Probability - random variables, poisson/binomial distributions

  • Thread starter Kate2010
  • Start date
  • #1
146
0

Homework Statement


A text file contains 1000 characters. When the file is sent by email from one machine to another, each character (independent of other characters) has probability 0.001 of being corrupted. Use a poisson random variable to estimate the probability that the file is transferred with no errors. Compare this to the answer you get when modelling the number of errors as a binomial random variable.


Homework Equations


p_x(k) = (a^k)(e^-a)/(k!), poisson
p_x(k) = (n choose k)(p^k)(q^{n-k}), binomial


The Attempt at a Solution


Poisson:
Let x be the number of corrupted characters.
E(X) = 0.001 = a
P(X=n) = (0.001^n)(e^-0.001)/(n!)
P(X=0) = e^-0.001

Binomial:
E(x) = np = 1000 x 0.001 = 1

I don't really think I'm tackling either of these problems in the correct way but don't know what else to do. Thanks for any help.
 

Answers and Replies

  • #2
42
0
I don't think you have to worry about expected values here. When you model the number of errors with a binomial distribution, you want to find the probability of 0 "successes." "Success" in this case means a character being corrupted, which happens with probability .001. So you want to find p(k=0) using the formula

[tex] p(k)={n \choose k} p^k (1-p)^{n-k} [/tex].

For the poisson distribution, notice here that n is large and p small. When this is the case, a binomial distribution can be approximated by poission with parameter [tex]\lambda =np.[/tex]
 
  • #3
146
0
I'm still a bit confused, when I use poisson I get P(X=n) = (0.001^n)(e^-0.001)/(n!), so P(X=0) is e^-0.001 which is 0.99999 recurring. However, when I try it as you suggested, again using poisson but with np instead I get e^-1 which is about 0.37.
 
  • #4
42
0
I'm still a bit confused, when I use poisson I get P(X=n) = (0.001^n)(e^-0.001)/(n!), so P(X=0) is e^-0.001 which is 0.99999 recurring. However, when I try it as you suggested, again using poisson but with np instead I get e^-1 which is about 0.37.
Apologies for the (very) late response.

When you write P(X=n) = (0.001^n)(e^-0.001)/(n!), you are saying that the Poisson parameter is .001, which is not true. In this problem, the Poisson parameter is supposed to be approximated by np, as mentioned above.

The value .001 is used for modeling the number of errors as a binomial random variable, that is, the probability of no "successes" (i.e. no characters corrupted) is

[tex] P(X=0)=p(0)={n\choose 0} .001^0 (1-.001)^{n-0} [/tex]

where n=1000, the number of trials.
 
  • #5
146
0
Thanks, I understand now :)
 

Related Threads on Probability - random variables, poisson/binomial distributions

  • Last Post
Replies
5
Views
3K
Replies
2
Views
3K
Replies
6
Views
2K
Replies
10
Views
202
Replies
3
Views
1K
Replies
7
Views
759
Replies
3
Views
2K
Replies
2
Views
7K
Replies
10
Views
14K
Top