Probability of correct message sent over network

Question

We wish to transmit an n-bit message to a receiving agent, which is the case for the low-level messaging on most computer networks. Unfortunately, no network technology is perfect, and it is possible for bits to be corrupted (i.e., flipped from the intended value). We want to analyze the impact of this, and understand how using parity can protect against these errors. 

Suppose that the bits in the message are independently corrupted (flipped) during transmission with probability e.  With an extra parity bit sent along with the original information, a message can be corrected by the receiver if at most one bit in the entire message (including the parity bit) has been corrupted. Suppose we want to ensure that the correct message is received with probability at least 1 – d. What is the maximum feasible value of n? Calculate this value for the case e =0.0002 and d = 0.01. Note that the total length of the message including the parity bit is n+1.

Answer 1

Well if there are n+1 bits including the parity bit, the chance that no bits have been corrupted is

(1-e)ⁿ⁺¹

Probability of 1 bit being corrupted out of (n+1) = (n+1) e (1-e) ⁿ

This is because we can choose the 1 bit in (n+1) ways, and having done that the probability of that being corrupted is e, and probability of the other n being non corrupted is (1-e)ⁿ

So probability of message received correctly = (n+1) e (1-e)ⁿ + (1-e) ⁿ⁺¹ > 1-d

e = 0.0002, d = 0.01

(n+1) 0.0002 (0.9998)ⁿ + (0.9998)ⁿ⁺¹ > 0.99

You will have to experiment with some numbers here

Mike