Reply to queries

[Aritra Mandal]

Hello learners,

I am Aritra Mandal, your TA. I have tried to address all the queries. 

Please go through the following. Feel free to reply if there is any doubt.

Question 1>  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..

Answer>Take n=5,k=2 and p=1/2

Then, probability of  Ck = |Ck|*(1-p)^(k-1)*p = 2^(n-k)*(1-p)^(k-1)*p=(2^3)*((1/2)^1)*(1/2)=8*(1/2)*(1/2)=2, which is not possible. 

I think the probability of Ck is well explained in the video. I suggest you to go through it again.

Question 2> I have a question about what is happening in this video at 00:02:27 -

http://www.youtube.com/watch?v=ejbPM1ehP0A#t=147s.

I have this doubt in the previous video also. Here our sample space is {1,2,...,n}, where each k in S represent the k successes in the n-tuple. So k is a set containing n-tuples with all of them having exactly k successes. I don't understand why the cardinality of this set is nCk. nCk is basically the number of groups of k candidates tht can be made out of n candidates. In our case, out of n candidates, k are successes and n-k are failures. I don't understand like how nCk ensures the cardinality of this set. May you please explain..

Answer> k is a set containing n-tuples with all of them having exactly k successes

=nCk

=nC(n-k)

=k is a set containing n-tuples with all of them having exactly  n-k failures

=k is a set containing n-tuples with all of them having exactly k successes and n-k failures

I hope it clarifies your doubt.

Thanks and regards