1. **Problem statement:** A biased tetrahedral die with faces 0, 1, 2, 3 has probabilities $P(X=0)=\frac{1}{6}$, $P(X=1)=\frac{1}{6}$, $P(X=2)=\frac{1}{6}$, and $P(X=3)=\frac{1}{2}$. The final score $T$ is defined as: - If $X=3$, then $T=3$. - If $X \neq 3$, roll again and $T$ is the sum of the two rolls. We need to find $P(T=2)$. 2. **Formula and rules:** - Total probability rule. - Conditional probability. - Since $T$ depends on whether $X=3$ or not, split cases. 3. **Calculate $P(T=2)$:** - If $T=2$ and $X=3$, then $T=3$ always, so no contribution. - If $X \neq 3$, then $T = X + Y$ where $Y$ is the second roll independent with same distribution. So, $$P(T=2) = P(X \neq 3) \times P(X + Y = 2 \mid X \neq 3)$$ Since $P(X \neq 3) = 1 - \frac{1}{2} = \frac{1}{2}$. 4. **Find $P(X + Y = 2 \mid X \neq 3)$:** Possible $X$ values are 0,1,2 each with probability $\frac{1}{6}$, but conditional on $X \neq 3$, their conditional probabilities are: $$P(X=x \mid X \neq 3) = \frac{P(X=x)}{P(X \neq 3)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$ for $x=0,1,2$. 5. **Calculate $P(X+Y=2 \mid X \neq 3)$:** $$= \sum_{x=0}^2 P(X=x \mid X \neq 3) P(Y=2 - x)$$ Recall $P(Y=y) = P(X=y)$ since rolls are identical. Calculate each term: - For $x=0$: $P(X=0 \mid X \neq 3) = \frac{1}{3}$, $P(Y=2) = \frac{1}{6}$ - For $x=1$: $P(X=1 \mid X \neq 3) = \frac{1}{3}$, $P(Y=1) = \frac{1}{6}$ - For $x=2$: $P(X=2 \mid X \neq 3) = \frac{1}{3}$, $P(Y=0) = \frac{1}{6}$ Sum: $$\frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{6} = 3 \times \frac{1}{3} \times \frac{1}{6} = \frac{1}{6}$$ 6. **Final probability:** $$P(T=2) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$ **Answer:** $$\boxed{\frac{1}{12}}$$