Question

Suppose we are sending a digital signal which is a string of 0s and 1s of length five. (Example stringsare 00101, 11000, 10101 are all 5 bit strings.) When we send the message, each bit (0 or 1) is sentindependently and there is some chance that the bit is corrupted. Namely, each time we send a 0 thereis a 5% chance that a 1 is received and each time we send a 1 there is a 5% chance a 0 is received.Suppose we send a message of length 5, what is the probability that an incorrect message is received

Answer 1

0.7738 = 77.38% probability that an incorrect message is received.

Explanation:

For each bit, there are only two possible outcomes. Either it is corrupted, or it is not. The probability of a bit being corrupted is independent of any other bit. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

5% probability a bit is sent incorrectly:

This means that
`p = 0.05`

Message of length 5

This means that
`n = 5`

What is the probability that an incorrect message is received?

This is the probability of at least one incorrect bit, which is:

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(5,0).(0.05)^(0).(0.95)^(5) = 0.7738`

0.7738 = 77.38% probability that an incorrect message is received.