\section*{Problem 6A - 141236A}

A measurement complex \(M\) integrated into an industrial process transmits to a transmission unit \(A_{1}\) exactly one of the two signals \(S_{1}\) or \(S_{2}\), which is then transmitted from \(A_{1}\) to a transmission unit \(A_{2}\), from \(A_{2}\) to a transmission unit \(A_{3}\), and from \(A_{3}\) to an electronic computer \(R\).

Each transmission unit \(A_{i} (i=1,2,3)\) can transmit exactly the signals \(S_{1}\) or \(S_{2}\). The probability that \(A_{i}\) transmits the other signal instead of the one it received is 0.01.

It is now known that at the end of such a process, transmitted through \(A_{3}\) to the computer \(R\), the signal \(S_{1}\) was transmitted.

What is the probability that \(M\) transmitted \(S_{1}\) to \(A_{1}\) at the beginning of this process?

Hint:

If under certain conditions \(V\), in a large number \(n\) of cases, a total of \(g\) cases occur in which an event \(E\) occurs or has occurred, then the number \(p=\frac{g}{n}\) is the probability of the occurrence (or having occurred) of \(E\) under the conditions \(V\).

For the solution, the following theorems can also be used.

a) Addition rule of probability for independent events: The probability that one of two mutually exclusive events \(E_{1}\) and \(E_{2}\) occurs is equal to the sum \(p_{1}+p_{2}\) of the probability \(p_{1}\) of the occurrence of \(E_{1}\) and the probability \(p_{2}\) of the occurrence of \(E_{2}\).

b) Multiplication rule of probability: The probability that an event \(E\) and an event \(F\) occur is equal to the product \(p \cdot q\) of the probability \(p\) of the occurrence of \(E\) and the probability \(q\) that \(F\) occurs given \(E\).
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We assume that the events of erroneous transmission of the received signal for the three transmission units are stochastically independent of each other.

The output signal from \(M\) is received by \(R\) exactly as it is if either none of the transmission units transmit the received signal erroneously, or exactly two do.

The first case occurs with a probability of \((1-0.01)^{3}=0.99^{3}\).

For the second case, there are exactly three ways to choose the two transmission units that transmit erroneously, and for each of these possibilities, there is a corresponding transmission probability of \(0.99 \cdot 0.01^{2}\), resulting in a total probability of

\[
P=\frac{99^{3}+3 \cdot 99 \cdot 1^{2}}{100^{3}}=\frac{99 \cdot\left(99^{2}+3\right)}{10^{6}}=\frac{99 \cdot 9804}{10^{6}}=\frac{970596}{10^{6}}=0.970596=97.0596 \%
\]