Question

A communication channel transmits a signal as sequence of digits 0 and 1. The probability of incorrect reception of each digit is p. To reduce the probability of error at reception, 0 is transmitted as 00000 (five zeroes) and 1, as 11111. Assume that the digits are received independently and the majority decoding is used. Compute the probability of receiving the signal incorrectly if the original signal is (a) 0; (b) 101. Evaluate the probabilities when p D 0:2.

Answer 1

Answer:

Probability of receiving the signal incorrectly is ".05792"

Step-by-step explanation:

Since it is mentioned here that the majority decoding is used so it implies that if the message is decoded as zero if there have been minimum of 3 Zeros in the messgae.

As we can see that there have been 5 bits, so the incorrection will occur only if there have been atleast 3 incorrect bits. Here we are asked to find out the probability of receiveing the signal incorrectly so i will take incorrectly transmiited bit as "Success".

The binomial distribution gives the probability of exactly m successes in n trials where the probability of  each individual trial succeeding is p.

With this method, we get the following pattern,

Binomial at (m=3, n=5, p=0.2) for probability of incorrection due to exactly 3 fails,

Binomial at (m=4, n=5, p=0.2) for probability of incorrection due to exactly 4 fails,

Binomial at (m=5 ,n=5, p=0.2) for probability of incorrection due to exactly 5 fails.

hence the probability "P" of receiving the signal incorrectly will be,

           P=(⁵₃) p³ (1 − p)⁵⁻³   +     (⁵₄) p⁴(1 − p)⁴⁻³      +      (⁵₅) p³(1 − p)³⁻³

whereas,

(⁵₃)= 5! / (5-3)! = 10

(⁵₄)= 5! / (5-4)! = 5

(⁵₅)= 5! / (5-5)! = 1

Putting these values in above equation we get,

P= 10 (0.2)³(1-0.2)²  + 5(0.2)⁴(1-0.2)¹  + (0.2)⁵

P=.05792

So the probablility "P" of receiving the signal incorrectly becomes ".05792"