Question

Suppose that 10% of all bits transmitted through a digital communication channel are erroneously received and that whether any particular bit is erroneously received is independent of whether any other bit is erroneously received. Consider sending a very large number of messages, each consisting of 20 bits.

a. What proportion of these messages will have at most 2 erroneously received bits?

b. What proportion of these messages will have at least 5 erroneously received bits?

c. For what proportion of these messages will more than half the bits be erroneously received?

Answer 1

a) 0.677

b) 0.0431

c) 0

Explanation:

For each bit, there are only two possible outcomes. Either it is correctly received, or it is erroneously received. Whether any particular bit is erroneously received is independent of whether any other bit is erroneously received. This means that we use the binomial probability distribution 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.

10% of all bits transmitted through a digital communication channel are erroneously

This means that
`p = 0.1`

Messages of 20 bits:

This means that
`n = 20`

a. What proportion of these messages will have at most 2 erroneously received bits?

This is
`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)`. So

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

`P(X = 0) = C_(20,0).(0.1)^(0).(0.9)^(20) = 0.1216`

`P(X = 1) = C_(20,1).(0.1)^(1).(0.9)^(19) = 0.2702`

`P(X = 2) = C_(20,2).(0.1)^(2).(0.9)^(18) = 0.2852`

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1216 + 0.2702 + 0.2852 = 0.677`

b. What proportion of these messages will have at least 5 erroneously received bits?

This is

`P(X \geq 5) = 1 - P(X < 5)`

In which

`P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)`. So

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

`P(X = 0) = C_(20,0).(0.1)^(0).(0.9)^(20) = 0.1216`

`P(X = 1) = C_(20,1).(0.1)^(1).(0.9)^(19) = 0.2702`

`P(X = 2) = C_(20,2).(0.1)^(2).(0.9)^(18) = 0.2852`

`P(X = 3) = C_(20,3).(0.1)^(3).(0.9)^(17) = 0.1901`

`P(X = 4) = C_(20,4).(0.1)^(4).(0.9)^(16) = 0.0898`

`P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1216 + 0.2702 + 0.2852 + 0.1901 + 0.0898 = 0.9569`

Then

`P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9569 = 0.0431`

c. For what proportion of these messages will more than half the bits be erroneously received?

This is
`P(X \geq 11)`

`P(X = 11) = C_(20,11).(0.1)^(11).(0.9)^(9) \approx 0`

If 11 is approximately 0, the rest(11, 12, 13,...) will also be approximately 0. So this proportion is 0.