Question

Suppose there exists a Hamming code of length 14 that corrects 2 errors. Assuming that the probability of a correct transmission of an individual symbol is 0.9, find the probability that a message transmitted using this code will be correctly received.

Answer 1

Answer:

0.8432

Step-by-step explanation:

The probability of a correct transmission of individual character = 0.9

Let's consider the three possible scenarios:

1) Probability of all 14 characters transmitted correctly = 0.9^14 = 0.2287

2)Probability of only one character being wrong: (0.9 ^ 13) * 0.1

This equals 0.02541    To account for the order at which the incorrect character is produced, lets consider the permutation: $\frac{14!}{13!*1!}$

This accounts for same objects or results being repeated, and is an integral part of permutation questions. The result is 14. Thus there are 14 ways for the probability of 0.02541 to occur. That means a total of 0.02541 * 14 probability. This equals 0.3559

3) Probability of both characters being wrong:

   Similar steps to (2). Probability = (0.9^12) * (0.1 ^ 2) =0.002824

   Number of ways this could occur: $\frac{14!}{12!*2!}$ = 91

   Total probability for this event: 0.002842 * 91 = 0.2586

   The answer to our question is thus all these probabilities combined, which makes: 0.2287 + 0.3559 + 0.2586 = 0.8432