Example 6.12. Two signals $A$ and $B$ are transmitted over a communication line with probabilities 0.72 and 0.28, respectively. Due to interference, $\frac{1}{6}$ of the $A$-signals are distorted and received as $B$-signals, and $\frac{1}{7}$ of the transmitted $B$-signals are received as $A$-signals.

a) Determine the probability that an $A$-signal will be received at the receiving point.

b) Given that an $A$-signal has been received, what is the probability that it was also transmitted?
## Solution:

a) Let event $A$ be the appearance of an $A$-signal at the receiving point. Introduce hypotheses: $H_{A}$ - signal $A$ is transmitted, $H_{B}$ - signal $B$ is transmitted. According to the problem, $P\left(H_{A}\right)=0.72 ; P\left(H_{B}\right)=0.28$.

The probability that an $A$-signal is received given that it was sent is:

$$
P\left(A / H_{A}\right)=1-\frac{1}{6}=\frac{5}{6}
$$

The probability that an $A$-signal is received given that a $B$-signal was sent is:

$$
P\left(A / H_{B}\right)=\frac{1}{7}
$$

From this, by the formula of total probability, we get

$$
P(A)=P\left(H_{A}\right) \cdot P\left(A / H_{A}\right)+P\left(H_{B}\right) \cdot P\left(A / H_{B}\right)=0.72 \cdot \frac{5}{6}+0.28 \cdot \frac{1}{7}=0.64
$$

b) The probability of receiving an $A$-signal given that it was transmitted, we find using Bayes' formula:

$$
P\left(H_{A} / A\right)=\frac{P\left(H_{A}\right) \cdot P\left(A / H_{A}\right)}{P(A)}=\frac{0.72 \cdot \frac{5}{6}}{0.64}=\frac{0.6}{0.64}
$$