There would be 138 students in 6 classes because first you'd divide 207 from 9 and get 23 then you'd multiply 23 to 6 and get 138.
Well I don't know.
Let's think about it:
-- There are 6 possibilities for each role.
So 36 possibilities for 2 rolls.
Doesn't take us anywhere.
New direction:
-- If the first roll is odd, then you need another odd on the second one.
-- If the first roll is even, then you need another even on the second one.
This may be the key, right here !
-- The die has 3 odds and 3 evens.
-- Probability of an odd followed by another odd = (1/2) x (1/2) = 1/4
-- Probability of an even followed by another even = (1/2) x (1/2) = 1/4
I'm sure this is it. I'm a little shaky on how to combine those 2 probs.
Ah hah !
Try this:
Probability of either 1 sequence or the other one is (1/4) + (1/4) = 1/2 .
That means ... Regardless of what the first roll is, the probability of
the second roll matching it in oddness or evenness is 1/2 .
So the probability of 2 rolls that sum to an even number is 1/2 = 50% .
Is this reasonable, or sleazy ?
