Final answer:
The probability of rolling exactly two 1's on a fair six-sided die three times is 5/72, based on the binomial probability formula considering the number of ways to choose two successes and the probabilities of rolling or not rolling a 1.
Step-by-step explanation:
To calculate the probability of rolling exactly two 1's on a fair six-sided die three times, we can use the concept of binomial probability. The formula for the binomial probability of getting exactly k successes in n trials is P(X=k) = (n choose k)(p)^k(1-p)^(n-k), where (n choose k) is a combination representing the number of ways to choose k successes from n trials, p is the probability of success on any given trial, and (1-p) is the probability of failure.
In this case, the probability of rolling a 1 (p) on a single roll is 1/6, and not rolling a 1 has a probability of 5/6. Since we want exactly two 1's in three rolls (k=2 and n=3), we need to consider the different orders in which this can occur (for example, 1-1-Not1, 1-Not1-1, Not1-1-1). There are 3 ways (or (3 choose 2)) to roll two 1's in three rolls.
The binomial probability is therefore: P(2,3) = (3 choose 2)(1/6)^2(5/6) = 3 * (1/6)^2 * (5/6) = 3/36 * 5/6 = 5/72.
Thus, the correct answer is (d) 5/72.