scopes in python3: lists as parameters of a function?

I'm studying for a python3 exam and came across this question:

Given this snippet of code:

def my_function(my_list_1):
    print("Print #1:", my_list_1)
    print("Print #2:", my_list_2)
    del my_list_1[0]  # Pay attention to this line.
    print("Print #3:", my_list_1)
    print("Print #4:", my_list_2)


my_list_2 = [2, 3]
my_function(my_list_2)
print("Print #5:", my_list_2)

explain the output #3-#5:

Print #1: [2, 3] 
Print #2: [2, 3] 
Print #3: [3] 
Print #4: [3] 
Print #5: [3]

I'm a little perplexed by these questions:

1. how does python know who 'my_list_2' is, within the 'my_function' def block of code?

2. Suppose I changed the last 3 lines of code from my_list_2=[2,3] to pluto=[2,3]:

    pluto = [2, 3]
    my_function(pluto)
    print("Print #5:", pluto)
   

this would generate a NameError which I think makes sense -- so why does it not throw an Error in the initial code?

3. inside the def block, how does Print #4 for my_list_2 reflect the changes applied to my_list_1?

my_list_2 is a global variable. So it's accessible in your function. It's called LEGB in python which you can read about it here .

And list in python is a mutable variable type. When you pass my_list_2 as param to your function, if you make any changes to that, that would apply all over your code. Because it's mutable changes on it would happen in place.

When you put the list as a parameter, the interpreter puts the link to the object "list."

So, if you call

del my_list_1[0]

You modify the actual list you passed. So, as a result, both my_list_1 and my_list_2 lists are related to the same object, and you see one dropped element there.

An alternative code represent that is:

my_list_1 = [2, 3]
my_list_2 = my_list_1
del my_list_2[0]
print(my_list_1, my_list_2)
# [3] [3]