Suppose two fair dice are rolled, find the probability that getting the first die is 3 or the sum is 10?

4&6, 5&5, and 6&4 are only three possible rolls (not ten)

so 6/36 + 3/36 = 9/36 or 1/4

mutually exclusive event - if the first is a three then you cannot get a sum of 10

P(A or B) = P(A) + P(B)

P(A) = 1/6 (only one 3 on standard 6-sided die)

P(B) = 10/36 (6,4 or 4,6 or 5,5 are the only ten rolls out of (6*6) 36 possible outcomes)

P(A or B) = 6/36 + 10/36 = 16/36 = 4/9 = 0.44....

I won't do your school work for you.

Can you imagine going for treatment to a doctor who cheated his way through medical school?

If you cannot do the work, you do not deserve the grade.