Fly's Problem (Advanced)

[Phlembac Adib Hasan]

A fly starts jumping on a straight line.At each jump it crosses $5\; cm$ either left or right.What's the probability that the fly will be $30\; cm$ apart from starting point after $20^{th}$ jump?
Additional part :generalize it for every even number and distance $5k$.

[sowmitra]

Adib, could you clarify which data we are supposed to assume to be even??
I couldn't be clear which data has to be even. So, I am posting my solution without
generalization:

First, there are $20$ jumps. In each jump, the fly can move both forward and backward. So, $2$ choices for each jump. Therefore total no. of distance it can cover in $20$ jumps = $2^{20}$
Now in order to to be at $30$ cm, suppose, the fly has to jump $n$ times to the right and $20-n$ times to the left.
So, we have, $$5*n-5*(20-n)=30\Rightarrow n=13$$
No. of ways to jump these $13$ times= $\binom{20}{13}$. Here, the fly may remain $30$cm to the right as well as to the left of the line. So, total no. of ways=$2*\binom{20}{13}$
Therefore, the probability that the fly will remain $30$cm apart from the starting point after $20$ jumps= $\displaystyle\frac{2*\binom{20}{13}}{2^{20}}$

[Phlembac Adib Hasan]

I asked to find the probability for every even number like the probability after 2nd jump,after 4th jump etc.I hope it's clear now.

[sowmitra]

Adib, thank you for making it clear . I think this is the generalization :

Since, the no. of jumps are even, suppose no. of jumps=$2n$
So, total no. of distances the fly can be at after the $2n$ jumps=$2^{2n}$
Now, let, the fly has to jump $a$ times to the right to be at $5k$ distance from the starting point. Then, it has to jump $2n-a$ times to the left.
Therefore, we get, $$5a-5(2n-a)=5k \Rightarrow a=\frac{k}{2}+n$$
No. of ways to jump these $a$ times= $\binom{2n}{\frac{k}{2}+n}$
Again, the fly may remain at $5k$ distance from the starting point both at the left and at the right of the line. So, total no. of ways =$2*{\binom{2n}{\frac{k}{2}+n}}$
Hence, the probability that the fly will remain at $5k$ distance from the starting point after $2n$ jumps =$\displaystyle\frac{2*{\binom{2n}{\frac{k}{2}+n}}}{2^{2n}}$

I am not quite sure if this the correct solution (mostly because there is a fraction $\frac{k}{2}$) .
I would be very grateful if you could point out the bug in the solution (if there is any).

[nafistiham]

Adib, thank you for making it clear . I think this is the generalization :

Since, the no. of jumps are even, suppose no. of jumps=$2n$
So, total no. of distances the fly can be at after the $2n$ jumps=$2^{2n}$
Now, let, the fly has to jump $a$ times to the right to be at $5k$ distance from the starting point. Then, it has to jump $2n-a$ times to the left.
Therefore, we get, $$5a-5(2n-a)=5k \Rightarrow a=\frac{k}{2}+n$$
No. of ways to jump these $a$ times= $2nC_{\frac{k}{2}+n}$
Again, the fly may remain at $5k$ distance from the starting point both at the left and at the right of the line. So, total no. of ways =$2*{2nC_{\frac{k}{2}+n}}$
Hence, the probability that the fly will remain at $5k$ distance from the starting point after $2n$ jumps =$\displaystyle\frac{2*{2nC_{\frac{k}{2}+n}}}{2^{2n}}$

I am not quite sure if this the correct solution (mostly because there is a fraction $\frac{k}{2}$) .
I would be very grateful if you could point out the bug in the solution (if there is any).

for combination or permutation you should use these I think $_{}^{n}\textrm{C}_k $ or $\binom{n}{k}$
(just click on them to see the code,or there is equation editor)

[sowmitra]

@Tiham: Thank you for pointing it out.
I have changed the previous codes to the binomial codes. I am actually new at working with LATEX, so, I do not know all the codes yet. Besides, I couldn't find the code of Combinations in the pdf 'latexhelp'. Ei jonnoe oi choramita korechilam. Thanks again for helping me.

[nafistiham]

My pleasure.And, well done.I would not have the courage to re edit all those codes.

[sowmitra]

Hey, where is Adib ? Has he forgotten about this post ?!
Will someone please show me the bug in the solution ?

[sm.joty]

@Tiham: Thank you for pointing it out.
I have changed the previous codes to the binomial codes. I am actually new at working with LATEX, so, I do not know all the codes yet. Besides, I couldn't find the code of Combinations in the pdf 'latexhelp'. Ei jonnoe oi choramita korechilam. Thanks again for helping me.

Humm........
This may be helpful for you.
http://www.thestudentroom.co.uk/wiki/La ... rentiation