Use map with operator.add:

>>> from operator import add
>>> list( map(add, list1, list2) )
[5, 7, 9]

or zip with a list comprehension:

>>> [sum(x) for x in zip(list1, list2)]
[5, 7, 9]

Timing comparisons:

>>> list2 = [4, 5, 6]*10**5
>>> list1 = [1, 2, 3]*10**5
>>> %timeit from operator import add;map(add, list1, list2)
10 loops, best of 3: 44.6 ms per loop
>>> %timeit from itertools import izip; [a + b for a, b in izip(list1, list2)]
10 loops, best of 3: 71 ms per loop
>>> %timeit [a + b for a, b in zip(list1, list2)]
10 loops, best of 3: 112 ms per loop
>>> %timeit from itertools import izip;[sum(x) for x in izip(list1, list2)]
1 loops, best of 3: 139 ms per loop
>>> %timeit [sum(x) for x in zip(list1, list2)]
1 loops, best of 3: 177 ms per loop

You could just use zip like,

>>> list1
[[1, 2, 3], [4, 5, 6]]
>>> list2
[[10, 2, 3], [11, 5, 6]]
>>> [[x+y for x,y in zip(l1, l2)] for l1,l2 in zip(list1,list2)]
[[11, 4, 6], [15, 10, 12]]

or if you are not sure, if both list will be of same length, then you can use zip_longest (izip_longest in python2) from itertools and use the fillvalue like,

>>> import itertools
>>> y = itertools.zip_longest([1,2], [3,4,5], fillvalue=0)
>>> list(y)
[(1, 3), (2, 4), (0, 5)]

that then you can use it for unequal sized data like,

>>> from itertools import zip_longest
>>> list1=[[1, 2, 3], [4, 5]]
>>> list2=[[10, 2, 3], [11, 5, 6], [1,2,3]]
>>> [[x+y for x,y in zip_longest(l1, l2, fillvalue=0)] for l1,l2 in zip_longest(list1,list2, fillvalue=[])]
[[11, 4, 6], [15, 10, 6], [1, 2, 3]]

Just do this:

[sum(x) for x in zip(*C)]

In the above, C is the list of c_1...c_n. As explained in the link in the comments (thanks, @kevinsa5!):

* is the "splat" operator: It takes a list as input, and expands it into actual positional arguments in the function call.

For additional details, take a look at the documentation, under "unpacking argument lists" and also read about calls (thanks, @abarnert!)

You can do this using list comprehension https://www.programiz.com/python-programming/list-comprehension

list_7 = [sum(inner_list) for inner_list in list_6]

Within the brackets ([]) you are iterating over each item in list_6. Since each item in list_6 is itself a list (inner_list) we can call python's sum function on the list to get the sum of the values within inner_list https://www.w3schools.com/python/ref_func_sum.asp.

I see now that you were looking for the sums to be lists themselves ([[13],[4],[5],...]],) in which case you would want to do:

list_7 = [[sum(inner_list)] for inner_list in list_6]

Putting brackets around sum(inner_list) creates a new list whose only entry is the sum of the inner_list.

Are you using Numpy?

If you are, I think you simply want np.sum(weighted_prob, axis=0)

You can try this:

L1 = [[1,2,3], [4,5,6], [7,8,9]]
L2 = [[10,20,30], [40,50,60], [70,80,90]]
final_list = [[c+d for c, d in zip(a, b)] for a, b in zip(L1, L2)]

Output:

[[11, 22, 33], [44, 55, 66], [77, 88, 99]]

Yes, you can 'add up' the sub lists to form new lists in d:

a = [[1,2,3], [4,5,6], [7,8,9]]
b = [[11,12,13], [14,15,16], [17,18,19]]
c = [[21,22,23], [24,25,26], [27,28,29]]

d = []
for a_i,b_i,c_i in zip(a,b,c):
    d.append(a_i + b_i + c_i)

print(d)

Output as requested.

In fact, you can use the built-in sum():

d = []
for items in zip(a, b, c):
    d.append(sum(items, start=[]))

print(d)