To calculate the probability of rolling two consecutive threes in three rolls of a standard six-sided die, we can identify the successful outcomes out of all possible outcomes. ### 1. Total Possible Outcomes Since each die has 6 sides and you are rolling it 3 times, the total number of possible outcomes is: $6 \times 6 \times 6 = 216$ ### 2. Successful Outcomes We are looking for sequences where a "3" appears at least twice in a row. Let "X" be any number other than 3 (there are 5 such numbers: 1, 2, 4, 5, 6). The possible sequences are: * **3, 3, X**: The first two are 3s, and the third is not a 3. There are 5 such outcomes (3,3,1; 3,3,2; 3,3,4; 3,3,5; 3,3,6). * **X, 3, 3**: The first is not a 3, and the last two are 3s. There are 5 such outcomes (1,3,3; 2,3,3; 4,3,3; 5,3,3; 6,3,3). * **3, 3, 3**: All three are 3s. There is 1 such outcome. Note: We count (3,3,3) only once. If we simply added the first two groups (5 + 5 = 10), we would be counting (3,3,3) twice (once as "3,3,X" where X=3, and once as "X,3,3" where X=3). Total successful outcomes = $5 + 5 + 1 = 11$. ### 3. Probability The probability is the number of successful outcomes divided by the total number of outcomes: $\frac{11}{216}$ As a decimal, this is approximately **0.0509**, or about **5.09%**.