Q.What possible output from the given options is expected to be displayed when the following code is executed?

import random  
Cards=["Heart","Spade","Club","Diamond"]  
for i in range(2):  
    print(Cards[random.randint(1,i+2)],end="#")  

(A) Spade#Diamond# (B) Spade#Heart# (C) Diamond#Club# (D) Heart#Spade#

The code uses random.randint(1, i+2) inside a loop that runs twice. In the first iteration, i = 0, so random.randint(1, 2) picks either index 1 or 2. That means the first card printed is always either "Spade" (index 1) or "Club" (index 2). In the second iteration, i = 1, so random.randint(1, 3) picks any index from 1 to 3 — that is, "Spade", "Club", or "Diamond". The key point: index 0 ("Heart") is never chosen in either iteration because the lower bound of randint is 1, not 0.

Now check the options:

• (A) Spade#Diamond# — possible (first pick Spade, second pick Diamond).
• (B) Spade#Heart# — impossible, because Heart (index 0) is never selected. …

Let’s walk through what this code actually does, step by thought.

The list Cards has four elements:

"Heart" at index 0, "Spade" at index 1, "Club" at index 2, "Diamond" at index 3.

The loop runs exactly twice (for i in range(2)). Inside, the code picks a random integer using random.randint(1, i+2) and uses that integer as an index into Cards.

The critical detail is the range of indices that randint can produce in each iteration.

First iteration (i = 0):

random.randint(1, 0+2) = random.randint(1, 2).

So the possible indices are 1 or 2 only.

Index 1 → "Spade"

Index 2 → "Club"

Index 0 ("Heart") and index 3 ("Diamond") are impossible in the first draw.

Second iteration (i = 1):

random.randint(1, 1+2) = random.randint(1, 3).

Now the possible indices are 1, 2, or 3.

Index 1 → "Spade"

Index 2 → "Club"

Index 3 → "Diamond"

Again, index 0 ("Heart") is impossible.

Now look at the options:

(A) Spade#Diamond# — possible? First draw Spade (index 1) is allowed. Second draw Diamond (index 3) is allowed. So this is possible. …