You can put this solution on YOUR website! For mutually exclusive events (a die cannot show a 3 and a 4 at the same time):
P(A or B) = P(A) + P(B), so
P(3 or 4) = P(3) + P(4)
P(3) = 2/12 = 1/6 Since there are two 3's in 12 numbers on two dice.
P(4) = 2/12 = 1/6 Since there are two 4's in 12 numbers on two dice.
P(3) + P(4) = 1/6 + 1/6 = 2/6 = 1/3
The probability of throwing a 3 or a 4 on a single throw of two dice is 1/3