Working with percentages

J

Jbalchi

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Mar 23, 2006
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Hello
I wasn't quite sure what to call the subject but hopefully this will do. I am looking for a method to figure out the chances of an event happening, that has a predefined chance of happening, when it's done multiple times. To calrify this..

Let's say I had a gun that had a 20% chance of firing, every time I pull the trigger. Is there a way to find out the chances of it happening if I were to pull it more than once? It always has a 20% chance of happening, but I want to know what the overall chances are if I pulled it 5 times, or 8 times.

The example may sound kind of strange but it'll give me the type of thing I'm looking for. I don't need a specific answer, I want to know how to figure it out so I could find the solution if it was 12% and I pull it 4 times, or if it was 47% and I pull it 3 times.

But just to make 1 problem, I will use the first example. If the gun has a 20% chance of firing, what are the chances it fires at least 1 time if I pull it 3 times.

If someone can come up with an answer to that specific problem and show me how they got their answer I'd appreciate it. Let me know if I should clarify it further. And keep in mind I only want to know the chances of it hapening at least once, not 2 or 3 times. (We will pretend that once it fires I won't be pulling the trigger again anyway). Thanks in advance for any help.
 
20% = .2
Pulling it once it is .2^1 = .2
(^ means to the power.)
Pulling it twice it is .2^2 = .04 that it fires both times.
Pulling it three times it is .2^3 = .008 that it fires all three times.
If you were interested in it's NOT firing you would use (1-.2) = .8.
Pulling it once it is .8^1 = .8 it does not.
Pulling it twice it is .8^2 = .64 that it fails both times.
Pulling it three times it is .8^3 = .512 that it fails all three times.
Pulling it three times it is .8^2*.2^1 = .128 that it fires exactly once in the three times.
Better get a new gun. :)
 
Thanks Gene,
Although that information is nice to know, it's just a small step from what I wanted. I don't want to know the chances of it firing once if I pulled the trigger 3 times. I am looking for the chances that it will fire at least one of the times.

Maybe an easier way to put it is, pulling the trigger gives a 20% chance I'd die. If I pull it 3 times what are the chances I'll die? Sorry for the seemingly violent message here lol, but it gives more of an idea because you can't really die twice. If I pull it once, there's a 20% chance I'll be dead, twice I'd have like 30% chance, 3 times maybe a 35.

I don't know the chances that's what I'm trying to find out, but those are estimates, to show I'm looking for an increasing value not a decreasing.

Once again, thanks for what you gave the first time, it's still nice information to know.
 
It seeem to me that my last line fits the problem. True you don't know which try it fired on but once is enough. If you are such a poor shot that you might need three "successes"...
Any how, you know the odds of it not firing. Firing at 1,2 or three times is one minus the odds of its not firing.
1-.512 = .488
 
Thanks, yes that's what I wanted to know. I knew the answer was probably in the help you gave me before but my head is hurting too bad to do much thinking now.

I'm off to bed, and thanks for the quick response on that.
 
Hello, Jbalchi!

We're talkin' Russian Roulette . . . right?

You want the probability that the gun fires for the first time on the \(\displaystyle n^{\text{th}}\) try?

Let \(\displaystyle F\) = fire, \(\displaystyle M\) = misfire.

Then, for example: \(\displaystyle \,P(F)\,=\,0.2,\;P(M)\,=\,0.8\)


The gun must misfire the first \(\displaystyle n-1\) tries and fire on the \(\displaystyle n^{\text{th}}\).

The probability is: \(\displaystyle \,(0.8)^{n-1}(0.2)\)
 
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