2023 AMC 10A Problems/Problem 7: Difference between revisions
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Case 1: | Case 1: | ||
The chance of rolling a running total of 3 in one roll is 1/6. | The chance of rolling a running total of <math>3</math> in one roll is <math>1/6</math>. | ||
Case 2: | 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. | 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: | 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. | 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, 1/6 + 1/18 + 1/216 = 49/216. | Using the rule of sum, <math>1/6 + 1/18 + 1/216 = 49/216</math>. | ||
Revision as of 16:34, 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 
in one roll is 
.
Case 2:
The chance of rolling a running total of in two rolls is 
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 
since the dice values would have to be three ones.
Using the rule of sum, 
.
