Bitwise AND(&) for numpy array - [True,True] and [True,False]

Question

Why should we use & (bitwise AND operator) when working with two numpy arrays? At the same time logical AND is enough when working with two lists.

For example:
a=[True,True]
b=[True,False]
>> a and b:
[True, False]
>> b and a:
[True, True] # why??

If we create the same numpy arrays:
a_arr=np.array(a)
b_arr=np.array(b)
>> a_arr and b_arr
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>a_arr & b_arr
array([ True, False])
>>b_arr & a_arr
array([ True, False])

Answer 1

1) a and b will print b, b and a will print b
When you use and between lists, then it will return the last list.
For example:
e = a and b # assigns b to e
d = b and a # assigns a to d

Boolean operations work differently on lists, but there are built in functions like the any() and all(), for example:
all(a) and all(b) == all(b) and all(a)

2) I don't know much about numpy, but maybe it doesn't even allow boolean comparison between it array objects, to remove confusion