Sending a dot or dash when signal might get switched: Bayes' rule question

Question

Suppose there is some machine at the other end of a wire that sends either a “dot”or “dash” to you. The machine chooses a dot with probability $3/7$ and a dash with probability $4/7$. But as the message travels down the wire, there is a $1/8$ probability that it becomes switched: a dot becomes a dash, or a dash becomes a dot.

My question is how can I calculate $P(\text{ Dot Received}\mid \text{Dot Sent})$

I can see that it will be $\frac{7}{8}$, because conditional on sending a dot there is a $\frac{1}{8}$ chance that we don't receive a dot.

However, if I want to calculate it using $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$, how can I know that $P(A\cap B) = \frac{3}{8}$. That is what is not obvious to me.

Thank you.

Answer 1

You don't calculate $\mathsf P(A\mid B)$ from $\mathsf P(A\cap B)$.   That is trying to run the race the wrong way.

Don't try to race to the starting line from the finish line when you are already at the start and don't know where the finish even is.   Follow the track.

You are given $\mathsf P(A^\complement\mid B)$ and $\mathsf P(B)$.   That tells you $\mathsf P(A\mid B)$ and from there you can calculate $\mathsf P(A\cap B)$.

$$\mathsf P(A\cap B) = \mathsf P(B)~\mathsf P(A\mid B)$$

Answer 2

If the question is to ask what is

$\mathsf P(\textrm{Dot sent} \mid \textrm{Dot received})$

then you need Bayesian theorem

$\mathsf P(\textrm{Dot sent} \mid \textrm{Dot received}) = \dfrac{\mathsf P(\textrm{Dot received} \mid \textrm{Dot sent})\cdot\mathrm P(\textrm{Dot sent})}{\mathsf P(\textrm{Dot received})}$

and the law of Total Probability

$\mathsf P(\textrm{Dot sent} \mid \textrm{Dot received}) = \tfrac{\mathsf P(\textrm{Dot received} \mid \textrm{Dot sent})\cdot\mathrm P(\textrm{Dot sent})}{\mathsf P(\textrm{Dot received} ~\mid~ \textrm{Dot sent})\cdot\mathrm P(\textrm{Dot sent})+\mathsf P(\textrm{Dot received} ~\mid~ \textrm{Dash sent})\cdot\mathrm P(\textrm{Dash sent})}$