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, what is the probability that the message will be incorrectly decoded? What independence assumptions are you making?
(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 answers: Independence assumptions that each bit is transmitted independently, where each bit is sent as 5 bits for error reduction (0 = 00000), majority decoding.
- Probability of error with this transmission method will only be if
more than 2 bits were incorrectly transmitted. In which case, a transmission for 0 transmits as 00111 (3 errors), will be incorrectly decoded as a 1.
$q=0.2$, $p = 1-q=0.8$
There fore probability for transmitting incorrectly is given by binomial :
$$\begin{align} P(X>2) &= 1 - P(X \le 2)\\ &= 1 - {P(X=0) + P(X=1) + P(X=2)} \end{align} $$ My doubt here since we are supposed to calculate probability for transmitting incorrectly, we use swap positions for q and p right ?
$$p(X=0)= \binom{5}{0} 0.2^0 0.8^5$$
p(x=1)= 5C1 (0.2)^1 X (0.8)^4
p(x=2)= 5C2 (0.2)^2 X (0.8)^3
{P(x=0) + P(x=1) + P(x=2)} = 0.94208
1 - 0.94208 = 0.05792 is this right ?