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

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Solution 1
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==Solution 1==
  
 
There are 3 cases where the running total will equal 3; one roll; two rolls; or three rolls:
 
There are 3 cases where the running total will equal 3; one roll; two rolls; or three rolls:

Revision as of 16:58, 9 November 2023

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

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