Difference between revisions of "2007 AMC 10B Problems/Problem 19"
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==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
When dividing each number on the wheel by <math>4,</math> the remainders are <math>1, 1, 2, 2, 3,</math> and <math>3.</math> Each column on the checkerboard is equally likely to be chosen. | When dividing each number on the wheel by <math>4,</math> the remainders are <math>1, 1, 2, 2, 3,</math> and <math>3.</math> Each column on the checkerboard is equally likely to be chosen. | ||
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<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> | ||
| − | ==Solution 2== | + | ==+Solution 2+== |
Alternatively, we may analyze this problem a little further. | Alternatively, we may analyze this problem a little further. | ||
Revision as of 22:40, 27 November 2011
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)
Solution
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
![\[\frac{2}{3} \times \frac{3}{6} = \frac{1}{3}\]](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
![\[\frac{1}{3} \times \frac{3}{6} = \frac{1}{6}\]](http://latex.artofproblemsolving.com/7/6/d/76da33bad7bfda26ad29f4bbdf0278c7bde0d552.png)
Add those two together
![\[\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \boxed{\textbf{(C)} \frac{1}{2}}\]](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)
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 | ||
