Why are those two events not independent?

cdux
Messages
187
Reaction score
0
Provided I know that after 10 attempts I will get 8 successes with a probability of a success per try being 'p', does that affect the success of the first try?

My first instinct was to say it's independent because each attempt is independent but now I know this is a special case where "I know the future" so to speak hence the first try can be expressed conditionally based on that information after 10 tries.

But what I am asking is, what is the formal way to express that dependency, is it only per definition a vague "the occurrence of those 10 tries affect the 1st" or something more explicit?
 
Physics news on Phys.org
cdux said:
Provided I know that after 10 attempts I will get 8 successes with a probability of a success per try being 'p', does that affect the success of the first try?
No. What you're describing is some event in which "success" has a probability of .8. This says that if you run a large number of trials, about 80% of them will result in "success", but that doesn't tell you what the outcome will be on any individual trial.

For example, flipping a fair coin should result in heads half the time and tails the other half, so p = .5. This doesn't give you information about an individual coin flip, though.
cdux said:
My first instinct was to say it's independent because each attempt is independent but now I know this is a special case where "I know the future" so to speak hence the first try can be expressed conditionally based on that information after 10 tries.
You don't really know the future, other than to be able to say something about the number of successes in the long run.
cdux said:
But what I am asking is, what is the formal way to express that dependency, is it only per definition a vague "the occurrence of those 10 tries affect the 1st" or something more explicit?
 
Mark44 said:
No. What you're describing is some event in which "success" has a probability of .8.
The OP can be read another way. Maybe it has been reported after the fact that 8 of the 10 trials were successful, but we'e not told which. Each try is known to have had an a priori probability of p. Assuming the trials were independent, we can now say the probability that the first trial was successful was .8, regardless of p.
cdux, which is the correct interpretation?
 
cdux said:
Provided I know that after 10 attempts I will get 8 successes with a probability of a success per try being 'p', does that affect the success of the first try?

My first instinct was to say it's independent because each attempt is independent but now I know this is a special case where "I know the future" so to speak hence the first try can be expressed conditionally based on that information after 10 tries.

But what I am asking is, what is the formal way to express that dependency, is it only per definition a vague "the occurrence of those 10 tries affect the 1st" or something more explicit?

You need the conditional probability that the first trial is a success, given 8 successes in 10 trials. That is, if##X_1 = 1## is the event that the first outcome is a success (with ##P\{X_1=1\} = p##) and if ##\{S_{10} = 8\}## is the event that the sum of the first 10 outcomes is 8 (the other 2 being 0), you want ##P\{X_1 = 1|S_{10} = 8\},## which you can get via Bayes' formulas.
 
Ray Vickson said:
you want ##P\{X_1 = 1|S_{10} = 8\},## which you can get via Bayes' formulas.
but it's much simpler just to ask what is the probability that the first trial is one of the eight successes.
 
haruspex said:
but it's much simpler just to ask what is the probability that the first trial is one of the eight successes.

I agree---it is simpler---after one is sure that that is a correct way of viewing the problem. Sometimes, to gain familiarity with a method/result/insight, it helps to start with formalism. (One can come up with somewhat similar scenarios in which that view would be questionable, so the binomial nature of the current problem plays an important role.)
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top