2023 AMC 10A Problems/Problem 7: Difference between revisions
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~walmartbrian ~andyluo | ~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 | |||
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?
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, 
.
~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 
1/36 * 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
