Stanton D. answered • 12/11/19

Tutor to Pique Your Sciences Interest

Hi Tony F.,

A fairly standard probability problem. The difficult part is deciding exactly which tickets to use as you calculate along.

For the first part, you need the probability of NOT drawing __all three__ of your tickets BEFORE the last 4 remain. [That's fairly standard for these types of probability problems; you have to go at them "backwards", so to speak, b/c the calculations for drawing various combinations of cards including some of yours over a long series of choices becomes otherwise mathematically too cumbersome.] But it's OK for this if 2 of your tickets ARE drawn before that, b/c you would still have 1 left. So on the first 2 drawings, you don't care if your first 2 tickets go! But then, you have to keep your remaining ticket out of the pot! The combined probability is a sequential multiplication of the odds at each drawing step:

So that is: (19/19)*(18/18)*(16/17)*(15/16)* .... *(4/5). :There are now 4 tickets left, one of which is one of yours. Total odds are 4/17 (all the other numbers cancel each other out).

You should be able from the reasoning above, to continue the multiplication series over the next 3 terms, to answer the second part. There is NO drawing for the last bottle per se (although you might think, from the progression of terms, that it should be *(0/1)), you have the last ticket, and it is the only one left!

So if chance favors you (and the laws of the state permit), enjoy your prize(s).

-- Cheers, -- Mr. d.