Difference between revisions of "2023 AMC 12A Problems/Problem 5"

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Problem

Janet rolls a standard 6-sided die 4 times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal 3?

$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{49}{216}\qquad\textbf{(C) }\frac{25}{108}\qquad\textbf{(D) }\frac{17}{72}\qquad\textbf{(E) }\frac{13}{54}$

Solution 1 (Casework)

There are $3$ cases where the running total will equal $3$ ; one roll; two rolls; or three rolls:

Case 1: The chance of rolling a running total of $3$ in one roll is $\frac{1}{6}$ .

Case 2: The chance of rolling a running total of $3$ in two rolls is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$ since the dice rolls are a 2 and a 1 and vice versa.

Case 3: The chance of rolling a running total of 3 in three rolls is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$ since the dice values would have to be three ones.

Using the rule of sum, $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$ .

~walmartbrian ~andyluo ~DRBStudent

Solution 2 (Brute Force)

Because there is only a maximum of 3 rolls we must count (running total = 3 means there can't be a fourth roll counted), we can simply list out all of the probabilities.

If we roll a 1 on the first, the rolls that follow must be 2 or {1,1}, with the following results not mattering. This leaves a probability of $\frac{1}{6}\times\frac{1}{6} + \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} = \frac{7}{216}$ .

If we roll a 2 on the first, the roll that follows must be 2, resulting in a probability of $\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$ .

If we roll a 3 on the first, the rolls that follow do not matter, resulting in a probability of $\frac{1}{6}$ . Any roll greater than three will result in a running total greater than 3 no matter what, so those cases can be ignored. Summing the answers, we have $\frac{7}{36} + \frac{1}{216} + \frac{1}{36} = \frac{42+1+6}{216} = \boxed{\textbf{(B) }\frac{49}{216}}$ .

~Failure.net

Solution 4

Consider sequences of 4 integers with each integer between 1 and 6, the number of permutations of 6 numbers is $6^4=1296$ .

The following 4 types of sequences that might generate a running total of the numbers to be equal to 3 (x, y, or z denotes any integer between 1 and 6).

Sequence #1, (1, 1, 1, x): there are $6$ possible sequences.

Sequence #2, (1, 2, x, y): there are $6^2 = 36$ possible sequences.

Sequence #3, (2, 1, x, y): there are $6^2 = 36$ possible sequences.

Sequence #4, (3, x, y, z): there are $6^3 = 216$ possible sequences.

Out of 1296 possible sequences, there are a total of $6 + 36 + 36 + 216 = 294$ sequences that qualify. Hence, the probability is $294 / 1296 = \boxed{\textbf{(B) }\frac{49}{216}}$

~sqroot

2023 AMC 10A (Problems • Answer Key • Resources)
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2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 09:02, 10 November 2023 (view source) Andyluo (talk \| contribs) m (→‎Solution (casework)) ← Older edit		Revision as of 09:02, 10 November 2023 (view source) Andyluo (talk \| contribs) (→‎Solution 1 (Casework)) Newer edit →
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−	==Solution (Casework)==	+	==Solution 1 (Casework)==

	There are <math>3</math> cases where the running total will equal <math>3</math>; one roll; two rolls; or three rolls:		There are <math>3</math> cases where the running total will equal <math>3</math>; one roll; two rolls; or three rolls:

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