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Mathematics 20 Online
OpenStudy (anonymous):

Does anyone know how to calculate the odds of winning the powerball by using the formula c=n!/(n-r)!r! There are 59 white balls and 39 red (power) balls

OpenStudy (amistre64):

n = total number of items; r = how many you want

OpenStudy (amistre64):

what constitutes winning?

OpenStudy (anonymous):

what do you mean?

OpenStudy (amistre64):

we have 59 white and 39 red ; but we have no way to determine what it means to "win"

OpenStudy (amistre64):

39 power balls is what i think i read

OpenStudy (amistre64):

the "odds" of winning would then be: #wins : #losses. odds of winning are 39:59 but i got no idea what the nCr formula would tell us

OpenStudy (anonymous):

I think he wants to know how many combinations of numbers there are. He uploaded this problem under the "just for fun" folder...it's not fun...but I want to try and figure it out but I need help.

OpenStudy (anonymous):

http://screencast.com/t/1s0RbwwdnShL

OpenStudy (anonymous):

ya, me neither :-( I'm totally lost here.

OpenStudy (amistre64):

the most likely solution, even tho it doesnt match the terminology of the question, would be to have n=39+59 and r = 39

OpenStudy (anonymous):

ya, that's what Igot too but I don't think that's what he's looking for. His wording is "How to calculate the odds of winning the powerball"

OpenStudy (amistre64):

the movie makes it clearer to me :)

OpenStudy (anonymous):

I thought that might help out :-)

OpenStudy (anonymous):

I've been watching it over and over again trying to make sense of it. It's onyl worth a couple of extra points, but hey extra points are extra points.

OpenStudy (amistre64):

so, 59 numbers and 5 ways to pick the numbers \[\frac{n!}{(n-r)!r!}\to\ \frac{59!}{54!5!}\]ways of picking the correct 5 balls this is apart from the power ball there is only 1 way to pick from 39 possible red balls\[\frac{n!}{(n-r)!r!}\to\ \frac{39!}{38!1!}\]

OpenStudy (anonymous):

ok, starting to make sense more...

OpenStudy (amistre64):

so, in total, we have 59.58.57.56.55 -------------- 5.4.3.2.1 59.58.57.7.11 -------------- = 5 006 386 ways to pick 5 numbers from 59 3

OpenStudy (anonymous):

where did you get the second set of numbers?) 59,58,57,7, 11?

OpenStudy (amistre64):

each of those ways can have 39 different red balls applied to them so 5 006 386 * 39 = 195 249 054 ways to choose 5white and 1 red ball to play with

OpenStudy (amistre64):

i simplified the top by dividing off the numbers that i saw fit into the top

OpenStudy (amistre64):

55/5 = 11/1 56/2 = 28/1 ; 28/4 = 7/1

OpenStudy (amistre64):

so if i did my mathing right; the odds of winning are 1, and odds of losing are 195 249 053 1: 195 249 053

OpenStudy (anonymous):

is 55/5 = 11/1 just simplifiying?

OpenStudy (amistre64):

uh .... yes

OpenStudy (anonymous):

Ok thanks! This helps out a lot.

OpenStudy (amistre64):

youre welcome; with any luck, i thought it out correctly :)

OpenStudy (anonymous):

you probably did! You seem pretty darn sharp!

OpenStudy (amistre64):

59c5 ways to pick 5 from 59 and 39ways to modifiy each pick. 39(59c5) = 195 249 054 ways to pick an answer; and only 1 of them is good odds are 1:195 249 053 seems right to me :)

OpenStudy (anonymous):

how did you calculate 39(59c5)? Is that distributing?

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