Calculating average number of events with dropping probability

HaloedBane

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[FONT=&quot]Hi there, first post here. I have a problem that’s drivin me crazy.[/FONT][FONT=&quot]
Say a particular event always has a 2% probability of occurring on each occasion, then the average number of such events occurring after 100 occasions is 2, right?[/FONT]

[FONT=&quot]How do I calculate it when the probability goes down each time an event occurs. For example: if the event has never happened, then the probability is 2%, but if it already occurred once, then the probability of it happening on each subsequent occasion drops to 1%, and if it's already occurred twice, then the probability of a third occurrence is 0.1%. (After 3 occurrences, the probability drops to 0).[/FONT]
[FONT=&quot]Given that data, how would I go about calculating the average number of such events occurring after 100 occasions? I know it’ll be less than 2.0, but I can’t figure out a formula.

Any help would be much appreciated![/FONT]
 

tkhunny

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Well, lets see if you're really up to this.

If \(\displaystyle p = 0.02\;and\;q = 1-p = 0.98\)

Please calculate and interpret this expression: \(\displaystyle \sum\limits_{n=1}^{100}n\cdot q^{n-1}\cdot p\)
 

HaloedBane

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I'm getting 30.1, but that looks wrong.
 

JeffM

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Hi there, first post here. I have a problem that’s drivin me crazy.
Say a particular event always has a 2% probability of occurring on each occasion, then the average number of such events occurring after 100 occasions is 2, right?

How do I calculate it when the probability goes down each time an event occurs. For example: if the event has never happened, then the probability is 2%, but if it already occurred once, then the probability of it happening on each subsequent occasion drops to 1%, and if it's already occurred twice, then the probability of a third occurrence is 0.1%. (After 3 occurrences, the probability drops to 0).
Given that data, how would I go about calculating the average number of such events occurring after 100 occasions? I know it’ll be less than 2.0, but I can’t figure out a formula.

Any help would be much appreciated!
Let's start with a simpler example.

Suppose there are 4 trials.

What is the probability of getting no successes?

Clearly, exactly \(\displaystyle 0.98^4.\) Do you buy that?

What is the probability of exactly one success? That is more complex. That probability is the sum of the probabilities of success only on first trial, only on second trial, only on third trial, only on fourth trial. So

\(\displaystyle 0.02 * 0.99^3 + (0.98 * 0.02 * 0.99^2) + (0.98^2 * 0.02 * 0.99) + (0.98^3 * 0.02) =\)

\(\displaystyle \displaystyle 0.02 * \left ( \sum_{j=0}^{4-1} 0.98^j * 0.99^{(4-1-j)} \right ).\)

How about two successes? Well, each term will have 0.02 and 0.01 as factors, simplifying down to 0.0002.

We can have success on trials 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, and 3 and 4. So the probability of exactly two successes in four trials is

\(\displaystyle 0.0002(0.999^2 + 0.99 * 0.999 + 0.99^2 + 0.98 * 0.999 + 0.98 * 0.99 + 0.98^2).\)

How about three successes? Each term will have 0.02, 0.01, and 0.001 as factors, simplifying down to 0.0000002.

We can have failure on trial 1, 2, 3, or 4. So the probability of exactly three successes is

\(\displaystyle 0.0000002(0.98 + 0.99 + 0.999 + 1).\)

Now you can compute the expected value the normal way.

I do not even want to think about doing this for 100 trials.

What I can say is that as the number of trials gets larger, the ratio of successes to trials will approach zero, and the expected number of successes will approach 3.
 
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HaloedBane

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Thanks for the guidance :)
 

tkhunny

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I'm getting 30.1, but that looks wrong.
You didn't do the "interpret" part. That's just the trials to the FIRST occurrence. How about the other two?
 
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