Given:
- $X$ takes values from 0 to 9 with equal probability $\frac{1}{10}$.
- $Y=X \bmod 3$.
Calculate $P(Y=0)+P(Y=1)$.
Step 1: $P(Y=0)$
$$
\begin{gathered}
P(Y=0)=P(X=0)+P(X=3)+P(X=6)+P(X=9) \\
P(Y=0)=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{4}{10}
\end{gathered}
$$
Step 2: $P(Y=1)$
$$
\begin{gathered}
P(Y=1)=P(X=1)+P(X=4)+P(X=7) \\
P(Y=1)=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{3}{10}
\end{gathered}
$$
Step 3: $P(Y=0)+P(Y=1)$
$$
P(Y=0)+P(Y=1)=\frac{4}{10}+\frac{3}{10}=\frac{7}{10}
$$
Thus, the correct answer is $\frac{7}{10}$.
