# A binary digit is chosen at random to be sent through a transmission channel. $"0"$ is chosen with probability $0.4$ and $"1"$ is chosen with

A binary digit is chosen at random to be sent through a transmission channel. $$"0"$$ is chosen with probability $$0.4$$ and $$"1"$$ is chosen with probability $$0.6$$. The communication channel is noisy so that a $$"0"$$ is distorted by a $$"1"$$ with probability $$0.2$$ and a $$"1"$$ is distorted by a $$"0"$$ with probability $$0.1$$. Find the probability that

$$(a)$$ A $$"0"$$ is received.

$$(b)$$ A $$"1"$$ is received.

$$(c)$$ A $$"0"$$ was sent, since a $$"0"$$ was received.

$$(d)$$ A $$"1"$$ was sent, since a $$"1"$$ was received.

Attempt

$$R_0$$ is event that a zero is received

$$T_0$$ is event that a zero is transmitted

$$R_1$$ is event that a one is received

$$T_1$$ is the event that a one is transmitted,

With the information of the problem, we have $$P(R_0)=0.1, P(T_0)=0.4, P(R_1)=0.2, P(T_1)=0.6$$

I believe that $$(a)$$ and $$(b)$$ follow from this. For $$(c)$$, what is to be calculated is

$$P(T_0|R_0)=\frac{P(T_0)P(R_0|T_0)}{P(R_0)}$$

I don't know if I have defined the events correctly, or if the way I have been thinking about it is correct.

The information problem is $$P(T_0)=0.4$$, $$P(T_1)=0.6$$, $$P(R_1|T_0)=0.2$$ and $$P(R_0|T_1)=0.1$$. Then:
(a). $$P(R_0)=P(R_0|T_0)P(T_0)+P(R_0|T_1)P(T_1)=(1-0.2)\times 0.4+0.1\times 0.6=0.38$$
(b). $$P(R_1)=1-P(R_0)=0.62$$
(c). $$P(T_0|R_0)=\frac{P(R_0|T_0)P(T_0)}{P(R_0)}=\frac{(1-0.2)\times 0.4}{0.38}=\frac{16}{19}$$
(d). $$P(T_1|R_1)=\frac{P(R_1|T_1)P(T_1)}{P(R_1)}=\frac{(1-0.1)\times 0.6}{0.62}=\frac{27}{31}$$