Q.Ravi, a Python programmer, is working on a project in which he wants to write a function to count the number of even and odd values in the list. He has written the following code but his code is having errors. Rewrite the correct code and underline the corrections made. define EOCOUNT(L): even_no=odd_no=0 for i in range(0,len(L)) if L[i]%2=0: even_no+=1 Else: odd_no+=1 print(even_no, odd_no)

Concept: Python function syntax, conditional statements, and comparison operators.

The code has four syntax errors that prevent it from running:

1. Python functions are declared with def, not define.
2. The for loop statement must end with a colon (:).
3. The equality comparison operator is ==, not = (which is assignment).
4. Python keywords like else are lowercase, not capitalized.

Here is the corrected code with errors underlined:

<u>def</u> EOCOUNT(L):
    even_no = odd_no = 0
    for i in range(0, len(L))<u>:</u>
        if L[i] % 2 <u>==</u> 0: …

The problem tests your understanding of Python function syntax and control structures. A function definition requires the def keyword, control statements (for, if) need colons, comparison uses == (not =), and Python keywords are case-sensitive.

Let me walk through each error and its fix:

1. Function definition keyword

   The line define EOCOUNT(L): is incorrect because Python uses def, not define, to declare functions. The correct syntax is def EOCOUNT(L):.

2. Missing colon after for statement

   The line for i in range(0,len(L)) is missing a colon at the end. Every compound statement in Python (loops, conditionals, function definitions) must end with a colon. It should be for i in range(0,len(L)):.

3. Assignment vs. comparison operator

   The condition if L[i]%2=0: uses a single = (assignment operator) instead of == (comparison operator). In Python, = assigns values while == tests equality. The correct condition is if L[i]%2==0:.

4. Incorrect capitalization of else

   Python keywords are case-sensitive and must be lowercase. Else: should be else:. …