A fair die is tossed twice. We want to find the probability of getting a sum of at least 10.

1. Problem Description

A fair die is tossed twice. We want to find the probability of getting a sum of at least

1

0.

2. Solution Steps

Let the outcome of the two tosses be $(x, y)$, where $x$ and $y$ are integers from 1 to

6. The total number of possible outcomes is $6 \times 6 = 36$. We want to find the number of outcomes where $x + y \ge 10$. We can enumerate the possible outcomes:

\begin{itemize}

\item If $x = 4$, then $y \ge 6$, so $y = 6$. The outcome is $(4, 6)$.

\item If $x = 5$, then $y \ge 5$, so $y = 5, 6$. The outcomes are $(5, 5), (5, 6)$.

\item If $x = 6$, then $y \ge 4$, so $y = 4, 5, 6$. The outcomes are $(6, 4), (6, 5), (6, 6)$.

\end{itemize}

These are all the possible outcomes where the sum is at least

1

0. The total number of such outcomes is $1 + 2 + 3 = 6$.

The probability of the sum being at least 10 is the number of favorable outcomes divided by the total number of possible outcomes:

$P(x+y \ge 10) = \frac{6}{36} = \frac{1}{6}$.

3. Final Answer

D. $\frac{1}{6}$