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Difference between revisions of "2023 AMC 10A Problems/Problem 7"
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~walmartbrian ~andyluo
 
~walmartbrian ~andyluo
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==Solution 2 (Slightly different to Solution 1)==
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There are 3 cases where the running total will equal 3.
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Case 1: Rolling a one three times
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Case 2: Rolling a one then a two
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Case 3: Rolling a three immediately
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The probability of Case 1 is <math>1/216, the probability of Case 2 is (</math>1/36 * 2) = <math>1/18, and the probability of Case 3 is 1/6
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Using the rule of sums, adding every case gets </math>49/216
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 +
~DRBStudent

Revision as of 19:36, 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

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 gets$ (Error compiling LaTeX. Unknown error_msg)49/216

~DRBStudent

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