Probability of measuring one qubit from the state of two qubits

If we have the state $|psi rangle = a_{00}|00rangle +a_{01}|01rangle +a_{10}|10rangle +a_{11}|11rangle $ then the probability of the second qubit being in the state $|1rangle$ is the probability of the state $|psi rangle$ having $|1rangle$ on the second qubit. In this case, it is from the states $|01rangle$ and $|11rangle$. So The probability of measuring the second qubit in the state $|1rangle$ is $bigg| a_{01} bigg|^2 + bigg| a_{11} bigg|^2 $.

Similarly, the probability of the second qubit of the state $|psirangle$ being measured in the state $|0rangle$ is then $bigg| a_{00} bigg|^2 + bigg| a_{10} bigg|^2 $.

You can work this out explicitly as well. First, we have

$$ |psi rangle = a_{00}|00rangle +a_{01}|01rangle +a_{10}|10rangle +a_{11}|11rangle = begin{pmatrix} a_{00} a_{01} a_{10} a_{11} end{pmatrix} $$ since we taken $|0rangle = begin{pmatrix} 1 0end{pmatrix}$ and $|1rangle = begin{pmatrix} 0 1end{pmatrix}$. The computational basis. And now we are looking for the probability that the second qubit is in the state $|0rangle$ and ignore the first qubit, then the measurement $M$ can be described as

$$ M = I otimes |0rangle langle 0 | = begin{pmatrix} 1 & 0 0 & 1end{pmatrix} otimes begin{pmatrix} 1 & 0 0 & 0end{pmatrix}= begin{pmatrix} 1 & 0 & 0 & 0 0 & 0 & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 0 end{pmatrix} $$ where $I$ corresponds to the identity operator, and $|0rangle langle 0|$ 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 $1rangle$ and not measuring the first qubit then you can do construct it as $M = I otimes |1rangle langle 1|$. Where $|1rangle langle 1| = begin{pmatrix} 0 1 end{pmatrix} begin{pmatrix} 0 & 1 end{pmatrix} = begin{pmatrix} 0 & 0 0 & 1end{pmatrix} $