Difference between revisions of "1990 AIME Problems/Problem 9"

Revision as of 13:27, 15 October 2007

Problem

A fair coin is to be tossed $10_{}^{}$ times. Let $i/j^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$ .

Solution

Solution 1

Clearly, at least $5$ tails must be flipped; any less, then by the pigeonhole principle there will be heads that appear on consecutive tosses.

Consider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(H)$ :

$(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)$

There are six slots for the heads to be placed, but only $5$ heads remaining. Thus, there are ${6\choose5}$ possible combinations of 5 heads. Continuing this pattern, we find that there are $\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 6 + 35 + 56 + 36 + 10 + 1 = 144$ . There are a total of $2^{10}$ possible flips of $10$ coins, making the probability $\frac{144}{1024} = \frac{9}{64}$ . Thus, our solution is $9 + 64 = 73$ .

Solution 2

Call the number of ways of flipping $n$ coins and not receiving any consecutive heads $S_n$ . Notice that tails must be received in at least one of the first two flips.

If the first coin flipped is a T, then the remaining $n-1$ flips must fall under one of the configurations of $S_{n-1}$ .

If the first coin flipped is a H, then the second coin must be a T. There are then $S_{n-2}$ configurations.

Thus, $S_n = S_{n-1} + S_{n-2}$ . By counting, we can establish that $S_1 = 2$ and $S_2 = 3$ . Therefore, $S_3 = 5,\ S_4 = 8$ , forming the Fibonacci sequence. Listing them out, we get $2,3,5,8,13,21,34,55,89,144$ , and the 10th number is $144$ . Putting this over $2^{10}$ to find the probability, we get $\frac{9}{64}$ .

@@ Line 11: / Line 11: @@
 :<math>(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)</math>
-There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\displaystyle\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 6 + 35 + 56 + 36 + 10 + 1 = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = 73</math>.
+There are six slots for the heads to be placed, but only <math>5</math> heads remaining. Thus, there are <math>{6\choose5}</math> possible combinations of 5 heads. Continuing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 6 + 35 + 56 + 36 + 10 + 1 = 144</math>. There are a total of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = 73</math>.
 === Solution 2 ===

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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