Difference between revisions of "2007 AMC 10B Problems/Problem 19"

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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>
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==Solution 3==
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Similarly to Solution 1, we create two lists, one for mod 4 and one for mod 5.(<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 and you get your answer of <math> \boxed{\textbf{(C)} \frac{1}{2}} </math>
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~Dynosol
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:27, 24 January 2021

Problem

The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ 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?

[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]

$\textbf{(A) } \frac{1}{3} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{5}{9} \qquad\textbf{(E) } \frac{2}{3}$

Solutions

Solution 1

When dividing each number on the wheel by $4,$ the remainders are $1, 1, 2, 2, 3,$ and $3.$ Each column on the checkerboard is equally likely to be chosen.

When dividing each number on the wheel by $5,$ the remainders are $1, 1, 2, 2, 3,$ and $4.$

The probability that a shaded square in the $1$st or $3$rd row of the $1$st or $3$rd column is chosen is \[\frac{2}{3} \times \frac{3}{6} = \frac{1}{3}\]

The probability that a shaded square in the $2$nd or $4$th row of the $2$nd column is chosen is \[\frac{1}{3} \times \frac{3}{6} = \frac{1}{6}\]

Add those two together and you get \[\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \boxed{\textbf{(C)} \frac{1}{2}}\]

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 \[\frac{1}{2}\] 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 \[\frac{1}{2}\] Combining these we have a general probability of \[\boxed{\textbf{(C)} \frac{1}{2}}\]

Solution 3

Similarly to Solution 1, we create two lists, one for mod 4 and one for mod 5.($1, 2, 3, 2, 3, 1$) and ($1, 2, 3, 1, 2, 4$). Next, we make a list of the coordinates we want for the shaded boxes. ($1,1$), ($1,3$), ($2,2$), ($3,1$), ($3,3$), ($2,4$). For each coordinate pair, we find the probability for it to occur. In order, we get $\frac{1}{9}, \frac{1}{18}, \frac{1}{9}, \frac{1}{9}, \frac{1}{18}, \frac{1}{18}$. Sum these and you get your answer of $\boxed{\textbf{(C)} \frac{1}{2}}$

~Dynosol

See Also

2007 AMC 10B (ProblemsAnswer KeyResources)
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

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