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2023 AMC 12A Problems/Problem 5: Difference between revisions

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problem 5
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soln 2 is exactly the same as soln 1
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==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:
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:
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:
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


==See Also==
==See Also==

Revision as of 22:31, 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

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 10 Problems and Solutions
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

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