Consider a binary communication system transmitting 0s and 1s. Each 0 or 1 is called a bit. Due to interference, there may be transmission errors: a 0 can be received as a 1 and, likewise, a 1 can be received as a 0. For a bit chosen at random in the message, we note the events:
$E _ { 0 }$: ``the bit sent is a 0'';
$E _ { 1 }$: ``the bit sent is a 1'';
$R _ { 0 }$: ``the bit received is a 0''
$R _ { 1 }$: ``the bit received is a 1''.
We know that: $p \left( E _ { 0 } \right) = 0{,}4 ; \quad p _ { E _ { 0 } } \left( R _ { 1 } \right) = 0{,}01 ; \quad p _ { E _ { 1 } } \left( R _ { 0 } \right) = 0{,}02$. Recall that the conditional probability of $A$ given $B$ is denoted $p _ { B } ( A )$.
The probability that the bit sent is a 0 and the bit received is a 0 is equal to: a. 0,99 b. 0,396 c. 0,01 d. 0,4
The probability $p \left( R _ { 0 } \right)$ is equal to: a. 0,99 b. 0,02 c. 0,408 d. 0,931
A value, approximated to the nearest thousandth, of the probability $p _ { R _ { 1 } } \left( E _ { 0 } \right)$ is equal to: a. 0,004 b. 0,001 c. 0,007 d. 0,010
The probability of the event ``there is a transmission error'' is equal to: a. 0,03 b. 0,016 c. 0,16 d. 0,015
A message of length eight bits is called a byte. It is admitted that the probability that a byte is transmitted without error is equal to 0,88.
10 bytes are transmitted successively in an independent manner. The probability, to $10 ^ { - 3 }$ near, that exactly 7 bytes are transmitted without error is equal to: a. 0,915 b. 0,109 c. 0,976 d. 0,085
10 bytes are transmitted successively in an independent manner. The probability that at least 1 byte is transmitted without error is equal to: a. $1 - 0{,}12 ^ { 10 }$ b. $0{,}12 ^ { 10 }$ c. $0{,}88 ^ { 10 }$ d. $1 - 0{,}88 ^ { 10 }$
Let $N$ be a natural integer. $N$ bytes are transmitted successively in an independent manner. Let $N _ { 0 }$ be the largest value of $N$ for which the probability that all $N$ bytes are transmitted without error is greater than or equal to 0,1. We can affirm that: a. $N _ { 0 } = 17$ b. $N _ { 0 } = 18$ c. $N _ { 0 } = 19$ d. $N _ { 0 } = 20$
Consider a binary communication system transmitting 0s and 1s. Each 0 or 1 is called a bit. Due to interference, there may be transmission errors: a 0 can be received as a 1 and, likewise, a 1 can be received as a 0. For a bit chosen at random in the message, we note the events:
\begin{itemize}
\item $E _ { 0 }$: ``the bit sent is a 0'';
\item $E _ { 1 }$: ``the bit sent is a 1'';
\item $R _ { 0 }$: ``the bit received is a 0''
\item $R _ { 1 }$: ``the bit received is a 1''.
\end{itemize}
We know that:\\
$p \left( E _ { 0 } \right) = 0{,}4 ; \quad p _ { E _ { 0 } } \left( R _ { 1 } \right) = 0{,}01 ; \quad p _ { E _ { 1 } } \left( R _ { 0 } \right) = 0{,}02$.\\
Recall that the conditional probability of $A$ given $B$ is denoted $p _ { B } ( A )$.
\begin{enumerate}
\item The probability that the bit sent is a 0 and the bit received is a 0 is equal to:\\
a. 0,99\\
b. 0,396\\
c. 0,01\\
d. 0,4
\item The probability $p \left( R _ { 0 } \right)$ is equal to:\\
a. 0,99\\
b. 0,02\\
c. 0,408\\
d. 0,931
\item A value, approximated to the nearest thousandth, of the probability $p _ { R _ { 1 } } \left( E _ { 0 } \right)$ is equal to:\\
a. 0,004\\
b. 0,001\\
c. 0,007\\
d. 0,010
\item The probability of the event ``there is a transmission error'' is equal to:\\
a. 0,03\\
b. 0,016\\
c. 0,16\\
d. 0,015
\end{enumerate}
A message of length eight bits is called a byte. It is admitted that the probability that a byte is transmitted without error is equal to 0,88.
\begin{enumerate}
\setcounter{enumi}{4}
\item 10 bytes are transmitted successively in an independent manner.
The probability, to $10 ^ { - 3 }$ near, that exactly 7 bytes are transmitted without error is equal to:\\
a. 0,915\\
b. 0,109\\
c. 0,976\\
d. 0,085
\item 10 bytes are transmitted successively in an independent manner.
The probability that at least 1 byte is transmitted without error is equal to:\\
a. $1 - 0{,}12 ^ { 10 }$\\
b. $0{,}12 ^ { 10 }$\\
c. $0{,}88 ^ { 10 }$\\
d. $1 - 0{,}88 ^ { 10 }$
\item Let $N$ be a natural integer. $N$ bytes are transmitted successively in an independent manner.\\
Let $N _ { 0 }$ be the largest value of $N$ for which the probability that all $N$ bytes are transmitted without error is greater than or equal to 0,1.\\
We can affirm that:\\
a. $N _ { 0 } = 17$\\
b. $N _ { 0 } = 18$\\
c. $N _ { 0 } = 19$\\
d. $N _ { 0 } = 20$
\end{enumerate}