2007 AMC 10B Problem 19

Attempt Problem 19 of the 2007 AMC 10B 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 2007 AMC 10B solutions, or check the answer key.

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19.

The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by 4,4, and the second number is divided by 5.5. The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?

13\dfrac{1}{3}

49\dfrac{4}{9}

12\dfrac{1}{2}

59\dfrac{5}{9}

23\dfrac{2}{3}

Answer: C
Concepts:basic probabilityparitymodular arithmetic
Difficulty rating: 1490
Solution:

The shaded squares are those where the two remainders are both odd or both even. The first remainder is even (from the numbers 22 and 66) with probability 13\dfrac13 and odd with probability 23.\dfrac23.

The second remainder is even with probability 12\dfrac12 and odd with probability 12.\dfrac12.

The probability that they share parity is 1312+2312=12.\dfrac13\cdot\dfrac12+\dfrac23\cdot\dfrac12=\dfrac12.

Thus, the correct answer is C.

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