Let C be the ternary repetition code of length 4 over the alphabet $\{0,1,2\}$.
If the probability of each symbol being wrongly received is t and each symbol is likely the probability of a word being incorrectly decoded is $t^2 (3-2t)$
I do not understand how to get this answer and this is the working in the solutions which I dont understand.
I know the formula in the binary code of length n,
p(exactly i errors in specified positions ) = $t^i (1-t)^{n-1}$
Why is P$($ $1$ being received $)$ $=$ $1-t$? How is this calculated?
Why is P$($ $o$ or $1$ being received $)$ $=$ $t$ How is this calculated?
Why do we then say
P$($ $0$ being received)= p$($ $2$ being received$)$=$\frac{1}{2}t$ ?
We then get $(1-t)^4 + 8(\frac{1}{2} t) (1-t)^3 + 12 (\frac{1}{2}t^2 (1-t)^2$
Where do the coefficients come from?
I understand we then simplify the terms to get $(1-t)^2 (1+2t)$ and the final answer is obtained by doing $1-(1-t)^2 (1+2t)$
It appears you're looking at part (b) of Exercise 6 in this collection of solved problems. If so, you have to view your question in the context of part (a), which is about sending the word $1111$.
The solution to part (a) lists all the words that will be correctly decoded. There is one word ($1111$) with no errors, there are eight with exactly one error, and there are 12 with exactly 2 errors. To compute the probability of correct decoding, you add up the probabilities of each of these words. This explains the coefficients $1, 8, 12$.
As for your remaining questions: By assumption, $t$ is the probability of a symbol being wrongly received, so for any specified position in the word $1111$, $$P(\mbox{1 is received}) = 1 - P(\mbox{1 is wrongly received}) = 1-t.$$ Moreover, if $1$ is not received, then either $0$ or $2$ is received, so for any specified position in $1111$, $$P(\mbox{0 or 2 is received}) = P(\mbox{1 is wrongly received}) = t. $$ Now, assuming the author meant 'each incorrect symbol is equally likely', you argue that $P(\mbox{0 is received}) = P(\mbox{2 is received})$ and therefore each of them is equal to $t/2$.