First we need to assume each mole is independent? So if one becomes malignant then the probability goes up for the other moles.
If a certain type of skin mole has 1% probability of malignant change over a lifetime, what is the probability of malignant change if a person has 1000 such moles? I am a veterinarian. I took Intro. Statistics 101/102 as an undergraduate around 1980. I seem to have forgotten some basic concepts! Thanks.
First we need to assume each mole is independent? So if one becomes malignant then the probability goes up for the other moles.
Stop cowardice, ban guns!
No, let's not make that assumption. Let's say each mole acts independently. What I'm getting at is: By virtue of having 1000 of these moles on your body, do you have greater than a 1% chance over your lifetime that you will develop a malignant mole? Thanks.
I had a nice reply written and then hit the back arrow and lost it. Big picture, you are looking at a binomial distribution, since you have a binary outcome. You can think of your problem as repeated independent trials, such as flipping an unfair coin (1% chance of heads) and looking for at least a heads in 1000 flips, or exactly 1 case, or greater than 0 cases. It all depends on your hypothesis.
Stop cowardice, ban guns!
And thus...? Let's take one of your models and go towards a solution. What do I take away from this? That the chance of malignancy occurring in the person with 1000 independent moles is somewhere north of 1% over his/her lifetime? Thank you.
I thought you may like to investigate it first, and then try to get the answer.
I believe the answer is: 0.9999568
Which I just took the probability of getting exactly 0 heads and subtracted it from 1.0. So this is the probability of 1 or more moles being malignant.
Some extra information on it: http://msemac.redwoods.edu/~darnold/...binomial1.html
Stop cowardice, ban guns!
Ronald U. (02-17-2016)
Bayes' rule?
Thank you. As I mentioned, I am over three decades out from my first and only statistics course (2 semesters Introduction). Thus I have no means of investigating without taking a refresher course! What is the formula for deriving the probability you found? Thanks again for your time and attention.
Veterinarian
Texas
Assuming hlsmith is correct, there are exact methods for problems of this kind. Go here:
http://home.ptd.net/~artnpeg/recurse.htm
and put in 1000 trials, length of streak of 1, prob success each trial of .01
You'll get 100% probability. Dropping prob success each trial down to .001 gives
63.2% probability. At prob success of .0001 you get 9.5%
Art
Ronald U. (02-17-2016)
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