Difference between revisions of "2020 AMC 10A Problems/Problem 13"
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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} * 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} * \frac{1}{2} = \frac{1}{8}</math>. | 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} * 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} * \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} * \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{(C) } \frac{5}{8}.}</math> ~Crypthes | + | |
| + | 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} * \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{(C) } \frac{5}{8}.}</math> ~ Crypthes | ||
==See Also== | ==See Also== | ||
Revision as of 21:41, 31 January 2020
- The following problem is from both the 2020 AMC 12A #11 and 2020 AMC 10A #13, so both problems redirect to this page.
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
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, 
~ Crypthes
See Also
| 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. 
