2020 AMC 10A Problems/Problem 13

Revision as of 21:41, 31 January 2020 by Crypthes (talk | contribs) (Solution)
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 $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$, 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 $(0,0), (0,4), (4,4),$ and $(4,0)$. What is the probability that the sequence of jumps ends on a vertical side of the square$?$

$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$

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 $\frac{1}{4} * 1 = \frac{1}{4}$. If the frog goes to the right, it will be in the center of the square at $(2,2)$, 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 $\frac{1}{2}$. The probability of this happening is $\frac{1}{4}$ * \frac{1}{2} = \frac{1}{8}$.

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$ (Error compiling LaTeX. ! Missing $ inserted.)\frac{1}{2}$. Because there's a$\frac{1}{2}$chance of the frog going up and down, the total probability for this case is$\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$and summing up all the cases,$\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(C) } \frac{5}{8}.}$ ~Crypthes

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS