Sending bits and parity bits over noisy channel

Consider a sender is trying to send three information bits $a_1$, $a_2$, and $a_3$ over a noisy channel with error probability $$p = 0.001$$ That is with probability $p$ each bit may be flipped independently from 0 to 1 and vice versa.

The sender adds three parity bits to it as follows: $$a_4 = a_2 + a_3 \;(\text{mod } 2)$$ $$a_5 = a_1 + a_3 \;(\text{mod } 2)$$ $$a_6 = a_1 + a_2 \;(\text{mod } 2)$$

What happens if the sender only adds two parity bits $a_4$ and $a_5$?

Does this mean just a change in the probability? Meaning it will increase the chance of not recognizing if a bit was sent incorrectly?

• What matters is the probability of getting the original bits right, for computing that you need to specify how you plan to use the parity bits to decode. Of course, in general (for a rational coder-decoder pair) adding redundacy (parity bits) should decrease the overall probabiblity of error. Commented Apr 29, 2016 at 15:33
• @leonbloy So by what you are saying, the answer to the question (what happens if only $a_4$ and $a_5$ is sent) depends on the probability the original bits are right? Commented Apr 29, 2016 at 15:41

The probability of an uncorrectable error using all 3 bits is ${6\choose2} p^2 \cdot(1-p)^4\approx1.5\cdot10^{-5}$
The probability of an undetected error using just 2 bits is ${4\choose1} p (1-p)^4\approx0.004$