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

Revision as of 00:40, 21 November 2018

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, using stars-and-bars there are ${6\choose5}$ possible combinations of 5 heads. Continuing this pattern, we find that there are $\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144$ . 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 $9 + 64 = \boxed{073}$ .

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 $\frac{9}{64}$ . Our solution is $9+64=\boxed{073}$ .

Solution 3

We can also split the problem into casework.

Case 1: 0 Heads

There is only one possibility.

Case 2: 1 Head

There are 10 possibilities.

Case 3: 2 Heads

There are 36 possibilities.

Case 4: 3 Heads

There are 56 possibilities.

Case 5: 4 Heads

There are 35 possibilities.

Case 6: 5 Heads

There are 6 possibilities.

We have $1+10+36+56+35+6=144$ , and there are $1024$ possible outcomes, so the probability is $\frac{144}{1024}=\frac{9}{64}$ , so the answer is $\boxed{073}$ .

$\textbf{-RootThreeOverTwo}$

@@ Line 24: / Line 24: @@
 === Solution 3 ===
 We can also split the problem into casework.
-\newline Case 1: 0 Heads
-\newline There is only one possibility.
+Case 1: 0 Heads
-\newline Case 2: 1 Head
-\newline There are 10 possibilities.
+There is only one possibility.
-\newline Case 3: 2 Heads
-\newline There are 36 possibilities.
+Case 2: 1 Head
-\newline Case 4: 3 Heads
-\newline There are 56 possibilities.
+There are 10 possibilities.
-\newline Case 5: 4 Heads
-\newline There are 35 possibilities.
+Case 3: 2 Heads
-\newline Case 6: 5 Heads
-\newline There are 6 possibilities.
+There are 36 possibilities.
-\newline We have <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, so the answer is <math>\boxed{073}</math>.
+Case 4: 3 Heads
+There are 56 possibilities.
+Case 5: 4 Heads
+There are 35 possibilities.
+Case 6: 5 Heads
+There are 6 possibilities.
+We have <math>1+10+36+56+35+6=144</math>, and there are <math>1024</math> possible outcomes, so the probability is <math>\frac{144}{1024}=\frac{9}{64}</math>, so the answer is <math>\boxed{073}</math>.
 <math>\textbf{-RootThreeOverTwo}</math>

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