AMC 12 questions #1

Since there are $2 + 4 + 6 = 12$ dots on the even-numbered faces of a standard die out of a total of $1 + 2 + 3 + 4 + 5 + 6 = 21$ dots, there is a probability of $\dfrac{12}{21} = \dfrac{4}{7}$ that the dot is removed from an even-numbered face and a probability of $\dfrac{9}{21} = \dfrac{3}{7}$ that it will be removed from an odd-numbered face.

If the dot is removed from an even-numbered face then there will be $4$ odd faces left, so the probability that an odd-numbered face is on top when the altered die is rolled is $\dfrac{4}{6} = \dfrac{2}{3}.$

If the dot is removed from an odd-numbered face then there will be just $2$ odd faces left, giving us a probability of $\dfrac{2}{6} = \dfrac{1}{3}$ that an odd number will be rolled.

Thus the desired probability is $(\dfrac{4}{7})(\dfrac{2}{3}) + (\dfrac{3}{7})(\dfrac{1}{3}) = \boxed{\dfrac{11}{21}}.$

Probability 1 will go from odd to blank = 1/21

Probability that 2 will go from even to odd = 2/21

Probability that 3 will go from odd to even = 3/21

Probability that 4 will go from even to odd = 4/21

Probability that 5 will go from odd to even = 5/21

Probability that 6 will go from even to odd = 6/21

If the dot is taken off the 1, 3 or 5 (P = 9/21 = 3/7) P of throwing odd side = 2/6 = 1/3

If the dot is taken off the 2, 4, or 6 (P = 12/21 = 4/7) P of throwing odd side = 4/6 = 2/3

P of throwing odd side = (3/7 * 1/3) + (4/7 * 2/3) = 3/21 + 8/21 = 11/21