2011 AMC 8 Problem 18

Attempt Problem 18 of the 2011 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

A fair six-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?
16\dfrac{1}{6}
512\dfrac{5}{12}
12\dfrac{1}{2}
712\dfrac{7}{12}
56\dfrac{5}{6}
Answer: D
Concepts:dice (probability)symmetry
Small hint:
By symmetry, half of the unequal rolls have the first roll larger.
Big hint:
Separate the equal rolls from the unequal rolls.
Solution:
There are 33 possible outcomes when rolling a die twice: the first number is greater than the second, both numbers are equal, or the first number is less than the second number. The first and third outcomes have the same probability since they are symmetric.
The second outcome has a 16\dfrac{1}{6} chance of happening, since the first number can be anything, and the second number must equal the first number. The other two outcomes have a combined probability of 1−16=56.1 - \dfrac{1}{6} = \dfrac{5}{6}. This means that each outcome has a 56÷2=512\dfrac{5}{6} \div 2 = \dfrac{5}{12} chance of happening.
The desired probability is the first outcome plus the second outcome, for a total probability of 512+16=712.\dfrac{5}{12} + \dfrac{1}{6} = \dfrac{7}{12}.
Thus, D is the correct answer.
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