Final answer:
The probability of rolling at least one 3 when rolling a fair six-sided die twice is found by calculating the opposite event's probability (no 3s in two rolls) and subtracting it from 1. The probability of no 3s is (5/6) x (5/6) = 25/36, so the desired probability is 1 - 25/36 = 11/36.
Step-by-step explanation:
The student has asked about the probability of observing at least one 3 when rolling a die twice. To solve this problem, we'll look at the scenario where a 3 does not occur in either roll, which is easier to calculate, and then subtract that probability from 1 to find the probability of the opposite event, which is rolling at least one 3.
For one die roll, the probability of not rolling a 3, which we denote as q, is 5/6 because 5 outcomes are not a 3 out of 6 possible outcomes. When we roll the die twice, we multiply the individual probabilities of each roll to find the combined probability of not rolling a 3 both times. So q = (5/6) x (5/6) = 25/36. The probability of rolling at least one 3, denoted as 1 - q, is then 1 - 25/36 = 11/36. Therefore, the probability of rolling at least one 3 when rolling a die twice is 11/36.