Parity bit relation with message bits

E.g.  Consider the parity bit P₁ and we have to find the position of message bits which we’ll cover with this parity bit.

Firstly write the binary equivalents of positions of message bit

Bit1        bit2       bit3        bit4        bit5        bit6        bit7

Parity    parity                    parity

P₁                  P₂             M₁                P₃           M₂               M₃          M₄

 001          010         011          100         101         110         111

Now let’s see in the binary equivalent of position of parity bit P₁ that at which position we have 1and we see 1 is at LSB so we select the message bits which have positions with 1 at LSB which are M_1, M₂ and M₄ So P₁ bit would check the parity for M_1, M₂ and M₄

E.g.  Consider the parity bit P₂ and we have to find the position of message bits which we’ll cover with this parity bit

We have 1 at second position from left so we choose message bits which have 1 at 2^nd position n their position’s binary equivalent. Hence we get message bits M₁ M₃ and M₄. So P₂ checks parity for message bits of M₁ M₃ and M₄

Similarly we have 1 at 3^rd position of P₃ message bits with 1 at 3^rd position are  M₂M₃ M₄

So we now have

P₁ Checks bit number 1,3,5,7

P₂ Checks bit number 2,3,6,7

P₃ Checks bit number 4,5,6,7

These parity bits can be either even or odd parity bits but all  parity bits must be same i.e. all odd or all even