Generating random samples in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of numpy random choice with replace=False

What is the output of this code snippet?

import numpy as np
np.random.seed(0)
sample = np.random.choice([10, 20, 30, 40, 50], size=3, replace=False)
print(sample)

NumPy

import numpy as np
np.random.seed(0)
sample = np.random.choice([10, 20, 30, 40, 50], size=3, replace=False)
print(sample)

A [20 40 50]

B [50 20 10]

C [10 20 30]

D [40 10 30]

💡 Hint

Remember that setting the seed fixes the random output.

With seed 0 and replace=False, numpy picks 3 unique elements in a fixed order. The output is [20 40 50].

❓ data_output

intermediate

Shape of random normal samples array

What is the shape of the array produced by this code?

import numpy as np
samples = np.random.normal(loc=5, scale=2, size=(4,3))
print(samples.shape)

NumPy

import numpy as np
samples = np.random.normal(loc=5, scale=2, size=(4,3))
print(samples.shape)

A (4, 3)

B (3, 4)

C (12,)

D (4,)

💡 Hint

The size parameter defines the shape of the output array.

The size=(4,3) means 4 rows and 3 columns, so shape is (4, 3).

❓ visualization

advanced

Histogram of uniform random samples

Which option shows the correct histogram plot for 1000 samples from a uniform distribution between 0 and 1?

NumPy

import numpy as np
import matplotlib.pyplot as plt
np.random.seed(1)
samples = np.random.uniform(0, 1, 1000)
plt.hist(samples, bins=10, edgecolor='black')
plt.show()

A Histogram with bars increasing from left to right

B Histogram with one tall bar at 0 and others near zero

C Histogram with a bell shape centered at 0.5

D Histogram with bars roughly equal height across bins

💡 Hint

Uniform distribution means equal chance for all values in the range.

Uniform samples spread evenly, so histogram bars are roughly equal height.

🧠 Conceptual

advanced

Effect of random seed on reproducibility

Why do we set a random seed before generating random samples in numpy?

A To limit the range of random numbers generated

B To ensure the random samples are the same every time the code runs

C To speed up the random number generation

D To make the samples more random

💡 Hint

Think about reproducibility in experiments.

Setting a seed fixes the random number generator state, so results repeat exactly.

🔧 Debug

expert

Identify the error in random sample generation

What error does this code raise?

import numpy as np
np.random.seed(2)
sample = np.random.choice([1, 2, 3], size=5, replace=False)
print(sample)

NumPy

import numpy as np
np.random.seed(2)
sample = np.random.choice([1, 2, 3], size=5, replace=False)
print(sample)

A TypeError: 'replace' argument must be boolean

B SyntaxError: invalid syntax

C ValueError: Cannot take a larger sample than population when 'replace=False'

D No error, prints 5 unique samples

💡 Hint

Check if sample size is larger than population when replace=False.

Sampling 5 unique items from a list of 3 without replacement is impossible, causing ValueError.

Practice

1. What does the numpy.random.choice function do?

easy

A. It calculates the mean of an array.

B. It selects random elements from a given array or list.

C. It sorts an array in ascending order.

D. It reshapes an array into a new shape.

Solution

1. Step 1: Understand the function purpose

   numpy.random.choice is designed to pick random elements from a given array or list.

2. Step 2: Compare with other options

   Sorting, calculating mean, and reshaping are different numpy functions, not related to random sampling.

3. Final Answer:

   It selects random elements from a given array or list. -> Option B

4. Quick Check:

   Random sampling = selecting elements randomly [OK]

Hint: Remember: choice means picking randomly from data [OK]

Common Mistakes:

• Confusing choice with sorting or reshaping functions
• Thinking it calculates statistics like mean
• Assuming it modifies array shape

2. Which of the following is the correct syntax to randomly select 3 elements from array arr without replacement using numpy?

easy

A. numpy.random.choice(arr, 3, replace=True)

B. numpy.choice(arr, size=3, replace=False)

C. numpy.random.choice(arr, size=3, replace=False)

D. numpy.random.choice(arr, size=3, replace=True)

Solution

1. Step 1: Identify correct function and parameters

   The function is numpy.random.choice. To select 3 elements without replacement, use size=3 and replace=False.

2. Step 2: Check each option

   numpy.random.choice(arr, size=3, replace=False) uses correct function and parameters. numpy.random.choice(arr, 3, replace=True) uses replacement True (wrong). numpy.choice(arr, size=3, replace=False) uses wrong function name. numpy.random.choice(arr, size=3, replace=True) uses replacement True (wrong).

3. Final Answer:

   numpy.random.choice(arr, size=3, replace=False) -> Option C

4. Quick Check:

   Correct syntax = choice + size + replace=False [OK]

Hint: Use replace=False to avoid repeated picks [OK]

Common Mistakes:

• Using replace=True when no repeats wanted
• Misspelling function name as numpy.choice
• Passing size as positional without keyword

3. What is the output of this code?

import numpy as np
np.random.seed(0)
arr = np.array([10, 20, 30, 40])
sample = np.random.choice(arr, size=2, replace=False)
sample_sorted = np.sort(sample)
sample_sorted.tolist()

medium

A. [10, 40]

B. [10, 20]

C. [20, 40]

D. [30, 40]

Solution

1. Step 1: Understand random seed and choice

   Setting seed to 0 fixes randomness. Using choice with size=2 and replace=False picks 2 unique elements from [10,20,30,40].

2. Step 2: Determine chosen elements and sort

   With seed 0, the chosen elements are [10, 40]. Sorting gives [10, 40].

3. Final Answer:

   [10, 40] -> Option A

4. Quick Check:

   Seed 0 + choice + sort = [10, 40] [OK]

Hint: Seed fixes output; sort to order chosen elements [OK]

Common Mistakes:

• Ignoring seed and expecting different output
• Not sorting before converting to list
• Assuming replacement allows duplicates

4. The following code throws an error. What is the cause?

import numpy as np
arr = np.array([1, 2, 3])
sample = np.random.choice(arr, size=5, replace=False)

medium

A. Incorrect function name used.

B. Array contains integers instead of floats.

C. Missing import statement for numpy.

D. Size is larger than array length without replacement.

Solution

1. Step 1: Analyze parameters and array size

   The array has 3 elements, but size=5 is requested without replacement.

2. Step 2: Understand replacement=False effect

   Without replacement, you cannot pick more elements than exist. This causes a ValueError.

3. Final Answer:

   Size is larger than array length without replacement. -> Option D

4. Quick Check:

   Sampling more than available without replace=False causes error [OK]

Hint: Check if sample size > array length when replace=False [OK]

Common Mistakes:

• Assuming replacement=True by default
• Ignoring array length vs sample size
• Thinking data type causes error

5. You want to simulate rolling a weighted 6-sided die 10 times using numpy, where side 6 is twice as likely as others. Which code correctly generates this sample?

hard

A. np.random.choice([1,2,3,4,5,6], size=10, replace=True, p=[1/7,1/7,1/7,1/7,1/7,2/7])

B. np.random.choice([1,2,3,4,5,6], size=10, replace=False, p=[1/6]*6)

C. np.random.choice([1,2,3,4,5,6], size=10, replace=True)

D. np.random.choice([1,2,3,4,5,6], size=10, replace=True, p=[1/6,1/6,1/6,1/6,1/6,1/6])

Solution

1. Step 1: Understand weighted probabilities

   Side 6 should be twice as likely, so probabilities sum to 1 with side 6 having weight 2/7 and others 1/7 each.

2. Step 2: Check sampling parameters

   Sampling 10 times with replacement is needed to allow repeats. np.random.choice([1,2,3,4,5,6], size=10, replace=True, p=[1/7,1/7,1/7,1/7,1/7,2/7]) uses correct probabilities and replace=True.

3. Step 3: Verify other options

   The code with replace=False, p=[1/6]*6 incorrectly prevents repeats needed for multiple rolls. The codes with uniform probabilities (explicit [1/6,1/6,1/6,1/6,1/6,1/6] or none specified) do not weight side 6 twice as likely.

4. Final Answer:

   np.random.choice([1,2,3,4,5,6], size=10, replace=True, p=[1/7,1/7,1/7,1/7,1/7,2/7]) -> Option A

5. Quick Check:

   Weighted probabilities + replace=True for repeated rolls [OK]

Hint: Use p= with weights summing to 1 and replace=True [OK]

Common Mistakes:

• Using replace=False for multiple rolls
• Not setting probabilities for weighted sides
• Using equal probabilities when weights differ