Probability to get the correct message

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In summary: So the probability is : $$P(X\leq n−x−1)=\sum_{m=0}^{n-x−1}\binom{n}{m}W^m(1-W)^{n-m}$$ right? In summary, The technique being discussed is using repetition of symbols to improve the reliability of a noisy communication channel. Each symbol is transmitted multiple times and decoded using the rule of majority. The probability of a symbol being decoded correctly depends on the number of repetitions and the probability of correct transmission. The probability of the entire message being decoded correctly is found by taking the product of the probabilities of each symbol being transmitted correctly. To find the probability of not more than a certain number of symbols being altered, bin
  • #1
mathmari
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Hey! :giggle:

One of the techniques we are using at the digital communications to improve the reliability of a noisy communication channel, is to repeat a symbol many times.
For example, we can send each symbol $0$ or $1$ say three times. More precisely, applying the rule of majority, a $0$ (or $1$) is sent as $000$ (or $111$ respectively) and is decoded at the receiver with $0$ (or $1$) if and only if the received sequence of three symbols contains at least two $0$ (or $1$ respectively).

In this case we consider a message that is sent through a noisy communication channel and consists of $n$ symbols ($0$ or $1$). At the transmission each symbol can be altered (independent from the other ones) due to the noise. To increase the reliability of the transmission each symbol is repeated $r$ times. The probability of correct transmission of each symbol ($0$ or $1$) is $p$. Applying the rule of majority, a symbol $0$ (or $1$) is decoded correctly at the receiver if and only if the received sequence of $r$ symbols contains at least $k$ $\ 0$ (or $1$ respectively), with $k>\frac{r}{2}$.
Calculate:
(a) the probability a sent symbol to be decoded correctly.
(b) the probability the whole message to be decoded correctly.
(c) the probability to be altered not more than $x$ symbols of the message, with $x<n$.For (a) do we have to consider that at least $k$ symbols have to be decoded correctly with probability $p$ ? :unsure:
 
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  • #2
Hey mathmari!

I think we should look at just $1$ symbol to be sent and to be decoded.
When we send $1$ symbol, we transmit that symbol $r$ times.
We want to know the probability that the decoded symbol is the same as the sent symbol.
That is the case if at least $k$ of the $r$ transmitted symbols are transmitted correctly, which happens with probability $p$. 🤔
 
  • #3
Klaas van Aarsen said:
I think we should look at just $1$ symbol to be sent and to be decoded.
When we send $1$ symbol, we transmit that symbol $r$ times.
We want to know the probability that the decoded symbol is the same as the sent symbol.
That is the case if at least $k$ of the $r$ transmitted symbols are transmitted correctly, which happens with probability $p$. 🤔

Since each symbols independent from the other ones, if they will be altered or not, is the probability that at least $k$ symbols are transitted correctly equal to $$\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}$$ ? :unsure:
 
  • #4
mathmari said:
Since each symbols independent from the other ones, if they will be altered or not, is the probability that at least $k$ symbols are transitted correctly equal to $$\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}$$ ?
Yep. (Nod)
 
  • #5
Klaas van Aarsen said:
Yep. (Nod)

Great!

At (b) we want that all symbols are transmitted correctly, so the probability is then equal to $p^r$, or not? :unsure:
 
  • #6
mathmari said:
At (b) we want that all symbols are transmitted correctly, so the probability is then equal to $p^r$, or not?
Isn't that the probability that each transmission of a $1$ symbol was received correctly?
I think (b) asks for the probability that $n$ symbols are sent and decoded correctly. :unsure:
 
  • #7
Klaas van Aarsen said:
Isn't that the probability that each transmission of a $1$ symbol was received correctly?
I think (b) asks for the probability that $n$ symbols are sent and decoded correctly. :unsure:

Ahh so the probability that $1$ symbol was received correctly is $p^r$. Then is the probability that $n$ symbols are sent and decoded correctly equal to $\left (p^r\right )^n=p^{rn}$ ? :unsure:
 
  • #8
mathmari said:
Ahh so the probability that $1$ symbol was received correctly is $p^r$. Then is the probability that $n$ symbols are sent and decoded correctly equal to $\left (p^r\right )^n=p^{rn}$ ?
Didn't you find in (a) that the probability that $1$ symbol was sent and decoded correctly was $\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}$? :unsure:
 
  • #9
Klaas van Aarsen said:
Didn't you find in (a) that the probability that $1$ symbol was sent and decoded correctly was $\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}$? :unsure:

Ahh I was confused.. I thought that all symbols of the message have to be correctly transmitted, but for each symbol of the message it must be that at least $k$ symbols must be correct at the sequence, right?

The probability that one sent symbol is decoded correctly is equal to $\displaystyle{\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}}$.
We consider $n$ symbols of the message. So each of the $n$ symbols has the probability $\displaystyle{\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}}$ to be transimtted correctly. Since each symbol is independent from each other we take the product of the probabilities, i.e. $\displaystyle{\left (\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}\right )^n}$.

Is that correct? :unsure:

 
  • #10
mathmari said:
$\displaystyle{\left (\sum_{i=k}^r\binom{r}{i}p^i(1-p)^{r-i}\right )^n}$.

Is that correct?
Yep. (Nod)
 
  • #11
Klaas van Aarsen said:
Yep. (Nod)

Great!

For (c) do we consider binomial distribution?

Let X be the number of symbols that are correctly transmitted.
Then we have that $$P(X\geq n−x)=1−P(X<n−x)=1−\sum_{m=0}^{n-x}\binom{n}{m}W^m(1-W)^{n-m}$$ where $W$ is the probability we calculated at (a).
And this probability is equal to the probability that not more than $x$ symbols of the message will be altered.

Is that correct? :unsure:
 
  • #12
mathmari said:
Let X be the number of symbols that are correctly transmitted.
Then we have that $$P(X\geq n−x)=1−P(X<n−x)=1−\sum_{m=0}^{n-x}\binom{n}{m}W^m(1-W)^{n-m}$$ where $W$ is the probability we calculated at (a).
And this probability is equal to the probability that not more than $x$ symbols of the message will be altered.
Binomial distribution sound good.

But don't we have $P(X<n−x)=P(X\le n−x-1)=\sum\limits_{m=0}^{n-x-1}\binom{n}{m}W^m(1-W)^{n-m}$? :unsure:
 
  • #13
Klaas van Aarsen said:
Binomial distribution sound good.

But don't we have $P(X<n−x)=P(X\le n−x-1)=\sum\limits_{m=0}^{n-x-1}\binom{n}{m}W^m(1-W)^{n-m}$? :unsure:

Ahh because we miss the equality :oops:
 
  • #14
But we could write it also as follows, or not?
$$ \sum\limits_{m=0}^{x}\binom{n}{m}W^{n-m}(1-W)^{m} $$ which is equal to the probability that at most $x$ from $n$ symbols are wrongly transmitted (altered), or isn't that equivalent to the formula above? :unsure:
 
  • #15
mathmari said:
But we could write it also as follows, or not?
$$ \sum\limits_{m=0}^{x}\binom{n}{m}W^{n-m}(1-W)^{m} $$ which is equal to the probability that at most $x$ from $n$ symbols are wrongly transmitted (altered), or isn't that equivalent to the formula above?
I believe that is equivalent yes. (Nod)
 
  • #16
Klaas van Aarsen said:
I believe that is equivalent yes. (Nod)

Great! Many thanks! (Malthe)
 

Related to Probability to get the correct message

What is probability and how does it relate to getting the correct message?

Probability is a measure of the likelihood of an event occurring. In the context of getting the correct message, probability refers to the chance that a message will be received and understood correctly by the intended recipient.

How is the probability of getting the correct message calculated?

The probability of getting the correct message can be calculated by dividing the number of successful outcomes (receiving and understanding the message) by the total number of possible outcomes.

What factors can affect the probability of getting the correct message?

There are several factors that can affect the probability of getting the correct message, such as the clarity of the message, the communication channel used, the language and cultural barriers, and the receiver's level of attention and understanding.

Can the probability of getting the correct message be increased?

Yes, the probability of getting the correct message can be increased by using clear and concise language, choosing an appropriate communication channel, and addressing any potential barriers or distractions that may affect the receiver's understanding.

Why is it important to consider the probability of getting the correct message in scientific research?

In scientific research, clear and accurate communication is crucial for the dissemination and understanding of results. Considering the probability of getting the correct message ensures that the intended information is conveyed accurately, reducing the risk of misinterpretation and potential errors in data analysis and conclusions.

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