my question is on below:
Q. A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received with probability 0.2. Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we transmit 00000 instead of 0 and 11111 instead of 1. If the receiver of the message uses “majority” decoding,
a. What is the probability that the message will be incorrectly decoded?
b. A message consisting of 118 bytes (944 bits) is encoded and send over the channel. What is the probability that the message contains error after it is decoded?
(By majority decoding we mean that the message is decoded as “0” if there are at least three zeros in the message received and as “1” otherwise.)
My solution for part a is:
$$\begin{align}
P(X>2) &= 1 – P(X \le 2) \\
& = 1 – {P(X=0) + P(X=1) + P(X=2)}
\end{align}
$$
or
$$\begin{align}
P(X\ge 3) &= P(X=3) + P(X=4) + P(X=5) \\
\end{align}
$$
Both solutions return answer 0.05792
My confusion for part b is:
Since I already get the probability that the message will be incorrectly decoded, how can I calculate the probability of the message which contains 944 bits that decode error? Does this chains into a conditional probability?
Best Answer
Each bit is decoded independently.
It contain an error if one of the bit is decoded wrongly.
Hence the probability that there is an error is equal to $1$ subtract the probability that there is no error.
Probability of error for a single bit is $q$.
Probability of a bit is decoded correctly is $1-q$.
The probability that there is no error is $(1-q)^n$ where $n$ is the number of bits.
The probability that there is an error is hence $1-(1-q)^n$.