I remember doing this question a couple of years ago now, looking back at my old textbook they've updated the question as they changed the rules slightly for the game. Anyways...
Lets look at division 2.
What you need to do to find the probability is to flip their formula, but now to explain where the formula comes from.
We have
numbers, of which we want to pick
from. The possible number of ways to pick
numbers from
is
=\frac{45!}{7!38!})
, so that will be our denominator.
Now to win division 2, we need at least
of those winning numbers and at least
supp number. So how many ways can we pick
winning numbers from those
?
=\frac{7!}{6!1!})
Now for the second part the key here is that the supplementary numbers can't be any of those
winning numbers. That is we only have
numbers left that can be the supp numbers. Now the
-nCr(36,1))
part is basically the number of options left we have, take away the number of options left that are wrong (that is that don't give us either of the two supps). So there is
numbers besides the winning numbers, and of those we are picking
, since we only have one number left to make our
, take away the number of ways we can pick
from the
remaining numbers that aren't supps.
Then the number of ways those two situations work together is the product of the two, which becomes our numerator.
The logic works through for the other options, I'll explain say division 5 & 7 though.
Division 5:
Total number of ways we can pick the
numbers
=\frac{45}{7!38!})
. The number of ways we can pick
correct numbers from the
correct numbers is
=\frac{7!}{5!2!})
, and then we can pick any two other incorrect numbers that are not supps as well, that is we have
numbers remaining to pick that are basically useless, and from those we want
numbers, i.e.
=\frac{36!}{2!34!})
. Multiplying those together and the dividing by the total number of ways that we found previously, gives the probability.
Division 7:
is the number of options we can have picking
numbers from the
. The
=\frac{7!}{3!4!})
is the number of ways we can pick
correct numbers from the
correct numbers.
=\frac{38!}{4!34!})
is the number of ways we can pick
numbers from the remaining
, but to get the amount of options that are correct, we need to take away the number of options of picking the numbers that aren't supps, there are
of those left, and we have
spots to fill. Then again to get the combination of both, we multiply them together, and then divide by the total number of options.
I hope that helps, and makes sense....
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