The transfer of a bit can be described by the following events:
$S_0$:= $0$ sent, $E_0$: $0$ received.
$S_1$:= $1$ sent, $E_1$: $1$ received.
The probability of a transmission error is 1%. Furthermore, we know that $P(S_1)=p$ and thus $P(S_0)=1-p$.
I now have to find the conditional probabilities $P(S_0|E_0)$, $P(S_1|E_0)$, $P(S_1|E_1)$ and $P(S_0|E_1)$.
I know the formulas and I decoded the probability of a transmission error as follows:$$P((S_0\cap E_1)\cup (S_1\cap E_0))$$ I also figured that $P(E_0|S_0)+P(E_1|S_0)=P(E_0|S_1)+P(E_1|S_1)=1$, but I can't find the probabilities. Are the probabilities dependent on $p$? Or is the information given enough to figure out what $p$ is?
Best Answer
It says the probability of a transmission error is $1 \%$ and that should mean,
$P(E_0 \mid S_0) = 0.99, P(E_1 \mid S_0) = 0.01$ $P(E_0 \mid S_1) = 0.01, P(E_1 \mid S_1) = 0.99$
By law of total probability,
$P(E_0) = P(E_0 \mid S_0) \cdot P(S_0) + P(E_0 \mid S_1) \cdot P(S_1)$ $~~~~~~~~~~= 0.99 (1-p) + 0.01p = 0.99 - 0.98p$
Similarly,
$P(E_1) = 0.01 (1-p) + 0.99p = 0.01 + 0.98p$
Now you have everything you need to calculate the conditional probabilities asked in the question. For example, for $P(S_0 \mid E_0)$,
$P(S_0 \mid E_0) = \dfrac{P(E_0 \mid S_0) \cdot P(S_0)}{P(E_0)} = \dfrac{0.99 (1-p)}{0.99 - 0.98p}$
You can find others similarly.