2023 AMC 10A Problems/Problem 7: Difference between revisions
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<math>\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}</math> | <math>\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}</math> | ||
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. | |||
Revision as of 16:32, 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?
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.
