Sam is rolling a number cube with the numbers 1, 1, 2, 2, 3, and 3 printed on it.

what is the probability that, if he rolls the cube twice, the sum of the numbers will add to 4?

0.3333

Step-by-step explanation:

You first need to see how many ways he can roll a sum of 4.  There are 12 ways. Treat each face as it's own value.  There are 2 of each number on the die, so each face of the die is it's own event.  

The 12 ways..

1-2.  Rolls a 1 with the one of the 1-faces, then rolls a 3 with either of the 3-faces on the second roll.   This counts for 2 different sums of 4.  

    1 (first 1 face) - 3 (first 3 face)

    1 (first 1 face) - 3 (second 3 face)

3-4.  He rolls a 1 but it's a different face than the other one than the one he rolled above.  This gives us 2 more sums of 4

    1 (other 1 face) - 3 (first 3 face)

    1 (other 1 face) - 3 (second 3 face)

That's 4 total ways, the same logic will apply for the two 2 faces, and the two 3 faces.  Each will have 4 more outcomes, giving us 8 more

There are 36 total outcomes  (6*6 = 36), so

P(sum of 4) = 12/36 = 0.3333

step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length