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 ?
The last awnser is 15x + 2 and the first one is explained below:
65 X 1.5 = 97.5
97.5 + 95 = 192.5
192.5 is awnser
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-14+3/2b-(1+2/8b), then take out the blanket become -14+3/2b-1-2/8b , get 5/4b-15=0 so 5/4b=15 ,get b=15×4÷5. b=12
-2(x + 3) = 8
Multiply -2 with x and 3
-2x -6 = 8
Now add 6 to both sides
-2x -6 = 8
+6 +8
-2x = 16
Lastly divide -2 from both sides to find x
-2x = 16
—— —-
-2x -2x
X = -8
