SOLUTION: a) Plot the pmf and cdf of the geometric random variable X with probability of success p = 1/3. b) Find E[X] and var(X). c) Find P(X > 4)

Algebra ->  Probability-and-statistics -> SOLUTION: a) Plot the pmf and cdf of the geometric random variable X with probability of success p = 1/3. b) Find E[X] and var(X). c) Find P(X > 4)       Log On


   

Question 1164546: a) Plot the pmf and cdf of the geometric random variable X with probability of success p = 1/3.
b) Find E[X] and var(X).
c) Find P(X > 4)
Found 2 solutions by amarjeeth123, mccravyedwin:
Answer by amarjeeth123(574) About Me  (Show Source):
You can put this solution on YOUR website!
E[X]=1/p=1/(1/3)=3
Var[X]=((1-p)/p^2)=(2/3)/(1/9)=(2/3)*9=2*3=6
Since the number of trials is not given we cannot calculate the desired probability.

Answer by mccravyedwin(420) About Me  (Show Source):
You can put this solution on YOUR website!
This concerns getting the probablity that we will get our first success on the
kth trial, with a probability of success on each trial being p = 1/3.

The other tutor was correct for part b), which was: 

b)
E%28X%29=1%2Fp=%281%5E%22%22%29%2F%281%2F3%29=3
Var%28X%29=%28%281-p%29%2Fp%5E2%29=%282%2F3%29%2F%281%2F9%29=%282%2F3%29%2A9=2%2A3=6

a) Instead of plotting, I'll just list the values:
PMF(probability mass function) P%28X=k%29=%281-p%29%5E%28k-1%29%2Ap
CDF(cumulative distribution function) P%28X%3C=k%29=1-%281-p%29%5Ek

 k|PMF(k) |CDF(k)
 1|0.3333 |0.3333    
 2|0.2222 |0.5556
 3|0.1481 |0.7037
 4|0.0988 |0.8025
 5|0.0658 |0.8683
 6|0.0439 |0.9122
 7|0.0293 |0.9415
 8|0.0195 |0.9610
 9|0.0130 |0.9740
10|0.0087 |0.9827

c)
P%28X%3Ek%29=%281-p%29%5Ek  
P%28X%3E4%29=%281-1%2F3%29%5E4=%282%2F3%29%5E4=%282%2F3%29%5E4=16%2F81=0.1975 (approximately)
 
Edwin