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

Revision as of 22:16, 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 1

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 $1/6$ .

Case 2: The chance of rolling a running total of $3$ in two rolls is $1/6\cdot 1/6\cdot 2$ 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 $1/6\cdot 1/6\cdot 1/6$ since the dice values would have to be three ones.

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

~walmartbrian ~andyluo

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 $1/216$ , the probability of Case 2 is ( $1/36 * 2) = 1/18$ , and the probability of Case 3 is $1/6$

Using the rule of sums, adding every case gives the answer $\boxed{\textbf{(B) }\frac{49}{216}}$

~DRBStudent

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

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

@@ Line 41: / Line 41: @@
 ==See Also==
 {{AMC10 box|year=2023|ab=A|num-b=6|num-a=8}}
-{{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
+{{AMC12 box|year=2023|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}

Page

Toolbox

Search

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

Revision as of 22:16, 9 November 2023

Contents

Problem

Solution 1

Solution 2 (Slightly different to Solution 1)

See Also