Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

Let $X_1$ be the outcome of the first roll, with $P(X_1 = k) = \frac{1}{6}$ for $k \in \{1, 2, 3, 4, 5, 6\}$.

• Case 1: $X_1 \in \{4, 5, 6\}$ The die is rolled only once. The total sum is $S = X_1$.

  • $S \ge 6$ occurs only when $X_1 = 6$.

  • $P(S \ge 6 \cap X_1 \in \{4, 5, 6\}) = P(X_1 = 6) = \frac{1}{6} = \frac{6}{36}$.

• Case 2: $X_1 \in \{1, 2, 3\}$ The die is rolled a second time. Let $X_2$ be the second roll outcome ($X_2 \in \{1, 2, 3, 4, 5, 6\}$). The total sum is $S = X_1 + X_2$.

  • If $X_1 = 1$: $S \ge 6 \implies X_2 \ge 5 \implies X_2 \in \{5, 6\}$ (2 outcomes). Probability = $\frac{1}{6} \times \frac{2}{6} = \frac{2}{36}$.

  • If $X_1 = 2$: $S \ge 6 \implies X_2 \ge 4 \implies X_2 \in \{4, 5, 6\}$ (3 outcomes). Probability = $\frac{1}{6} \times \frac{3}{6} = \frac{3}{36}$.

  • If $X_1 = 3$: $S \ge 6 \implies X_2 \ge 3 \implies X_2 \in \{3, 4, 5, 6\}$ (4 outcomes). Probability = $\frac{1}{6} \times \frac{4}{6} = \frac{4}{36}$.

Adding all disjoint cases together:

Hence, (B) is the correct option.