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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.

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1 Answer 1

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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). $P(T_1|R_1)=\frac{P(R_1|T_1)P(T_1)}{P(R_1)}=\frac{(1-0.1)\times 0.6}{0.62}=\frac{27}{31}$

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