Respuesta :
Answer:
a) 11.40% probability of 5 bits being in error during the transmission of 1 kb
b) 11.60% probability of 8 bits being in error during the transmission of 2 kb
c) 0.01% probability of no error bits in 3kb
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the giveninterval.
Poisson distribution with mean of 3.2 bits/kb (per kilobyte).
This means that [tex]\mu = 3.2kb[/tex], in which kb is the number of kilobytes.
(a) What is the probability of 5 bits being in error during the transmission of 1 kb?
This is P(X = 5) when [tex]\mu = 3.2[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 5) = \frac{e^{-3.2}*3.2^{5}}{(5)!} = 0.1140[/tex]
11.40% probability of 5 bits being in error during the transmission of 1 kb
(b) What is the probability of 8 bits being in error during the transmission of 2 kb?
This is P(X = 8) when [tex]\mu = 2*3.2 = 6.4[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 8) = \frac{e^{-6.4}*6.4^{8}}{(8)!} = 0.1160[/tex]
11.60% probability of 8 bits being in error during the transmission of 2 kb
(c) What is the probability of no error bits in 3kb?
This is P(X = 0) when [tex]\mu = 3*3.2 = 9.6[/tex]
Then
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-9.6}*9.6^{0}}{(0)!} = 0.0001[/tex]
0.01% probability of no error bits in 3kb
