Challenge (1 Viewer)

[doodydo]

hey everyone , this is killing me!
ok: (i was advised to put this in the 4U forum)

there are 45 balls and 6 are picked,
the odds for drawing 5 numbers that you have on the ticket are 2621:1 based on 14 games .
the odds for 5 winning numbers plus either supplementary number are 48483:1! (based on 14 games)
can someone tell me how to find these!
it's so damn annoying!

 

[Euler]

what does the : mean??

could you give an example of a simple situation, like the odds of a heads on a fair coin using this : symbol? the odds of getting 4 on a fair dice?

 

[CM_Tutor]

the odds of a heads on a fair coin using the : symbol? the odds of getting 4 on a fair dice?

are 2:1 and 6:1, respectively.

So, doodydo is saying, for example, that in a 45 ball lottery, where 6 numbers are chosen, if you play 14 games the probability of correctly picking 5 numbers is 1 / 2621.

 

[turtle_2468]

Lol, so do you want me to answer it? I could... but then calculon is paying us out So someone doing 4U might get good practice doing it.. I'll wait

 

[hatty]

Originally posted by turtle_2468
Lol, so do you want me to answer it? I could... but then calculon is paying us out So someone doing 4U might get good practice doing it.. I'll wait

it appears turtle can't do it. lol

 

[Calculon]

Originally posted by turtle_2468
Lol, so do you want me to answer it? I could... but then calculon is paying us out So someone doing 4U might get good practice doing it.. I'll wait

I haven't done permutations/3u probablility yet sorry, although I did solve a 3u probability question the other day in the ext. 1 forum with year 10 knowledge

 

[Calculon]

Originally posted by hatty
it appears turtle can't do it. lol

You'll have more than me paying you out if you don't solve it

 

[Giant Lobster]

damn u turtle... i cant do it by inspection

actually i cant do it at all.

 

[turtle_2468]

The first one is wrong...
take the probability of losing (ie not getting 5 correct numbers) in 1 game.
There are 6 ways to win (pick 5 out of the 6 numbers)...
Also, there are 45C6 ways to pick 6 out of 45. So P(losing a game)=1-(6/45C6).
Let that number be x.
P(losing 14 games)=x^14
So P(winning at least once in 14 games)=1-x^14
=1:96966.

And, the second question is quite silly... I mean, if you pick 6 numbers how can you have "5 winning numbers plus either supplementary number"? That would imply 7 numbers drawn... make the question more obvious please...
(I'm assuming in my answer above that you draw 6 balls, and you need 5 of them to win..)

 

[Archman]

lol thats because the lotto ppl dont use 14*(1-x) (..which is quite wrong) rather than 1-x^14.

Mind u, it makes the odds look better than they actually are.