Measurement of 1 qubit in a two qubit system

If you had run this on the statevector_simulator then you will only see $|00rangle$. However, due to hardware noise, which in this case coming out from the measurement error, you will have some $|1rangle$ in your measurement on $q_1$. Depends on the hardware noise level, you might see 'more' or 'less' of $|1rangle$ appearing in your result. If you want to improve your output probabilities result, just add more shots.

Update: Why would we would we only see the state $|00rangle$ in the ideal world, like if we have used the statevector_simulator instead of the hardware?

The reason is because the starting/initial state on a quantum computer always start in the state $|000cdots 0rangle$. That is, every qubit initialize with $|0rangle$. And this state can be prepared with very high fidelity. In your case, you have two qubit, so your system is initialize at the state $|00rangle$. That is $|q_0rangle = |0rangle$ and $|q_1 rangle = |0 rangle$. So now without doing any operation on this qubit, you ask the quantum computer, what state is the second qubit, $q_1$ in? The answer should be $|0rangle$ if you have a measurement process with no noise! Because the state the qubit $q_1$, $|q_1rangle = |0rangle$. There is no other answer it can give you. However, if there is noise presence, it might change this answer from a $|0rangle$ to a $|1rangle$ every so often, depending on the noise level of course. Say you ask 100 times, 95 of the time you will get the correct answer of $|0rangle$ and 5 time it will return a $|1rangle$ due to error. Think of this as like when you are driving in car through a scenic mountain area, then your friend call and ask if you want to come over to hangout. Because of where you are, the signal is not very clear, so eventhough you say "YUP", he/she might hear as a "NOPE" instead.... so if you have say"YUP", "YUP",.... "YUP" 100 times, he/she might pick up a "NOPE" 5 time out of the 100 time. This is what you see in your histogram plot... sometime the measurement process pick-up a $|1rangle$ because of the noise in the measurement process as your friend has picked up a "NOPE" from bad signal.

Note that this should not be confusing with the state $|q_1 rangle = dfrac{|0rangle + |1 rangle}{sqrt{2}}$... In this case, when you ask the quantum computer, "Hey QC, what state is $|q_1rangle$ in? It will return you with a $|0rangle$ or a $|1rangle$ with $1/2$ probability. This is not because of the noise in your system. If you run this experiment $100,000$ time, you might get $49820$ time it will returns a $|0rangle$ and $50,180$ time it will returns a $|1rangle$. This is because your state $|q_1rangle$ is in superposition of both $|0rangle$ and $|1rangle$, so when making a measurement, it will collapse onto either one of these state. And in this case, where $|q_1 rangle = dfrac{|0rangle + |1 rangle}{sqrt{2}}$, the probability of collapsing onto the state $|0rangle$ is 50% probability and the state $|1rangle$ is also 50%. This is similar to tossing a coin if you think about it...