We wish to modify the cellular telephone coding system in order to reduce the number of errors. In particular, if there are two or three zeroes in the received sequence of 5 bits, we will say that a deletion (event D) occurs. Otherwise, if at least 4 zeroes are received, then the receiver decides a zero was sent. Similarly, if at least 4 ones are received, then the receiver decides a one was sent. We say that an error occurs if either a one was sent and the receiver decides zero was sent or if a zero was sent and the receiver decides a one was sent. For this modified protocol, what is the probability P[E] of an error? What is the probability P[D] of a deletion?

Here, we have 5 trails that corresponds to the 5 times the binary symbol is sent. On each trail, success is said to occur when the binary symbol is received correctly. The probability of a success is p = 1 - q = 0.9 and q = 0.1

Now if the 0 is being transmitted, then the 0 is sent 5 number of times and then we call the decoding a 0 success. Let the repeated transmission are independent. In this case, we have 5 trails. Let each repeated bit the receiver decodes as 1 a success, using the binomial probabilities. We using the [tex]$S_{k,5}$[/tex] to denote an event of [tex]$k$[/tex] successes in the 5 trails, then the probability k are decoded at the receiver is:

[tex]$P[S_{k,5}]=\binom{5}{k}p^k(1-p)^{5-k}$[/tex]

[tex]$=\binom{5}{k}(0.9)^k(1-0.9)^{5-k}$[/tex]

[tex]$=\binom{5}{k}(0.9)^k(0.1)^{5-k}$[/tex]

Here, k = 1, 2, 3, 4, 5

The probability of the transmitted bit to be decoded correctly is:

[tex]$P[C]=P[S_{5,5}]+P[S_{2,5}]$[/tex]

[tex]$=\binom{5}{3}(0.9)^5(0.1)^{0}+\binom{5}{4}(0.9)^4(0.1)^1$[/tex]

[tex]$=(0.9)^5+5(0.9)^4(0.1)^1$[/tex]

= 0.91854

The probability that the deletion occurs is

[tex]$P[D]=P[S_{3,5}]+P[S_{2,5}]$[/tex]

[tex]$=\binom{5}{3}(0.9)^3(0.1)^{2}+\binom{5}{2}(0.9)^2(0.1)^3$[/tex]

[tex]$=10(0.9)^3(0.1)^2+10(0.9)^2(0.1)^3$[/tex]

= 0.081

The probability of an error is

[tex]$P[E]=P[S_{1,5}]+P[S_{0,5}]$[/tex]

[tex]$=\binom{5}{1}(0.9)^1(0.1)^{4}+\binom{5}{0}(0.9)^0(0.1)^5$[/tex]

[tex]$=5(0.9)^1(0.1)^4+(0.1)^5$[/tex]

= 0.00046