Engineering Statistics: Binomial Distribution

In summary: This makes sense, because as k gets larger and larger, the number of available digits (0,1,2) gets smaller and smaller, making it less likely for none of them to appear. In summary, the probability of a k-digit number containing at least one 0, one 1, and one 2 is (7/9)^1 * (7/10)^{k-1} where k >= 3 is an integer.
  • #1
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Homework Statement



Let k >= 3 be any integer. What is the probability that a random k-digit number will have at least one 0, one 1 and one 2? (as usual every number starts with either 1,2,...9 and NOT zero)

Homework Equations



b(x : n,p) = (n x)p^x*(1-p)^(n-x) where x = 0, 1, 2, ... ,n
0 otherwise

This equation represents a binomial probability distribution. The outcome is either a success or a failure. In this case, a success would be a number containing at least a 0, 1 or 2. A failure would be otherwise.

The Attempt at a Solution



I'm not sure how to apply this equation in this situation, especially with the constraint put on k. I know that the number of "trials" must be 1. I also know that the outcome is either a success or a failure. I'm assuming that I'm looking for p in the above equation, which is the probability of success. Perhaps I'm missing something? I'm not sure, I've been attempting this for a while now with no luck.

Any advice to point me in the right direction would be great. Thanks
 
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  • #2
First of all, I don't like "at least" questions, I prefer them formulated as "at most". So you could use the complement rule to reformulate the question to "What is the probability that the chosen integer will contain no 0, 1 or 2".

Hint: consider the first digit and the remaining (k - 1) digits separately. Then build the k-digit number by randomly choosing from the set {1, 2, ..., 9} for the first digit and {0, 1, ..., 9} for the second one. Look at the event "a chosen numebr is not 0, 1 or 2".
 
  • #3
First of all, thanks CompuChip, I'm getting closer at an answer.
This is what I've come up with so far:

P(1st digit not being 1 or 2) = 7/9
P(2nd digit not being 0,1 or 2) = 7/10
P(3rd '' '' '' '' ) = 7/10
.
.
.
P(Kth " " " " ) = 7/10

In this set, at MOST 2/k digits can be 0,1, or 2

I'm not sure how to formulate this however. My intuition tells me that eventually (at infinity) the probability of NOT having 0, 1 and 2 is zero. I'm having a real hard time grasping this.
 
  • #4
aeroguy77 said:
First of all, thanks CompuChip, I'm getting closer at an answer.
This is what I've come up with so far:

P(1st digit not being 1 or 2) = 7/9
P(2nd digit not being 0,1 or 2) = 7/10
P(3rd '' '' '' '' ) = 7/10
.
.
.
P(Kth " " " " ) = 7/10

Looks very good.
Now, what is the probability that all of these events happen at the same time.
That is, that if you choose a k-digit number, that none of the digits will be (0), 1 or 2.

Your intuition is correct by the way, if you take the probability you are about to calculate and take the limit as k goes to infinity, the probability goes to 0.
 

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