Conditional probability, noisy channel.

Question

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How could I go about this one? What we already know :

${P(R=1|S=1)+P(R=0|S=0)=\dfrac{1}{3}\\P(S=0) =\dfrac{6}{10}, P(S=1)=\dfrac{4}{10}}$

What I got is this:

${P(S =0|R=0) = \dfrac{P(S=0 \cap R=0)}{P(R=0)}= \dfrac{P(S=0)\times P(R=0|S=0)}{P(R=0)}}$ How do I find the values of the denominator and numerator or either side? They are all unknown, I tried to find them, but got stuck at my first expression. ${ }$

I tried to make a decision tree, but I think I failed miserably. I got that the answer was $\dfrac12\cdots$ Could I get some hints? Thanks!

Answer 1

The subtlety here is that you are told that the conditional probability of receiving a value that is incorrect is $1/3$. That is to say, given the true value of a signal, the probability it is incorrectly received is $1/3$; thus the question implies $$\Pr[R = 1 \mid S = 0] = \Pr[R = 0 \mid S = 1] = \frac{1}{3}.$$ It is poorly worded, no doubt, but this is the only way the question has a unique solution.

With this in mind, the computation is trivial: $$\begin{align*} Pr[S = 0 \mid R = 0] &= \frac{\Pr[R = 0 \mid S = 0]\Pr[S = 0]}{\Pr[R = 0]} \\ &= \frac{(1 - \Pr[R = 1 \mid S = 0])\Pr[S = 0]}{(1 - \Pr[R = 1 \mid S = 0])\Pr[S = 0] + \Pr[R = 0 \mid S = 1]\Pr[S = 1]} \\ &= \frac{(1 - 1/3)(6/10)}{(1 - 1/3)(6/10) + (1/3)(4/10)} \\ &= \frac{3}{4}. \end{align*}$$

Answer 2

The expression you have written:

$$P(S=0|R=0) = \frac{P(S=0)\cdot P(R=0|S=0)}{P(R=0)}$$

is correct. It is called Baye's theorem. To calculate $P(R=0|S=0)$, we know that: $$P(R=0|S=0) + P(R=1|S=0) = 1$$ To see why this is true, if it is already known that $S=0$, then the possibilities of $R$ are just two of them: $R=0$ or $R=1$. (This is also the case even if we did not "see" the value of $S$. $R$ in that case can only take values $1$ or $0$. The way it would differ from the case if we "saw" the value of $S$ is that the probabilities that we would see $R=0$ or $R=1$ would be different in the two cases). Since the sample space of the event, "Value of $R$ given that $S=0$" is just $\{0,1\}$, we have that the sum of the respective probabilities in the sample space must be 1.

Since the value of $P(R=1|S=0) = 1/3$, we have that $P(R=0|S=0)=2/3$.

Now, to find $P(R=0)$, use the following expression, which is called the law of total probability:

$$P(R=0) = P(R=0|S=0) \cdot P(S=0) + P(R=0|S=1)\cdot P(S=1)$$ $$ = \frac{2}{3} \cdot \frac{6}{10} + \frac{1}{3} \cdot \frac{4}{10}$$ Substituting the above computed values, the answer comes out to be $\frac{3}{4}$.