Question

On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

$\dfrac{5}{11}$

$\dfrac{10}{21}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

$\dfrac{6}{11}$

Solution:

The die has $21$ dots, so a dot is removed from the face with $n$ dots with probability $\dfrac{n}{21}.$

If a dot is removed from an odd face, the top is odd with probability $\dfrac{1}{3}$ (any of the three odd faces on top); if from an even face, the top is odd with probability $\dfrac{2}{3}.$ The removed dot lies on an odd face with probability $\dfrac{1 + 3 + 5}{21}$ and an even face with probability $\dfrac{2 + 4 + 6}{21}.$

Hence the answer is $\dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}.$

Thus, the correct answer is D.

2005 AMC 12A Problem 14

Solution:

Problem 14 in Other Years