Difference between revisions of "2019 AIME II Problems/Problem 2"

Revision as of 20:18, 22 March 2019

Problem 2

Lily pads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$ . Then $P_7 = 1$ , $P_6 = \frac12$ , and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$ . Working our way down, we find $P_5 = \frac{3}{4}$ $P_4 = \frac{5}{8}$ $P_3 = \frac{11}{16}$ $P_2 = \frac{21}{32}$ $P_1 = \frac{43}{64}$ $43 + 64 = \boxed{107}$ .

Solution 2(Casework)

Define a one jump to be a jump from k to K + 1 and a two jump to be a jump from k to k + 2. Case 1: (6 one jumps) (1/2)^6 = 1/64 Case 2: (4 one jumps and 1 two jumps) 5C1 x (1/2)^5 = 5/32 Case 3: (2 one jumps and 2 two jumps) 4C2 x (1/2)^4 = 3/8 Case 4: (3 two jumps) (1/2)^3 = 1/8 Summing the probabilities gives us 43/64 so the answer is \boxed{107}.

@@ Line 18: / Line 18: @@
 Case 3: (2 one jumps and 2 two jumps) 4C2 x (1/2)^4 = 3/8
 Case 4: (3 two jumps) (1/2)^3 = 1/8
-Summing the probabilities gives us 43/64 so the answer is \boxed{107}$.
+Summing the probabilities gives us 43/64 so the answer is \boxed{107}.
 ==See Also==
 {{AIME box|year=2019|n=II|num-b=1|num-a=3}}
 {{MAA Notice}}

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Difference between revisions of "2019 AIME II Problems/Problem 2"

Revision as of 20:18, 22 March 2019

Contents

Problem 2

Solution

Solution 2(Casework)

See Also