Find The Number Of Permutation

Question

I was solving one probability question, and to solve the problem I could relate it to binary and want solution.

The question was that in a fair toss repeated 10 times, what's the probability of 10th try to be the 5'th head.

I used brute-force and calculated it to be 63/512. But I can only brute force for definite numbers. What if I get n and x. So my problem is as stated below.

For all possible combination of a $n$ -bit binary number to have $x $ number of 1's where $x < n$.

Any Helps Will be Appreciated

score 2 · Accepted Answer · answered Sep 04 '15 at 17:57

2

The number of $n$-bit binary numbers with $x$ ones is given by the binomial coefficient

$$ \binom nx=\frac{n!}{x!(n-x)!}\;. $$

answered Sep 04 '15 at 17:57

joriki

238,052

Can I find a proof for it – Kishan Kumar Sep 04 '15 at 18:06
@KishanKumar: You certainly can -- e.g. in the sections titled "Definition and interpretations", "Multiplicative formula" and "Factorial formula" in the article I linked to. – joriki Sep 04 '15 at 18:21

score 1 · Answer 2 · answered Sep 05 '15 at 06:49

1

Actually, when you switched from the specific question to a generalised one, you changed the question. Look closely at them !

The specific question involves what is called a negative binomial distribution, whose details you can look up.

However, as the name suggests, it is linked to the binomial distribution, and you can easily derive a formula. For your specific case, you need 4 heads in 9 tries followed by a head, thus

$${9\choose4}\left(\frac{1}{2}\right)^9\left(\frac12\right) = \frac{63}{512}$$

answered Sep 05 '15 at 06:49

true blue anil

41,295

Are you referring that I did not considering the last event – Kishan Kumar Sep 05 '15 at 09:46
No, no, last event needs to be considered, and I have also considered it. It is the final term $\frac12$ in my expression. What I was saying is that there is a difference between getting the 5th head on the 10th toss (concrete example of yours), and getting 5 heads in 10 tosses (akin to the general problem you posed) – true blue anil Sep 05 '15 at 12:16
I am solving the question as 4 heads in 9 tosses. And then multiplying it with 1/2 to get the actual answer – Kishan Kumar Sep 05 '15 at 14:01
That is fine for the original question. – true blue anil Sep 05 '15 at 14:07
Is there any better approach – Kishan Kumar Sep 05 '15 at 14:08
No. Why do you find it difficult ? You just make a commonsense modification to the binomial distribution. – true blue anil Sep 05 '15 at 14:50
I just changed the question from coin toss to binary numbers. So, its bit hard to explain to someone. For example, if it would have been rolling a dice – Kishan Kumar Sep 06 '15 at 03:03

Find The Number Of Permutation

2 Answers2