Respuesta :
Answer:
0.0296 = 2.96% probability that at least one bit is received incorrectly.
Step-by-step explanation:
For each bit, there are only two possible outcomes. Either it is received correctly, or it its not. Each bit is received independently of the others, which 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The probability of each bit being received incorrectly is 0.001
This means that [tex]p = 0.001[/tex]
30-bit message
This means that [tex]n = 30[/tex]
What is the probability that at least one bit is received incorrectly?
This is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{30,0}.(0.001)^{0}.(0.999)^{30} = 0.9704[/tex]
Then
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9704 = 0.0296[/tex]
0.0296 = 2.96% probability that at least one bit is received incorrectly.
