Question

Bit transmission errors between computers sometimes occur, where one computer sends a 0 but the other computer receives a 1 (or vice versa). Because of this, the computer sending a message repeats each bit three times, so a 0 is sent as 000 and a 1 as 111. The receiving computer "decodes" each triplet by majority rule: whichever number, 0 or 1, appears more often in a triplet is declared to be the intended bit. For example, both 000 and 100 are decoded as 0, while 101 and 011 are decoded as 1. Suppose that 6% of bits are switched (0 to 1, or 1 to 0) during transmission between two particular computers, and that these errors occur independently during transmission.(a) Find the probability that a triplet is decoded incorrectly by the receiving computer.(b) Using your answer to part (a), explain how using triplets reduces communication errors.(c) How does your answer to part (a) change if each bit is repeated five times (instead of three)?(d) Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?

Answer 1

Answer:

(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010

(b)

(1 – p) = 0.010

(c)

E(x) = 25000 x 0.010

     = 259.2

Explanation has given below.

Step-by-step explanation:

Solution:

(a) Probability that a triplet is decoded.

2 out of three

P = 0.94, n = 3

m= no of correct bits

m   bit (3, 0.94)

At p(m≤1) = B (1; 3, 0.94)

 = 0.010

(b) Using your answer to part (a),

(1 – p) = 0.010

Error for 1 bit transmission error.

(c)  How does your answer to part (a) change if each bit is repeated five times (instead of three?

P( m ≤ 2 )

L = Bit (5, 0.94)

   = B (2; 5, 0.94)

   = 0.002

(d)  Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?

Given:

h = 25000

Bits are switched during transmission between two computers = 6% = 0.06

m = Bit (25000, 0.06)

E(m) = np

        = 25000 x 0.06

         = 1500

m = Bit (25000, 0.01)

E(m) = 25000 x 0.010

     = 259.2