Doubt in probability of first success in kth trial

[jman...@gmail.com]

Hi,

I have a question about what is happening in this video at 00:10:50 -

http://www.youtube.com/watch?v=04r6R_62nBU#t=650s.

In this problem, Ck is the set that first success occurs in kth trial. So, upto k-1 trial there are all failure and at kth level there is success. From k+1-th trial to nth trial there can either be success or failure (2 choices). So by permutation (law of counting). the cardinality of Ck is 2^(n-(k+1)). So there are total 2^(n-k) equally likely outcomes. Probability of each outcome in Ck = (1-p)^(k-1)*p So shouldn't the probability of Ck = |Ck|*(1-p)^(k-1)*p = 2^(n-k)*(1-p)^(k-1)*p. Like we did in calculating the probability of k successes where n chooses k comes..

Thanks!

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Reference Key - Key('StudentQuestionEntity', 5989406199513088, namespace='ns_noc21_ma31')

[Vinay M]

Good question, I think the professor did not emphasize on the original question of How many trails are required to obtain the first success? 

Let us consider n total trials, Ck is the set of all possible distribution of outcomes such that first success is at kth trial, our total number of outcomes are 2^n, and total number of outcomes with first success is at kth trial is as you found out 2^(n-k). So now why not say that the probability of having success at kth trial is the cardinality the above set Ck divided by 2^n, i.e, 2^(n-k) /2^n = 2^(-k). Notice the in Ck all the outcomes are NOT equally likely and same for other Ck sets.  Also important thing is different Cks themselves, the set themselves have different collective probabilities and this is all because of the success probability of a single trial 'p' being general. It could be other than 0.5 and therefore different Ck sets are having different collective probabilities.

Let's see why the outcomes inside Ck are not equally likely given p is in a general setting here i.e. it is not 0.5. 

Consider a particular k say 10 and let our total trial n be 20. C10 is the set containing all the outcomes with first success at 10th trial. One of these outcomes could be as follows:- FFFFFFFFFSFFFFFFFFFF, other outcome could be this - FFFFFFFFFSFFSSFFSSSS. The former outcome could have different probability than the later because trial success 'p' could be other than 0.5 also. Therefore you cannot consider them equally likely and then just multiply the cardinality of the set to get its total probability of this set Ck. So, how to get the total probability of this set Ck then? Should we compute all its elements/outcomes probability individually and add them? Nope that is too cumbersome. The trick is you just have to calculate the probability of set itself singly and not focus on its elements. The set's probability is just as calculated in the video, which is just (1-p)^k-1 x p. Here seems that we just calculated the specific probability of arragngements upto kth trial only but not further upto n. Here is that subtle process and we do not need to worry about what happens after kth trial, just calculate upto k and this is the probability of the whole set. And again why is that? I'll leave this as an excercise for you to work out. It's interesting, one hint that i could give you is take common among the sum of probabilities of all the outcomes in Ck. Good luck. 

[Himanshu]

Yes, if we have total of n trials and we want to compute the probability of the event (Ck) that success first appear at the k-th trial. Then upto k-1 there is failure and at k there is success, it is common in all the elements of the event. After that we can have success/failure at any place.          

If we take into account the last (n-k) trials of all the elements of Ck, so we can relate it with binomial (n-p, p) distribution, as the result of any trial can be success/ failure. So sum of the probability of all the elements of Ck is 1 (considering trials from k+1 place).       

The probability of every element of Ck has (1-p)^(k-1)*p term in common.         

P(Ck) = sum over all the elements {(1-p)^(k-1)*p  X  terms coming after kth trial)}          

So, P(Ck) = (1-p)^(k-1)*p  X sum over all the elements {terms coming after kth trial)}     

Hence, p(Ck) =  (1-p)^(k-1)*p       

Thanks @Vinay, now I have understood.