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Difference between revisions of "2023 AMC 10A Problems/Problem 7"

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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>
 
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>
  
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Using the rule of sums, adding every case gives the answer <math>49/216</math> <math>\boxed{(B)}</math>.
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Using the rule of sums, adding every case gives the answer \boxed{(B)49/216}$
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~DRBStudent
 
~DRBStudent

Revision as of 20:33, 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 $1/6 * 1/6 * 2$ 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 $1/6 * 1/6 * 1/6$ since the dice values would have to be three ones.

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

~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{(B)49/216}$


~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

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