Difference between revisions of "2023 AMC 10A Problems/Problem 7"
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Case 3: Rolling a three immediately | Case 3: Rolling a three immediately | ||
− | The probability of Case 1 is <math>1/216, the probability of Case 2 is (< | + | 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 < | + | Using the rule of sums, adding every case gets <math>49/216</math> |
~DRBStudent | ~DRBStudent |
Revision as of 19:41, 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 , the probability of Case 2 is (, and the probability of Case 3 is
Using the rule of sums, adding every case gets
~DRBStudent