Difference between revisions of "2020 AMC 10A Problems/Problem 13"
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<math> \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 </math> | <math> \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 </math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is <math>\frac{1}{4} \cdot 1 = \frac{1}{4}</math>. If the frog goes to the right, it will be in the center of the square at <math>(2,2)</math>, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is <math>\frac{1}{2}</math>. The probability of this happening is <math>\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}</math>. | ||
+ | |||
+ | |||
+ | If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is <math>\frac{1}{2}</math>. Because there's a <math>\frac{1}{2}</math> chance of the frog going up and down, the total probability for this case is <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math> and summing up all the cases, <math>\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, we compute <math>1 \cdot C + \frac{1}{2} \cdot B</math>. | ||
+ | |||
+ | |||
+ | <cmath>1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}</cmath> | ||
+ | <cmath>\frac{1}{4} + \frac{3}{8}</cmath> | ||
+ | We get <math>\frac{5}{8}</math>, or <math>B</math> | ||
+ | <cmath>\text{O O O O O}</cmath> | ||
+ | <cmath>\text{O B O O O}</cmath> | ||
+ | <cmath>\text{C A B O O}</cmath> | ||
+ | <cmath>\text{O B O O O}</cmath> | ||
+ | <cmath>\text{O O O O O}</cmath> | ||
+ | -yeskay | ||
+ | |||
+ | ==Solution 3== | ||
+ | If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>. | ||
+ | - Lingjun | ||
+ | |||
+ | ==Solution 4 (Complete States)== | ||
+ | Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>. | ||
+ | Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: | ||
+ | <cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath> | ||
+ | <cmath>P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}</cmath> | ||
+ | <cmath>P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath> | ||
+ | <cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath> | ||
+ | We have a system of <math>4</math> equations in <math>4</math> variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives | ||
+ | <cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath> | ||
+ | <cmath>P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}</cmath> | ||
+ | Plugging in the third equation into this gives | ||
+ | <cmath>P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}</cmath> | ||
+ | <cmath>P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}</cmath> | ||
+ | Next, plugging in the second and third equation into the first equation yields | ||
+ | <cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath> | ||
+ | <cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath> | ||
+ | Now plugging in (*) into this, we get | ||
+ | <cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)</cmath> | ||
+ | <cmath>P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}</cmath> | ||
+ | -mathisawesome2169 | ||
+ | |||
+ | ==Solution 5 (Very fast)== | ||
+ | |||
+ | We can immediately note that the probability of landing on any lattice point is equal to the probability of landing on another. With this in mind, we see that each vertical "boundary" has 5 lattice points. (Remember, the "corners" count!) There are two boundary lines, so there are <math>2 \times 5 = 10</math> lattice points on our desired vertical boundary lines. The total amount of lattice points on the <math>4 \times 4</math> boundary is <math>16</math>. Using <math>\frac{P(desired)}{P(total)}</math>, we get <math>\frac{10}{16} = \boxed{\textbf{(B) }\frac{5}{8}}</math> <cmath>\phantom{}</cmath> | ||
+ | -hansenhe | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | [https://www.youtube.com/watch?v=ZGwAasE32Y4&t=280s IceMatrix's Solution (Starts at 4:40)] | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | |||
+ | https://youtu.be/qNaN0BlIsw0 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | On The Spot STEM | ||
+ | |||
+ | https://youtu.be/xGs7BjQbGYU | ||
+ | |||
+ | ==Video Solution 4== | ||
+ | https://youtu.be/0m4lbXSUV1I | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == Video Solution 5 == | ||
+ | https://youtu.be/IRyWOZQMTV8?t=5173 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
+ | {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}} | ||
{{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Introductory Probability Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:08, 17 January 2021
- The following problem is from both the 2020 AMC 12A #11 and 2020 AMC 10A #13, so both problems redirect to this page.
Contents
Problem 13
A frog sitting at the point begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices and . What is the probability that the sequence of jumps ends on a vertical side of the square
Solution 1
Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is . If the frog goes to the right, it will be in the center of the square at , and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is . The probability of this happening is .
If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is . Because there's a chance of the frog going up and down, the total probability for this case is and summing up all the cases, .
Solution 2
Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, we compute .
We get , or
-yeskay
Solution 3
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes . Since it starts on , there is a chance (up, down, or right) it will reach a diagonal on the first jump and chance (left) it will reach the vertical side. The probablity of landing on a vertical is . - Lingjun
Solution 4 (Complete States)
Let denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at . Note that by reflective symmetry over the line . Similarly, , and . Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: We have a system of equations in variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives Plugging in the third equation into this gives Next, plugging in the second and third equation into the first equation yields Now plugging in (*) into this, we get -mathisawesome2169
Solution 5 (Very fast)
We can immediately note that the probability of landing on any lattice point is equal to the probability of landing on another. With this in mind, we see that each vertical "boundary" has 5 lattice points. (Remember, the "corners" count!) There are two boundary lines, so there are lattice points on our desired vertical boundary lines. The total amount of lattice points on the boundary is . Using , we get -hansenhe
Video Solution 1
IceMatrix's Solution (Starts at 4:40)
Video Solution 2
Video Solution 3
On The Spot STEM
Video Solution 4
~savannahsolver
Video Solution 5
https://youtu.be/IRyWOZQMTV8?t=5173
~ pi_is_3.14
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.