If we have the state $|\psi \rangle = \dfrac{1}{\sqrt{2}}|00\rangle + \dfrac{1}{2}|10\rangle - \dfrac{1}{2}|11\rangle$ then the probability of the second qubit being in the state $|0\rangle$ is the probability of the state $|\psi \rangle$ having $|0\rangle$ on the second qubit. In this case, it is from the states $|00\rangle$ and $|10\rangle$. So The probability of measuring the second qubit in the state $|0\rangle$ is $\bigg| \dfrac{1}{\sqrt{2}} \bigg|^2 + \bigg| \dfrac{1}{2} \bigg|^2 = \dfrac{3}{4} $.
You can also work this out more explicitly as well. That is, we have
$$
|\psi \rangle = \begin{pmatrix} 1/\sqrt{2} \ \ \\ 0 \\ 1/2 \\ -1/2 \end{pmatrix}
$$
We are looking for the probability that the second qubit is in the state $|0\rangle$ so the projective measurement $M$ is
$$
M = I \otimes |0\rangle \langle 0 | = \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
$$
and so according to Born's rule we have that the probability to measure the second qubit in the state $|0\rangle$ is
$$
\langle \psi | M | \psi \rangle = \begin{bmatrix} 1/\sqrt{2} & 0 & 1/2 &-1/2 \end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{bmatrix} \begin{bmatrix} 1/\sqrt{2} \ \ \\ 0 \\ 1/2 \\ -1/2 \end{bmatrix} = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}
$$
Also, the state post measurement is $|\psi_{post} \rangle = \dfrac{M|\psi\rangle}{\sqrt{3/4}}$.
You can extend this to the case where the first qubit is mesured in the state $|1\rangle$ too. In this case, the projective measurement $M = |1\rangle \langle 1| \otimes I$
