is original, self-checked and does not award an official Cambridge grade. Full timed papers and human subject-teacher review remain outstanding.
Before you begin
- No programming language is required. Be comfortable with place value and multiplication; each lesson introduces the computer-science notation it uses.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does even parity require?
What you will learn
- Calculate even and odd parity bits
- Check a received byte against its parity
- Explain the limitations of one-dimensional parity
Parity sets a required count of 1 bits
For even parity, the data plus parity bit must contain an even number of 1s. For odd parity, the total must be odd. The sender adds the parity bit; the receiver counts the 1s and checks the rule.
Apply and check the rule
Parity can detect any odd number of bit changes in the protected group, but an even number of changes may preserve the required parity. A parity error signals a problem; it does not identify or repair every changed bit.
Pause and explain
A byte with even parity arrives with an odd number of 1s. What can the receiver conclude?
Worked example
Add an even-parity bit to 1011001.
Show the worked solution
- The data contains four 1s, already an even count.
- The parity bit must keep the total even.
- Append 0.
Answer 10110010
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Add an odd-parity bit to 1100100.
Give me a hint
Count the data 1s and make the total odd.
Compare my reasoning
- The data has three 1s.
- Three is already odd, so add parity bit 0.
11001000
Look for these in your work
- Three 1s counted
- Odd count preserved
- Bit appended
A received byte is 10010110 and uses even parity. Does it pass?
Give me a hint
Count all 1 bits including the parity bit.
Compare my reasoning
- The pattern has four 1s.
- Four is even, so it passes the parity rule.
Yes; it passes even parity.
Look for these in your work
- All bits counted
- Parity rule applied
A data block uses even parity. Two bits are inverted during transmission. Explain why the parity check might fail to notice.
Give me a hint
Compare the number of changed bits with the required parity.
Compare my reasoning
- Each inversion changes the count of 1s by one up or down.
- Two inversions change the total parity an even number of times.
- The overall count may remain even, so the one-bit parity check can pass despite corruption.
Two bit changes can preserve the even/odd status of the total count. Basic parity may therefore miss an even number of errors.
Look for these in your work
- Effect of each inversion
- Two changes considered
- Detection limit explained
Common mistakes
- Counting zeros instead of ones
- Assuming parity detects every possible number of bit errors
What can you explain now?
For even parity, what parity bit should be added to 1011?
Compare with the explanation
1
1011 contains three 1s, so adding 1 makes four, an even count.
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Make it stick
Which error patterns can a single parity bit miss?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.