If we have the state $|\psi \rangle = a_{00}|00\rangle +a_{01}|01\rangle +a_{10}|10\rangle +a_{11}|11\rangle $ then the probability of the second qubit being in the state $|1\rangle$ is the probability of the state $|\psi \rangle$ having $|1\rangle$ on the second qubit. In this case, it is from the states $|01\rangle$ and $|11\rangle$. So The probability of measuring the second qubit in the state $|1\rangle$ is $\bigg| a_{01} \bigg|^2 + \bigg| a_{11} \bigg|^2 $.
Similarly, the probability of the second qubit of the state $|\psi\rangle$ being measured in the state $|0\rangle$ is then $\bigg| a_{00} \bigg|^2 + \bigg| a_{10} \bigg|^2 $.
You can work this out explicitly as well. First, we have
$$
|\psi \rangle = a_{00}|00\rangle +a_{01}|01\rangle +a_{10}|10\rangle +a_{11}|11\rangle = \begin{pmatrix} a_{00} \ \ \\ a_{01} \\ a_{10} \\ a_{11} \end{pmatrix}
$$
since we taken $|0\rangle = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1\end{pmatrix}$. The computational basis. And now we are looking for the probability that the second qubit is in the state $|0\rangle$ and ignore the first qubit, then the measurement $M$ can be described as
$$
M = I \otimes |0\rangle \langle 0 | = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}= \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
$$
where $I$ corresponds to the identity operator, and $|0\rangle \langle 0|$ corresponds to the outer product operation.
And so according to Born's rule the probability to measure the second qubit in the state $|0 \rangle$ is
$$
\langle \psi | M | \psi \rangle = \begin{pmatrix} a_{00}^* & a_{01}^* & a_{10}^* & a_{11}^* \end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{pmatrix} \begin{pmatrix} a_{00} \ \ \\ a_{01} \\ a_{10} \\ a_{11} \end{pmatrix} = |a_{00}|^2 + |a_{10}|^2
$$
here $a^*$ indicates the conjugate of $a$ and hence $a^*a = |a|^2$.
Now if you want to construct $M$ for the second qubit being measured in the state $1\rangle$ and not measuring the first qubit then you can do construct it as $M = I \otimes |1\rangle \langle 1|$. Where $|1\rangle \langle 1| = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} $
