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"
