This question is a continuation from this problem. Let's say I want to calculate the probability of the number of units between 1 faulty unit to the next is at least $Y$. How exactly would I calculate this? From a combinatorics stand point, if I was asked how many ways can I get exactly 2 heads out of 4 flips it'd be easy as saying $4\choose2$ and if it was at least 2 heads, then I would do $P(2H) + P(3H) + P(4H)$. However, I am not sure how to apply this concept to my question. How should I proceed?

EDIT: The answer to this question is leveraging the formula: $1-(1-e^{\lambda})$. I don't quite understand the intuitive explanation behind this formula.


1 Answer 1


If each unit has probability $p = \frac{1}{500}$ of being a failure, then the number of non-faulty units we see before obtaining a failure can be modeled as a Geometric random variable. If $X \sim geometric(p)$, then the probability mass function for X is:

$P(X=x) = p(1-p)^x, \ for \ x = 0,1,2,3,...$

Also notice that $P(X>y) = 1 - F(y)$, where F is the geometric CDF. Thus:

$P(X>y) = 1 - (1 - (1-p)^y) = (1-p)^y$

UPDATE: (In response to OP's update)

First of all, the answer you provided only makes sense if there is a y in the equation. I'm fairly sure you meant to put $1 - (1- e^{-\lambda y})$. And if this is the case, then I can offer you some intuition.

The Geometric Distribution is a discrete random variable expressing "exponential" probability decay. There is of course a continuous version of this distribution as well, which is called the Exponential distribution with parameter $\lambda$. In theory, using either distribution should give you similar results (although the discrete version will be more exact in our case).

Above, I put the formula $P(X>y) = 1 - F(y)$. If we use the CDF of the exponential distribution instead, then we get:

$P(X>y) = 1 - (1 - e^{-\lambda y}) = e^{-\lambda y}$.

Which is precisely what you had above. Just set $\lambda$ to be the rate at which we see failures ($\lambda = p = 1/500$).

Suppose you want to know $P(X>50)$.

GEOMETRIC: $P(X>50) = .90484$

EXPONENTIAL: $P(X>50) = .90475$

  • $\begingroup$ Hmm, it looks i have typed my question in correctly. What I meant to say was, the probability of getting exactly $X$. That's one event in and of it self. Then I would also like to learn how to get the probability of $Y$. Not X given Y. I will edit my questin, sorry about that. $\endgroup$
    – Jonathan
    Commented Nov 18, 2016 at 23:17
  • $\begingroup$ So you're looking for two seperate probabilities? $\endgroup$
    – knrumsey
    Commented Nov 18, 2016 at 23:20
  • $\begingroup$ Right! Sorry about not making that clear... $\endgroup$
    – Jonathan
    Commented Nov 18, 2016 at 23:21
  • $\begingroup$ That makes it a little bit simpler.. Let me update the answer. $\endgroup$
    – knrumsey
    Commented Nov 18, 2016 at 23:21
  • 1
    $\begingroup$ Geometric models discrete data. When you are counting number of parts before failure. Exponential models continuous data. For example the amount of time it takes to see a failure. So technically Geometric is more appropriate here. But we see that exponential is doing basically the same thing. $\endgroup$
    – knrumsey
    Commented Nov 28, 2016 at 2:27

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