# Combinatorics problem

• November 11th 2009, 05:20 PM
minkyboodle
Combinatorics problem
The problem is a lottery problem. For the Mega Millions drawing, the official website says that there is a 1 in 306 chance of getting 3 out of the 5 regular numbers correct without matching the mega ball. The prof wants to see how we calculate this. I am stuck and getting nowhere near 306.

This is where I am at...

To figure out all the possible combinations of the 5 regular numbers is :

C(56,5) which is 3819816

from here I am lost
• November 11th 2009, 11:11 PM
Hello minkyboodle
Quote:

Originally Posted by minkyboodle
The problem is a lottery problem. For the Mega Millions drawing, the official website says that there is a 1 in 306 chance of getting 3 out of the 5 regular numbers correct without matching the mega ball. The prof wants to see how we calculate this. I am stuck and getting nowhere near 306.

This is where I am at...

To figure out all the possible combinations of the 5 regular numbers is :

C(56,5) which is 3819816

from here I am lost

For those of us not familiar with the Mega Millions lottery (this Forum has members all over the world!), it would help if you stated clearly what 'getting 3 out of the 5 regular numbers' means. You may even find that, in providing us with an exact formulation of the problem, you realise for yourself how to solve it.

• November 12th 2009, 02:22 AM
minkyboodle
you have to pick 5 numbers from 1-56
• November 12th 2009, 03:42 AM
Hello minkyboodle
Quote:

Originally Posted by minkyboodle
you have to pick 5 numbers from 1-56

Yes. This can be done in $\binom{56}{5}$ ways. And then what?

• November 12th 2009, 04:17 AM
minkyboodle
I already know that the odds of getting 3 out of 5 is 1/306. I am just trying to do the proof and I am struggling to set it up.
• November 12th 2009, 04:22 AM
minkyboodle
I am going to try and re-word everything.

in the lottery, the lottery people pick 5 numbers from 1-56. The odds of someone matching 3 of those 5 selected numbers is 1/306. How do we get there?
• November 12th 2009, 08:05 AM