Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum
from 2 to 9 separately.

P(2) = 1/18

P(3) = 1/9

P(4) = 1/6

P(5) = 1/6

P(6) = 1/6

P(7) = 1/6

P(8) = 1/9

P(9) = 1/18

Step-by-step explanation:

Each die has six possible outcomes, so both dice together have 6*6 = 36 outcomes:

(1,1), (1,1), (1,2), (1,2), (1,3), (1,3),

(2,1), (2,1), (2,2), (2,2), (2,3), (2,3),

(3,1), (3,1), (3,2), (3,2), (3,3), (3,3),

(4,1), (4,1), (4,2), (4,2), (4,3), (4,3),

(5,1), (5,1), (5,2), (5,2), (5,3), (5,3),

(6,1), (6,1), (6,2), (6,2), (6,3), (6,3).

The sum of the values for each outcome is:

2, 2, 3, 3, 4, 4,

3, 3, 4, 4, 5, 5,

4, 4, 5, 5, 6, 6,

5, 5, 6, 6, 7, 7,

6, 6, 7, 7, 8, 8,

7, 7, 8, 8, 9, 9.

To find the probability of each sum, we just need to divide the number of times this sum appears over the total number of possibilities (36), so we have:

P(2) = 2/36 = 1/18

P(3) = 4/36 = 1/9

P(4) = 6/36 = 1/6

P(5) = 6/36 = 1/6

P(6) = 6/36 = 1/6

P(7) = 6/36 = 1/6

P(8) = 4/36 = 1/9

P(9) = 2/36 = 1/18

3.how is solving radical equations different than solving linear equations?

step-by-step explanation: