|
|
| (7 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | 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?
| + | #redirect[[2023 AMC 12A Problems/Problem 5]] |
| − | | |
| − | <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 <math>3</math> in one roll is <math>1/6</math>.
| |
| − | | |
| − | Case 2:
| |
| − | The chance of rolling a running total of <math>3</math> in two rolls is <math>1/6 * 1/6 * 2</math> 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 <math>1/6 * 1/6 * 1/6</math> since the dice values would have to be three ones.
| |
| − | | |
| − | Using the rule of sum, <math>1/6 + 1/18 + 1/216 = 49/216</math> <math>\boxed{(B)}</math>.
| |
| − | | |
| − | ~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 <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
| |
| − | | |
| − | Using the rule of sums, adding every case gets </math>49/216
| |
| − | | |
| − | ~DRBStudent
| |
