2007 AMC 10B Problem 19

Below is the professionally curated solution for Problem 19 of the 2007 AMC 10B, 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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Concepts:basic probabilityparitymodular arithmetic

Difficulty rating: 1490

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}

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.

Problem 19 in Other Years