Difference between revisions of "2007 AMC 10B Problems/Problem 19"
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When dividing each number on the wheel by <math>5,</math> the remainders are <math>1, 1, 2, 2, 3,</math> and <math>4.</math> | When dividing each number on the wheel by <math>5,</math> the remainders are <math>1, 1, 2, 2, 3,</math> and <math>4.</math> | ||
| − | The probability that a shaded square in the <math>1</math>st or <math>3</math>rd row of the <math>1</math>st or <math>3</math>rd column is | + | The probability that a shaded square in the <math>1</math>st or <math>3</math>rd row of the <math>1</math>st or <math>3</math>rd column is chosen is |
<cmath> \frac{2}{3} \times \frac{3}{6} = \frac{1}{3} </cmath> | <cmath> \frac{2}{3} \times \frac{3}{6} = \frac{1}{3} </cmath> | ||
| − | The probability that a shaded square in the <math>2</math>nd or <math>4</math>th row of the <math>2</math>nd column is | + | The probability that a shaded square in the <math>2</math>nd or <math>4</math>th row of the <math>2</math>nd column is chosen is |
<cmath> \frac{1}{3} \times \frac{3}{6} = \frac{1}{6} </cmath> | <cmath> \frac{1}{3} \times \frac{3}{6} = \frac{1}{6} </cmath> | ||
| − | Add those two together | + | Add those two together to get |
<cmath> \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \boxed{\textbf{(C)} \frac{1}{2}}</cmath> | <cmath> \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \boxed{\textbf{(C)} \frac{1}{2}}</cmath> | ||
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Next we isolate the rows numbered 3 or 4. Note that the probability of picking the rows is same, because of our list up above. The columns, of course, still have the same probability. Because the number of shaded and non-shaded squares are equal, we have <cmath> \frac{1}{2} </cmath> | Next we isolate the rows numbered 3 or 4. Note that the probability of picking the rows is same, because of our list up above. The columns, of course, still have the same probability. Because the number of shaded and non-shaded squares are equal, we have <cmath> \frac{1}{2} </cmath> | ||
Combining these we have a general probability of <cmath> \boxed{\textbf{(C)} \frac{1}{2}} </cmath> | Combining these we have a general probability of <cmath> \boxed{\textbf{(C)} \frac{1}{2}} </cmath> | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | Similarly to Solution 1, we create two lists, one for <math>mod</math> <math>4</math> and one for <math>mod</math> <math>5</math> : (<math>1, 2, 3, 2, 3, 1</math>) and (<math>1, 2, 3, 1, 2, 4</math>). Next, we make a list of the coordinates we want for the shaded boxes. (<math>1,1</math>), (<math>1,3</math>), (<math>2,2</math>), (<math>3,1</math>), (<math>3,3</math>), (<math>2,4</math>). For each coordinate pair, we find the probability for it to occur. In order, we get <math>\frac{1}{9}, \frac{1}{18}, \frac{1}{9}, \frac{1}{9}, \frac{1}{18}, \frac{1}{18}</math>. Sum these to get the answer of <math> \boxed{\textbf{(C)} \frac{1}{2}} </math> | ||
| + | |||
| + | ~Dynosol | ||
| + | |||
| + | ==Solution 4== | ||
| + | We take everything modulo <math>4</math> first. We get <math>1</math> mod <math>4</math>, <math>2</math> mod <math>4</math>, <math>3</math> mod <math>4</math>, twice. Therefore, the probability of any given column is equal. We take everything modulo <math>5</math> now. We have <math>1, 2, 3, 1, 2, 4</math> mod <math>5</math>. <math>1</math> and <math>2</math> are different colored squares, and <math>3</math> and <math>4</math> are also different colored squares. Therefore, every single one of them has the same probability. | ||
| + | |||
| + | Therefore, our probability for a white square or a shaded square is equal, and therefore, the probability the square is shaded <math>\boxed {(C)\frac {1}{2}}</math> | ||
| + | |||
| + | ~Arcticturn | ||
== See Also == | == See Also == | ||
Latest revision as of 09:58, 27 October 2021
Problem
The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by 
and the second number is divided by 
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?
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label("$6$",(-0.6,1)); label("$1$",(0.6,1)); label("$2$",(0.6,-1)); label("$9$",(-0.6,-1)); label("$7$",(1.2,0)); label("$3$",(-1.2,0)); label("$pointer$",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label("$1$",midpoint((4,-1)--(4,-2)),W); label("$2$",midpoint((4,0)--(4,-1)),W); label("$3$",midpoint((4,1)--(4,0)),W); label("$4$",midpoint((4,2)--(4,1)),W); label("$1$",midpoint((4,-2)--(5,-2)),S); label("$2$",midpoint((5,-2)--(6,-2)),S); label("$3$",midpoint((7,-2)--(6,-2)),S); [/asy]](http://latex.artofproblemsolving.com/d/1/6/d16f5082e7ba12a9e32d1bd5aad7b6620c1985ae.png)
Solutions
Solution 1
When dividing each number on the wheel by the remainders are 
and 
Each column on the checkerboard is equally likely to be chosen.
When dividing each number on the wheel by 
the remainders are and 
The probability that a shaded square in the 
st or 
rd row of the st or rd column is chosen is
![\[\frac{2}{3} \times \frac{3}{6} = \frac{1}{3}\]](http://latex.artofproblemsolving.com/5/0/8/5085076484cb2b4c737dae329bacc077edc4922e.png)
The probability that a shaded square in the 
nd or 
th row of the nd column is chosen is
![\[\frac{1}{3} \times \frac{3}{6} = \frac{1}{6}\]](http://latex.artofproblemsolving.com/7/6/d/76da33bad7bfda26ad29f4bbdf0278c7bde0d552.png)
Add those two together to get
![\[\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \boxed{\textbf{(C)} \frac{1}{2}}\]](http://latex.artofproblemsolving.com/9/4/1/94112e565703159cbc8443b1d20984d0a6e60d34.png)
Solution 2
Alternatively, we may analyze this problem a little further.
First, we isolate the case where the rows are numbered 1 or 2. Notice that as listed before, the probability for picking a shaded square here is ![\[\frac{1}{2}\]](http://latex.artofproblemsolving.com/8/f/5/8f5f98380657e00df5d8281d93087ce901f6ec32.png)
because the column/row probabilities are the same, with the same number of shaded and non-shaded squares
Next we isolate the rows numbered 3 or 4. Note that the probability of picking the rows is same, because of our list up above. The columns, of course, still have the same probability. Because the number of shaded and non-shaded squares are equal, we have
Combining these we have a general probability of ![\[\boxed{\textbf{(C)} \frac{1}{2}}\]](http://latex.artofproblemsolving.com/b/c/f/bcfd67cc2601b2a56c84d402070819f0aa32614f.png)
Solution 3
Similarly to Solution 1, we create two lists, one for 
and one for 
: (
) and (
). Next, we make a list of the coordinates we want for the shaded boxes. (
), (
), (
), (
), (
), (
). For each coordinate pair, we find the probability for it to occur. In order, we get 
. Sum these to get the answer of 
~Dynosol
Solution 4
We take everything modulo first. We get mod , mod , mod , twice. Therefore, the probability of any given column is equal. We take everything modulo now. We have mod . and are different colored squares, and and are also different colored squares. Therefore, every single one of them has the same probability.
Therefore, our probability for a white square or a shaded square is equal, and therefore, the probability the square is shaded 
~Arcticturn
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
