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

Revision as of 20:42, 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 gives the answer $49/216$ $\boxed{(B)}$ .

~DRBStudent

@@ Line 33: / Line 33: @@
 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>
-Using the rule of sums, adding every case gets <math>49/216</math>
+Using the rule of sums, adding every case gives the answer <math>49/216</math> <math>\boxed{(B)}</math>.
 ~DRBStudent

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

Revision as of 20:42, 9 November 2023

Solution 1

Solution 2 (Slightly different to Solution 1)