2007 AMC 10B Problem 19

The wheel has six equal sectors labeled $1, 2, 3, 6, 7, 9$. The board has columns $1, 2, 3$ and rows $1, 2, 3, 4$, shaded like a checkerboard with square $(1,1)$ shaded: in columns $1$ and $3$ the shaded rows are $1$ and $3$; in column $2$ they are $2$ and $4$.

Column (first spin, remainder mod $4$): the numbers give remainders $1, 2, 3, 2, 3, 1$, so each column has probability $\frac26 = \frac13$.

Row (second spin, remainder mod $5$): the remainders are $1, 2, 3, 1, 2, 4$, so
$$P(\text{row }1) = \tfrac26,\quad P(\text{row }2) = \tfrac26,\quad P(\text{row }3) = \tfrac16,\quad P(\text{row }4) = \tfrac16.$$

Now combine. If the column is $1$ or $3$, a shaded square needs row $1$ or $3$: probability $\frac26 + \frac16 = \frac12$. If the column is $2$, it needs row $2$ or $4$: probability $\frac26 + \frac16 = \frac12$. In every column the chance of landing on a shaded square is $\frac12$, so the overall probability is $\frac12$.

The answer is $\boxed{\textbf{(C)}\ \frac{1}{2}}$.

The two spins are independent, so condition on the column and add the probabilities of the shaded rows in that column. The row distribution is not uniform, but the checkerboard pairs rows $\{1, 3\}$ against $\{2, 4\}$ and both pairs happen to carry probability $\frac12$, so the column never matters.

• Assuming the rows are equally likely (1/4 each); the remainders mod 5 of 1, 2, 3, 6, 7, 9 favor rows 1 and 2.
• Forgetting that a remainder of $0$ is impossible here (no wheel number is a multiple of $4$ or $5$) and inventing a fourth column or fifth row.