if no. of 1s r odd the put 0 as a odd parity and viceversa

Tabrez said:
(Sep 30, 2011)

First try for 2 bit then analyised then you will see each odd no given to last bit one. Then odd priority become 0, if it is not matched then it become 1.

Lakshmi V said:
(Jan 29, 2015)

Parity is nothing checking no.of 1's and in the given question it is given to follow odd parity i.e. no.of 1's in given byte should be odd and if it is not odd we need to add 1 in order to make it odd parity.

Sarika Gupta said:
(May 22, 2015)

But according to me answer must be 1st. Because it ask for odd parity that means if no. of once is odd, odd parity must be high = 1. Please clarify.

Sandeep said:
(Feb 11, 2016)

The total numbers of 1s (including parity bit) should ultimately remain even. Correct me if I am wrong?

Sandeep said:
(Feb 11, 2016)

It's odd parity, so the total number of 1s (including the parity bit) should be odd.

Mahesh said:
(Nov 1, 2016)

According to me odd parity is equals to number of ones = 5 so 5 is odd number p = 1;also remaining same.

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