Eg**. Find out the parity bit (odd) for message 1101 and show us how it helps in detecting errors**

As 1101 has 3 i.e. odd number of 1’s so P=0 so that we still have the odd number of 1’s in the combination of 5 bits(message(4 bits) and parity bit(1 bit))

So message we send with parity is 11010 (5^{th} bit from left is parity bit)

Let’s now see the effect of errors on this message

**1 error: **Suppose we have error in 3^{rd} bit from left so bit at 3^{rd} position would change from 0 to 1. Hence message received would be 11110 instead of 11010 and at the receiver we check the parity of message and we see message has even number 1’s in the received signal which should have been odd so we have detected the error although we are not sure about the position of the error. So as error is detected we can make a request for another send of the same message.

**2 errors:** Suppose we have error at positions 1 and 4 from left so we have a bit change from 1 to 0 at 1^{st} and 1 to 0 at 2^{nd} position. We messaged received is 01000 and we have odd number of 1’s in received message which is also the parity status of sent message so no detection of errors so we can see this method is unable to detect even number of errors

**3 errors: **Suppose we have error in 2^{nd} , 3^{rd} and 4^{th} bit from left so bit change at 2^{nd} position is 1 to 0, at 3^{rd} position would from 0 to 1 and at 4^{th} from 1 to 0. Hence message received would be 10100 instead of 11010 and at the receiver we check the parity of message and we see message has even number 1’s in the received signal which should have been odd so we have detected the error although we are not sure about the positions of the error. So as error is detected we can make a request for another send of the same message.

So from here one can easily conclude that PARITY BIT method can detect only odd number of errors independent on whether we have odd parity message or even parity message.