Adding probabilities (or percentages)

Factual Questions
1

What seemed at first to be a simple problem has started to seriously melt my brain.

I have an event “A” that has a 3% chance of occurring evertime I press a button. Each button press means an independent 3% of event “A” occuring.

If I press the button 20 times what are the chances that event “A” will occur?

Obviously you cannot just add the 3% chance one after the other else eventually you would get a 100% chance of event “A” happening while in reality it is possible a million button presses may not result in event “A” happening. Was looking up probability stuff and seemed to find an answer to every possible probability question there is except this one. Can only assume I am missing something so absurdly easy no one bothers to write about it.

Feel free to change the example as you see fit in an explanation (e.g. die roll …after 20 throws what are the chances a “1” will appear).

2

I’m a little confused here, if event A occurs on the second button press, do you continue to press the button or do you stop?
And then what, if you still do all twenty rolls are you adding up the times Event A happened or is it a yes/no thing?

I’ve got my Stats book here, but I think I’ll need more clarification to figure out where to look.

3

You are looking for the chance that it will occur sometime during the 20 presses at least once correct?

You mean that there is a 3% chance of being what you want each time the button is pressed. Not a 3% chance that the button will always be the result A.

4

Turn it around. there is a 97% chance of it not happening. Multiply by 0.97 for each trial; e.g., in three trials there is a .97*.97*.97 chance of it not happening or 100% - 91.27% = 8.73% chance of it happening.

5

That’s what I was looking for.

Thanks.

And to the others I would stop once the event occurred. I just wanted to know how my chances increased on each button press.

6

[sarcasm]Dammit, you made that way easier then it needed to be, I’m sitting here with my stats book trying to remember my probability class from 7 years ago waiting for answers to my questions and you pull that ‘do it backwards crap’ your worse then my stats teacher.[/sarcasm]

7

I messed up the difference I’m trying to explain and I can’t get the wording correct so nevermind for now.

8

To finish the problem:

DanBlather writes:

> Turn it around. there is a 97% chance of it not happening. Multiply by 0.97 for
> each trial; e.g., in three trials there is a .97*.97*.97 chance of it not happening
> or 100% - 91.27% = 8.73% chance of it happening.

The OP asked about what would happen if the button was pushed 20 times. There’s a .97 probability that it won’t happen if the button was pushed once, so there’s a .97 ** 20 = approximately .54 probability that it won’t happen if the button is pushed 20 times. So there’s approximately a .46 probability that it will happen at least once.

9

Also, as a rule of thumb for solving probability problems …

If the problem as stated “makes your brain melt”, turn it inside out. e.g. turn “What is the probability of it happening.” into “What is the probability of it NOT happening.”

That often makes it much less melty.

10

Your chances didn’t increase on each button press. You were still a 97% loser each time. Check out the law of large numbers.

11

Yeah, your chances with each press are always 3%. But if you are given 20 goes, then before you start those twenty goes you have a relatively high chance of success. In fact, your chances decrease with each unsuccessful press as the number of tries remaining diminishes.

12

The probability that the first success happens on the nth press of the button is .97[sup]n - 1[/sup] * .03, and the probability that it happens within 20 presses is a little less than .46, which agrees with what Wendell posted. This is an instance of the first form listed of the geometric distribution.

13

Dan_Blather and Wendell Wagner have already given the correct answer, but here’s another approach (albeit one that’s harder to calculate):

You can only add probabilities if the events in question are mutually exclusive.

It would be possible to answer the question by adding together:
the probability that A occurs on the first press

  • the prob that A doesn’t occur on the first press but does on the second
  • the prob that A doesn’t occur on either of the first two presses but does on the third
  • …
  • the prob that A doesn’t occur on any of the first 19 presses but does occur on the 20th.

This would be 0.03 + 0.97*0.03 + 0.97[sup]2[/sup]*0.03 + … + 0.97[sup]19[/sup]*0.03.

This adds up to about 0.4562, which agrees with the other answer.