# Probability problem! Does anyone know?

• Apr 7th 2008, 08:40 PM
GoDuke
Probability problem! Does anyone know?
I am unsure how to solve this problem: (Thinking)

A communications channel transmits the digits 1 and 0. However, due to static, there is a 1% chance probability that any digit transmitted will be incorrectly received. Suppose that we were to transmit an important message consisting of 10 digits. What is the probability that the message will contain an error? Give answer to four decimal places.
• Apr 7th 2008, 10:39 PM
roy_zhang
Quote:

Originally Posted by GoDuke
I am unsure how to solve this problem: (Thinking)

A communications channel transmits the digits 1 and 0. However, due to static, there is a 1% chance probability that any digit transmitted will be incorrectly received. Suppose that we were to transmit an important message consisting of 10 digits. What is the probability that the message will contain an error? Give answer to four decimal places.

You can find the probability that the message will contain an error by using 1-P(no error). From given we know that P(receive any digit correctly)=1-0.01=0.99, so the probability that the 10-digits message contains no error can be found as \$\displaystyle P(\mathtt{no\; error})=(.99)^{10}\$. Hence the probability that the message will contain an error is \$\displaystyle 1-(0.99)^{10}=0.0956\$

Roy
• Apr 8th 2008, 09:18 AM
GoDuke
More probability
Thanks Roy! I appreciate the help.
There is a continuation problem that I am also unsure how to complete:

To reduce the chance of error suppose that we transmit 000 instead of 0 and 111 instead of 1. If the receiver of the message uses "majority" decoding, what is the probablility that a message consisting of a single digit will be incorrrectly decoded? Suppose that we were to transmit an important message consisting of 10 digits.

What is the probabability that the message will contain an error?
• Apr 8th 2008, 10:43 AM
roy_zhang
Quote:

Originally Posted by GoDuke
There is a continuation problem that I am also unsure how to complete:
To reduce the chance of error suppose that we transmit 000 instead of 0 and 111 instead of 1. If the receiver of the message uses "majority" decoding, what is the probablility that a message consisting of a single digit will be incorrrectly decoded?

We need to know the meaning of majority decoding first, it is a method to decode repetition codes and it will pick the most frequently transmitted digit in the code as the digit to be transmitted. In our case, using 3 digits repetition codes, if we receive the codes for a single digit are 111, 110, 101 or 011, we will decode this single transmitted digit as 1 (since there are more 1s in each of the codes than 0s).

OK, suppose you want to transmit a message consisting a single digit 1, it will be incorrectly decoded (i.e. decoded as 0) if you receive the codes as 000, 001, 010 or 100 (since there are more 0s in each of the codes than 1s). Realize that the probabilities for each one of these four codes are:

\$\displaystyle P(000)=(0.01)^3;\;P(001)=P(010)=P(100)=(0.01)^2(0. 99)\$

hence the probability that a message consisting of a single digit will be incorrectly decoded is:

\$\displaystyle P(\mathtt{error})=P(000)+P(001)+P(010)+P(100)=(0.0 1)^3+3(0.01)^2(0.99)=0.000298\$

Quote:

Suppose that we were to transmit an important message consisting of 10 digits. What is the probabibility that the message will contain an error?
Since we found out P(receive any digit incorrectly)=0.000298, we have P(receive any digit correctly)=1-0.000298=0.999702. (Compare this with 0.99, we indeed reduced the chance of error!) Now you can follow the exactly procedure in our discussions of your first question to answer this part.

Roy
• Apr 8th 2008, 12:54 PM
TKHunny
Shall we assume the independence of the digits? I'll proceed on this assumption.

Does "an" error mean "at least one" error? I'll proceed on this assumption.

\$\displaystyle Pr(Error) = 1 - Pr(No Error) = 1 - (0.99)^{10}\$
• Apr 8th 2008, 02:13 PM
GoDuke
Thank you!
Wow. I understand it now. Thanks so much for all your help!(Hi)