I am having some trouble understanding why the answer to this problem is e^(-2) and not just 1.
Problem 17
The Model chosen for a discrete, integer-valued, non-negative random variable N with mean 2 is the binomial distribution with n trials and probability of success p on each trial. Various combinations of n and p are considered and P[N=0] is calculated. Find the limit as n --> infinite P[N=0].
A. e^(-2)
B. e^(-1)
C. 0
D. (1/2)
E. 1
I got to the point where E[X] = 2 = np so we get that p=2/n. This means that q=(1-(2/n)). When plugged into the Binomial for N=0, we get (1-(2/n))^n.
Where it disconnects for me is why they feel the need to apply the log to this and not just solve as n goes to infinite?
Thanks to all in advance for the help.
Problem 17
The Model chosen for a discrete, integer-valued, non-negative random variable N with mean 2 is the binomial distribution with n trials and probability of success p on each trial. Various combinations of n and p are considered and P[N=0] is calculated. Find the limit as n --> infinite P[N=0].
A. e^(-2)
B. e^(-1)
C. 0
D. (1/2)
E. 1
I got to the point where E[X] = 2 = np so we get that p=2/n. This means that q=(1-(2/n)). When plugged into the Binomial for N=0, we get (1-(2/n))^n.
Where it disconnects for me is why they feel the need to apply the log to this and not just solve as n goes to infinite?
Thanks to all in advance for the help.
Acted Practice Exam-5 Problem #17