A system transmits 11111 or 00000 when 1 or 0 are transmitted to decrease the probability of error. Supposing that the probability one of the 5 transmitted bits is corrupted from 1 to 0 is 0.3 and that a bit is corrupted from 0 to 1 is 0.1.
Given that it is equally likely that the system transmits a 0 or a 1, if 01101 was received what is the probability that a 1 was transmitted.
The initial scenario would be the probability of obtaining a singular 1 or 0 which is 0.4 and 0.6 respectively.
Would this give us (4 choose 2)(0.4)^4(0.6) = 0.09216 as the probability?
Best Answer
You should use Bayes' rule:
\begin{align} P(1\text{ is sent}|01101)&= \frac{P(01101|1)P(1)}{P(01101|1)P(1)+P(01101|0)P(0)}\\ &=\frac{P(01101|1)}{P(01101|1)+P(01101|0)} \end{align}
We have $P(01101|1)=(0.3)^2(0.7)^3$ and $P(01101|0)=(0.9)^2(0.1)^3$.
Remark: Binomial coefficient terms are useful when we are like given that there are exactly how many success event rather than the scenario where a particular pattern is given.