Final answer:
The probability of rolling a sum of 7 with two dice at least once in three rolls can be found using the complement rule, and it is 1 minus the probability of not rolling a sum of 7 in all three rolls (1 - (5/6)^3).
Step-by-step explanation:
The question pertains to the concept of the complement of an event in probability theory. We know that the probability of rolling a sum of 7 with two dice is 6 out of 36 or 1/6, since there are 6 favorable outcomes (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) out of a total of 36 possible outcomes when rolling two dice. To find the probability that at least one sum of 7 will occur when rolling the pair of dice 3 times, we'll use the complement rule.
The complement of rolling a sum of 7 at least once is not rolling a sum of 7 at all in the three rolls. The probability of not rolling a sum of 7 on one roll is 1 - 1/6 = 5/6. As the rolls are independent, the probability of not rolling a sum of 7 on three separate rolls is (5/6) x (5/6) x (5/6). Therefore, the probability of rolling a sum of 7 at least once in three rolls is 1 - (5/6)^3.