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?

Solution :

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 to denote an event of successes in the 5 trails, then the probability k are decoded at the receiver is:

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

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

         = 0.91854

The probability that the deletion occurs is

         = 0.081

The probability of an error is

         = 0.00046

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions