Rolling two dice to find the probability of the sum of the answers being less than one

Question

A die is thrown twice. Determine the probability that the sum of the rolls is less than $4$ given that at least one of the rolls is a $1$. I know that the answer is $\frac{3}{11}$. However I need to know how to get that.

Answer 1

The number of total possibilities is $11$ (you can count $(1, 1), (1, 2), \dots, (1, 6)$ and $(1, 1), (2, 1), \dots, (6, 1)$ which are $12$ total but we doublecount $(1,1)$ so we subtract $1$).

If we assign $1$ as the first roll of the dice, then the other must be either $1$ or $2$, or else the sum is $4$ or more. If we assign $1$ as the second roll, same thing, the first must be $1$ or $2$. However, we doublecount $(1, 1)$, so we subtract $1$. That's $3$ total.

Therefore the probability is $$\boxed{\frac{3}{11}.}$$

Answer 2

We are given that at least one of the rolls is a $1$, so assuming this is a six-sided die, we get that there are $11$ total possibilities for two rolls($5$ for if the first roll is a $1$, $5$ for if the second roll is a $1$, and $1$ for if both of them are a $1$). Then the number of possibilities that satisfy the condition is $3$: $(1,1)$, $(1,2)$, and $(2,1)$. Therefore the probability is $\frac{\text{# of possibilities that satisfy the condition}}{\text{# of total possibilities}}=\frac3{11}$.