# What is the probability that in 3 rolls of a pair of six-sided dice, exactly one total of 7 is rolled?

I tried to solve the exercise and it seems meaningful, but the problem is that in the book the result is $$25/72$$, while mine is $$1/7776$$. I don't know if maybe I didn't understand the problem, but by the text I understood that there are two dices rolled three times, and I have to calculate the probability to have 7 as result. Now, I could use the binomial probability density function, but I didn't know very well how to use the data I have (if someone can also explain me the resolution with this structure I'll thank him). By logic, throwing three times two dices, the total number of possibilities is 46'656, because I have $$36^3$$. Now, to have 7 as result and a dice must have at least 1 as value, the only possibility that I have to reach 7 is 11/11/12 (the first two dices are 1, the 3th and 4th 1, the 5th 1 and the 7th 2) and all the combination of this sequence: 11/11/12 11/11/21 11/21/11 11/12/11 21/11/11 12/11/11

Calculating the probability, I have $$3* (1/36 * 1/36 * 2/36)$$, and the probability of having 7 as result is $$6/7776$$

I don't know where I'm wrong.