Question 272 marksShort AnswerSection B

$What possible output(s) from the given options will NOT be displayed when the following code is executed ? Also, mention, for how many iterations the for loop in the given code will run ? `` python import random a = [1,2,3,4,5,6] for i in range(4): j = random.randrange(i,5) print(a[j],end='-') print() ``$

Q: What possible output(s) from the given options will NOT be displayed when the following code is executed ? Also, mention, for how many iterations the for loop in the given code will run ? ```python import random a = [1,2,3,4,5,6] for i in range(4): j = random.randrange(i,5) print(a[j],end='-') print() ```

Option **(D) 5-1-2-4-** will NOT be displayed, and the `for` loop runs for exactly **4 iterations**. **Why:** `range(4)` gives `i = 0,1,2,3` (4 iterations). For each `i`, `j = random.randrange(i,5)` can only produce values from `i` up to `4` (5 excluded). So the value printed at each position is restricted to a shrinking range as `i` increases, and option D's second value (1) violates the valid range for `i=1`.

Python Libraries (random module) and Iterative Constructs

Random Module and Loop Tracing

Options

(A)

3-4-5-4-

(B)

2-2-4-5-

(C)

4-3-3-5-

(D)

5-1-2-4-

Official answer

Option (D) 5-1-2-4- will NOT be displayed, and the for loop runs for exactly 4 iterations . Why: range(4) gives i = 0,1,2,3 (4 iterations). For each i, j = random.randrange(i,5) can only produce values from i up to 4 (5 excluded). So the value printed at each position is restricted to a shrinking range as i increases, and option D's second value (1) violates the valid range for i=1 .

random.randrange()

range()

for loop

list indexing

code tracing

iteration count

random module

Marking scheme

1. $1 mark: correctly identifying option (D) '5-1-2-4-' as the output that will NOT be displayed, with valid justification.$
2. $1 mark: correctly stating that the for loop runs for 4 iterations.$

Hint

$Work out the valid range of a[j] separately for each loop iteration (i=0,1,2,3) using randrange(i,5), which excludes 5.$

Quick oral answer

$The loop runs 4 times since range(4) gives i=0 to 3; at each i, randrange(i,5) restricts a[j] to a shrinking set of valid values, and option D fails because its second value, 1, falls outside the valid range for i=1.$

Common mistakes

1. $Assuming randrange(i,5) can produce 5 itself — forgetting the stop value in randrange() is always excluded.$
2. $Checking only the first value of each option against {1,2,3,4,5} and ignoring that the valid range narrows for later positions (i=1,2,3).$
3. $Miscounting loop iterations as 5 instead of 4 by confusing range(4) with range(1,5).$

Analysis and explanation

\{0,1,2,3,4\}

Interesting facts

$Python's random.randrange(start, stop) mimics the same half-open interval convention as range() — stop is always excluded — a deliberate design choice for consistency across the language.$

$The random module in CPython uses the Mersenne Twister algorithm internally, which has a period of 2^19937 - 1, making its output sequence statistically excellent for simulations though not cryptographically secure.$

Question details

Q27 of 48 questions

Board: CBSE
Class: Class 12
Subject: Computer Science
Year: 2026
Section: Section B
Marks: 2
Type: Short Answer
Chapter: Python Libraries (random module) and Iterative Constructs
Topic: $Random Module and Loop Tracing$
Bloom's level: analyze
Syllabus topic: $Unit 1: Computational Thinking and Programming - 2 — Python Libraries: random.randrange() combined with for loops and list indexing$
Author: CBSE
Reviewed by: Padhantu Academic Team

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Frequently asked questions

$Does random.randrange(i,5) ever return 5?$

$No. Like range(), randrange(start, stop) always excludes the stop value, so randrange(i,5) can return values only from i up to 4.$

$How many times does the for loop in this code execute?$

$It executes 4 times because range(4) generates i = 0, 1, 2, 3.$