Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

probability

two numbers

and

are chosen at random without replacement from the set 

.

Find the probabilty that

  is divisable by 

edit it

Last edited by juantheron (2011-11-06 13:29:59)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: probability

hi juantheron

Find the probabilty that ......... is divisable by 5

What is missing here?

Bob

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: probability

Hi juantheron;

What is missing here?

Yep!

Seems one interesting question is:

Find the probabilty that x+y  is divisible by 5?

That probability for that is 1 / 5.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: probability

There are a couple things to note here.  First is that a number's divisibility by 5 can be determined solely by the 1's digit.  If it is 0 or 5 then the entire number is divisible by 5, otherwise it is not.  Secondly, the 1's digit of x^4 can also be determined solely by the 1's digit of x.  So first let's make a list of the 1's digit of the fourth power of every number from 0 to 9:

0^4 = 0 -> 0
1^4 = 1 -> 1
2^4 = 16 -> 6
3^4 = 81 -> 1
4^4 = 256 -> 6
5^4 = 625 -> 5
6^4 = 1296 -> 6
7^4 = 2401 -> 1
8^4 = 4096 -> 6
9^4 = 6561 -> 1

Now we can group these numbers into 2 categories, those that end in either 0 or 5 and those that end in 1 or 6.  If the chosen numbers x and y are both from the same group then x^4 - y^4 will be divisible by 5.  Try to finish it yourself from there.

Last edited by TheDude (2011-11-07 00:35:42)

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