What is the probability that a 1 was sent, given that 1 was received? I have the following information:

$0$ is sent with probability $0.3$
$1$ is sent with probability $0.4$
$2$ is sent with probability $0.3$
Due to noise, $0$ is changed to $1$ during transmission with probability $0.2$
Due to noise, $0$ is changed to $2$ during transmission with probability $0.1$
Due to noise, $1$ is changed to $0$ during transmission with probability $0.2$
Due to noise, $1$ is changed to $2$ during transmission with probability $0.1$
Due to noise, $2$ is changed to $0$ during transmission with probability $0.2$
Due to noise, $2$ is changed to $1$ during transmission with probability $0.1$

Let A - the event that $\text{1 is received}$,
B - the event that $\text{1 is sent}$
I must find $P(B|A)$:
$$P(B|A) = \frac{P(A|B)P(B)}{P(A)}$$
$P(B)$ = 0.4

$P(A|B)$ = P(1 is sent) $\times$ P(1 does not change)
P(1 does not change) = [1 - P(1 becomes 0)]$\times$[1 - P(1 becomes 2)] (right?)
(1-0.2)(1-0.1) = 0.72
P(1 is sent) $\times$ 0.72 = (0.4)(0.72) = 0.288

P(A) = P(1 is sent and it stays 1) + P(0 is sent and it changes to 1) + P(2 is sent and it changes to 1)
P(A) = 0.288 + (0.3)(0.2) + (0.3)(0.1) = 0.378

Final answer:
$$\frac{0.288 \times 0.4}{0.378} = 0.30$$

Has this been done correctly?
 A: Let's start out simply:
The odds of a 0 being sent are 0.3.  The odds of a 1 being sent are 0.4.  The odds of a 2 being sent are 0.3.
The odds of a 1 being received given that a 0 was sent are 0.2.  Out of the total sample space, that's odds of 0.3 * 0.2 = 0.06.
Likewise, odds of a 2 being sent and a 1 being received (out of total sample space) is 0.3 * 0.1 = 0.03.
Total odds (out of total sample space) of a 1 being received when something else was sent is then 0.03 + 0.06 = 0.09.
Odds of a 1 being sent and received is 0.4 * 0.7 = 0.28.  (Note: Given that a 1 was sent, odds of a 0 being received are 0.2.  Given that a 1 was sent, odds of a 2 being received are 0.1.  Thus, given that a 1 was sent, odds of a 1 being received are 0.7.)
So given that a 1 was received, the odds that it was actually a 1 that was sent are 0.28 / (0.28 + 0.09) = 0.28 / 0.37 = 28/37 $\approx$ 0.757.

You can visualize the total sample space, proportionately, like so:

With that picture, all you have to do is count squares to see that the answer is 28/37.  :)
A: Let $S=k$ be the event that $k$ was sent and let $R=k$ be the event that
$k$ was received.
You want to compute $P[S=1|R=1] = {P[S=1 \text{ and } R=1] \over P[R=1]}$.
$P[R=1] = \sum_k P[R=1|S=k] P[S=k] = 0.3 \times 0.2 + 0.4 \times 0.7 + 0.3 \times 0.1 $.
$P[S=1 \text{ and } R=1]  = P[R=1|S=1] P[S=1] = 0.4 \times 0.7$.
Hence $P[S=1|R=1] = {28 \over 37}$.
A: By letting $S,R$ be the value Sent and Received, then we are given, and can immediately infer, that: $$\begin{align}\newcommand{\P}{\operatorname {\mathsf P}}
&\P(S{=}0)=0.3, \P(S{=}1)=0.4, \P(S{=}2)=0.3 \\
&\P(R{=}1\mid S{=}0)=0.2, \P(R{=}2\mid S{=}0)=0.1 &\implies \P(R{=}0\mid S{=}0)=0.7 \\ 
&\P(R{=}0\mid S{=}1)=0.2, \P(R{=}2\mid S{=}1)=0.1 &\implies \P(R{=}1\mid S{=}1)=0.7 \\ 
&\P(R{=}0\mid S{=}2)=0.2, \P(R{=}1\mid S{=}2)=0.1 &\implies \P(R{=}2\mid S{=}2)=0.7 \\ 
\end{align}$$
Then by Bayes' Rule, with the Law of Total Probability: $$\P(S{=}1\mid R{=}1) = \dfrac{\P(R{=}1\mid S{=}1)\P(S{=}1)}{\P(R{=}1\mid S{=}0)\P(S{=}0){+}\P(R{=}1\mid S{=}1)\P(S{=}1){+}\P(R{=}1\mid S{=}2)\P(S{=}2)}$$
The rest is just substitution of values.

Why does $\P(R{=}1\mid S{=}1)=0.7$ and so forth?   It is because by the definition of probability (which is also applicable to conditional probabilities): $$\P(R{=}0\mid S{=}1)+\P(R{=}1\mid S{=}1)+\P(R{=}2\mid S{=}1)~=~1 \\[2ex] \therefore~\P(R{=}1\mid S{=}1)~=~1-0.1-0.2$$
A: P(1 does not change) = [1 - P(1 becomes 0)]$\times$[1 - P(1 becomes 2)] (right?)
This assumes that P(1 becomes 0) and P(1 becomes 2) are independent events.
Consider what happens if 1 is transformed to 0 with probability 0.5, and 1 is transformed to 2 with probability 0.5 - what is the probability that 1 is correctly received? 
The correct calculation is
P(1 does not change) = 1 - [(P(1 becomes 0) + P(1 becomes 2)] 
This because received 1 and received 2 are mutually exlusive.
