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

Revision as of 23:31, 9 November 2023

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

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

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@@ Line 6: / Line 6: @@
-==Solution 1==
+==Solution==
-There are 3 cases where the running total will equal 3; 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:
 Case 1:
-The chance of rolling a running total of <math>3</math> in one roll is <math>1/6</math>.
+The chance of rolling a running total of <math>3</math> in one roll is <math>\frac{1}{6}</math>.
 Case 2:
-The chance of rolling a running total of <math>3</math> in two rolls is <math>1/6\cdot 1/6\cdot 2</math> since the dice rolls are a 2 and a 1 and vice versa.
+The chance of rolling a running total of <math>3</math> in two rolls is <math>\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}</math> 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 <math>1/6\cdot 1/6\cdot 1/6</math> since the dice values would have to be three ones.
+The chance of rolling a running total of 3 in three rolls is <math>\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}</math> since the dice values would have to be three ones.
-Using the rule of sum, <math>1/6 + 1/18 + 1/216 = 49/216</math> <math>\boxed{\textbf{(B) }\frac{49}{216}}</math>.
+Using the rule of sum, <math>\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}</math>.
-~walmartbrian ~andyluo
+~walmartbrian ~andyluo ~DRBStudent
-==Solution 2 (Slightly different to Solution 1)==
-There are 3 cases where the running total will equal 3.
-Case 1: Rolling a one three times
-Case 2: Rolling a one then a two
-Case 3: Rolling a three immediately
-The probability of Case 1 is <math>1/216</math>, the probability of Case 2 is (<math>1/36 * 2) = 1/18</math>, and the probability of Case 3 is <math>1/6</math>
-Using the rule of sums, adding every case gives the answer <math>\boxed{\textbf{(B) }\frac{49}{216}}</math>
-~DRBStudent
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Difference between revisions of "2023 AMC 12A Problems/Problem 5"

Revision as of 23:31, 9 November 2023

Problem

Solution

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