Math Is Fun Forum

Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

Probability question I created

Consider a set A that has [1,2,3,4,5,6,7,8,9,10,11,12,13,14], now 4 integers are taken from that set to create a new set, B.  What is the probability that the first 2 digits summed together is greater than the last 2 digits summed together?

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The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability question I created

Hi mathgogocart;

The first condition that you must always include in problems like these is whether there is replacement or not. I will assume that there is not.

This is similar to the "Lucky Tickets Problem" and is solved using generating functions or programming. I am getting 203 / 429

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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.

Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

Re: Probability question I created

Would Python be suitable for that?

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The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability question I created

Of course, but M is much better. M allows you to combine math and programming in a natural sort of way.

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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.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Probability question I created

I have verified bobbym's result indirectly. I do not have exotic resources but with pen,paper and to a very less extent calculator I did the work. I estimated sample space of 2 numbers with same total and paired them on first 2 and next to places. the result is 1288 cases out of 14P4 give equality of sum of 1st two numbers to sum of 3rd and 4th numbers.This is 23/429. The remaining probabilty is equally divided between > and < conditions. So the required probability =(429-23)/429 X 1/2 =203/429.

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