# Probability-Selections&Arrangement

• Mar 11th 2010, 08:37 AM
WannaBe
Probability-Selections&Arrangement
The Question:

Given a 5x5 Chess-Board (with 25 squares).

On each square, someone writes randomly one of the digits- 0 or 1.

A. What is the probability of that in at least on row, the sum of the digits will be excatly 3?

B.Someone scatters randomly 25 discs on the board. On 10 of the discs, the digit 1 is written and on 15 of the other discs, the digit 0 is written.

What is the probability that the sum of the digits in each and every one of the rows will the same?

I've no idea about these two parts...I'll be delighted to get some guidance

Thanks a lot!
• Mar 11th 2010, 09:59 AM
Plato
Quote:

Originally Posted by WannaBe
The Question:
Given a 5x5 Chess-Board (with 25 squares).
On each square, someone writes randomly one of the digits- 0 or 1.
A. What is the probability of that in at least on row, the sum of the digits will be excatly 3?

Each row can be any one of $2^5=32$ different bit-strings.
Only $\binom{5}{3}=10$ of those contain exactly three ones.
The probability that row one does not add to three is $\frac{22}{32}$.
So what is the probability that no row adds to three?
• Mar 11th 2010, 10:42 AM
WannaBe
I'm sorry ,but I think I didn't understand your guidance...
In your first sentence you've said there are 32 different "arranging" options of a row...
In 10 of these possibilities - the sum of the digits in the row is 3...
So, when looking at a single row- the probabity for that its sum will not be three is
$\frac{22}{32}$ . We have 5 rows . Does it means that the probability that no row adds to three is $\frac {\frac{22}{32} } { 2^{25} }$ ?

If it isn't what you meant, I don't think I understood your guidance...

Hope you'll be able to guide me... [ BTW-I tried doing something like you did, but got stuck excatly at the same point...]

Thanks!
• Mar 11th 2010, 10:48 AM
Plato
The probability that no row adds to three is $\left(\frac{22}{32}\right)^5$.
• Mar 11th 2010, 11:12 AM
WannaBe
So the needed probability is:

$1 - \left(\frac{22}{32}\right)^5$
?
Thanks a lot!
Can you guide me in the second part of the question plz?

Thanks!
• Mar 11th 2010, 11:46 AM
Plato
Quote:

Originally Posted by WannaBe
Can you guide me in the second part of the question plz?

No, not until you have posted some efforts on your part.
• Mar 11th 2010, 12:44 PM
WannaBe
Well... You are right... Here is what I've tried:
The only option to scatter the discs in a way the digits will add up to an identical sum is that the sum of each row will be 3...
So... We need to put 3 1's in each row and 2 0's in each row... Let's start with the 1's:
We have $( \binom{5}{3} ) ^ 5$ options to put them in a way that there will be 3 in each row... We'll put the 0's in the empty squres...
Now we need to calculate the number of options to scatter the 25 discs:
So, we'll choose 10 squares and put the 1-discs on them....We have
$\binom{25}{10}$ to do it... After we put the 1-squares, the 0-digits will be put instantly...

Hence the probability is:
$\frac {( \binom{5}{3} ) ^ 5 } {\binom{25}{10}}=\frac{2500}{81719}$

I realy doubt that my calculations are correct... I think I did something wrong in the last part...
I'll be delighjted to get a verification on this and on the first one...

Thanks a lot!
• Mar 11th 2010, 12:53 PM
Plato
BUT there are only 10 ones.
How canthere be three is each of five rows?
• Mar 12th 2010, 12:15 AM
WannaBe
Sry...So all of my steps are correct for scattering the 0's ... The sum of each row should be 2 ...

Am I right?

Thanks
• Mar 12th 2010, 07:36 AM
Plato
Yes.
• Mar 12th 2010, 07:44 AM
WannaBe
thanks a lot!