Answer:
a) X can take the values {0,1,2,3,...,n}
The probability mass function of x is defined for k copies received correctly as:
[tex]P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}[/tex]
b) N should be N=2 to ensure that with probability 0.99 at least one copy of the message is received.
Step-by-step explanation:
The variable X:number of copies received correctly by the end user can be modeled as a binomial variable, as it is a sum of n Bernoulli variables with probability p.
X can take the values {0,1,2,3,...,n}
The probability mass function of x is defined for k copies received correctly as:
[tex]P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}[/tex]
being p: the probability that a packet is received correctly by an end user, and q=1-p: the probability that a packet is not received correctly by an end user.
b) We have to calculate n so as to have a probability P(x≥1)=0.99, when p=0.9.
[tex]P(x\geq1)=1-P(x=0)=0.99\\\\\\P(x=0)=\dbinom{n}{0}p^0(1-p)^n=1*1*(1-p)^n=(1-0.9)^n=0.1^n\\\\\\P(x\geq1)=1-0.1^n=0.99\\\\0.1^n=1-0.99=0.01\\\\n\cdot ln(0.1)=ln(0.01)\\\\n=ln(0.01)/ln(0.1)=2[/tex]
Answer:
Answer : N=2
Step-by-step explanation:
X: R.V denoting # copies received correctly,
so, X = Bin (N,P)
= P (X-x) = (N/x) p× (1-p)N-x x=0,1,2 .............
(b) p= 0.9
p(X, 7, 1) = 0.99
1- P (X=0) = 0.99
P(X=0) = 0.01
(0.01)N = 0.01
N=2
