Monte Carlo simulation basics in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a simple Monte Carlo estimation of π

What is the output of this code that estimates π using Monte Carlo simulation with 1000 points?

NumPy

import numpy as np
np.random.seed(0)
points = 1000
x = np.random.uniform(-1, 1, points)
y = np.random.uniform(-1, 1, points)
inside_circle = (x**2 + y**2) <= 1
pi_estimate = 4 * np.sum(inside_circle) / points
print(round(pi_estimate, 2))

A 3.14

B 2.56

C 3.00

D 4.00

💡 Hint

Think about how many points fall inside the unit circle compared to total points.

The code generates 1000 random points in a square from -1 to 1 on both axes. It counts how many fall inside the unit circle. The ratio multiplied by 4 estimates π. With seed 0, the estimate rounds to 3.14.

❓ data_output

intermediate

Number of points inside circle in Monte Carlo simulation

After running this Monte Carlo simulation code with 5000 points, what is the number of points inside the unit circle?

NumPy

import numpy as np
np.random.seed(1)
points = 5000
x = np.random.uniform(-1, 1, points)
y = np.random.uniform(-1, 1, points)
inside_circle = (x**2 + y**2) <= 1
count_inside = np.sum(inside_circle)
print(count_inside)

A 5000

B 4000

C 3932

D 2500

💡 Hint

Count how many points satisfy the circle condition.

With seed 1 and 5000 points, the sum of points inside the circle is 3932.

❓ visualization

advanced

Visualizing Monte Carlo points inside and outside the circle

Which option shows the correct scatter plot of 1000 Monte Carlo points with points inside the unit circle in blue and outside in red?

NumPy

import numpy as np
import matplotlib.pyplot as plt
np.random.seed(2)
points = 1000
x = np.random.uniform(-1, 1, points)
y = np.random.uniform(-1, 1, points)
inside_circle = (x**2 + y**2) <= 1
plt.figure(figsize=(5,5))
plt.scatter(x[inside_circle], y[inside_circle], color='blue', s=5, label='Inside Circle')
plt.scatter(x[~inside_circle], y[~inside_circle], color='red', s=5, label='Outside Circle')
plt.legend()
plt.title('Monte Carlo Points Inside and Outside Unit Circle')
plt.show()

A Scatter plot with points outside circle in blue only

B Red points inside circle, blue points outside, no legend

C All points in one color, no distinction

D Blue points inside circle, red points outside, circle boundary visible

💡 Hint

Check colors and labels for inside vs outside points.

The code colors points inside the circle blue and outside red, with a legend and title. Option D matches this exactly.

🧠 Conceptual

advanced

Understanding Monte Carlo simulation error behavior

Which statement correctly describes how the error in Monte Carlo estimation changes as the number of random points increases?

A Error decreases proportionally to 1 divided by the square root of the number of points

B Error decreases linearly with the number of points

C Error increases as more points are added

D Error remains constant regardless of the number of points

💡 Hint

Think about how randomness averages out with more samples.

Monte Carlo error typically decreases with the square root of the number of samples, meaning more points reduce error but with diminishing returns.

🔧 Debug

expert

Identify the error in Monte Carlo simulation code

What error will this code raise when run, and why?

NumPy

import numpy as np
np.random.seed(3)
points = 1000
x = np.random.uniform(-1, 1, points)
y = np.random.uniform(-1, 1, points)
inside_circle = (x**2 + y**2) <= 1
count_inside = np.sum(inside_circle)
print(count_inside)

A SyntaxError due to missing parentheses

B No error, prints the count of points inside circle

C TypeError because y is a float, not an array, so x**2 + y**2 fails

D ValueError because points is not an integer

💡 Hint

Check the shape and type of y compared to x.

y is generated as an array like x. There is no error, and the code prints the count of points inside the circle.

Practice

1. What is the main purpose of a Monte Carlo simulation in data science?

easy

A. To create visual charts

B. To sort data quickly

C. To estimate results using random sampling

D. To clean missing data

Solution

1. Step 1: Understand Monte Carlo simulation basics

   Monte Carlo simulation uses random sampling to estimate results when exact calculation is hard.

2. Step 2: Identify the main purpose

   It helps estimate outcomes by simulating many random scenarios.

3. Final Answer:

   To estimate results using random sampling -> Option C

4. Quick Check:

   Monte Carlo = random sampling estimation [OK]

Hint: Monte Carlo = random sampling for estimation [OK]

Common Mistakes:

• Confusing Monte Carlo with data sorting
• Thinking it cleans data
• Assuming it creates charts

2. Which numpy function is commonly used to generate random samples for Monte Carlo simulations?

easy

A. numpy.random.rand()

B. numpy.array()

C. numpy.mean()

D. numpy.sort()

Solution

1. Step 1: Identify numpy functions for randomness

   numpy.random.rand() generates random numbers between 0 and 1, useful for sampling.

2. Step 2: Compare with other options

   numpy.array() creates arrays, numpy.mean() calculates average, numpy.sort() sorts data; none generate random samples.

3. Final Answer:

   numpy.random.rand() -> Option A

4. Quick Check:

   Random samples = numpy.random.rand() [OK]

Hint: Random samples come from numpy.random.rand() [OK]

Common Mistakes:

• Choosing numpy.array() instead of random function
• Confusing mean or sort with random sampling
• Using functions that don't generate randomness

3. What is the output of this code snippet?

import numpy as np
np.random.seed(0)
samples = np.random.rand(5)
mean_val = np.mean(samples)
print(round(mean_val, 2))

medium

A. 0.50

B. 0.57

C. 0.60

D. 0.45

Solution

1. Step 1: Generate 5 random samples with seed 0

   Using np.random.seed(0) fixes randomness. np.random.rand(5) produces [0.5488135, 0.71518937, 0.60276338, 0.54488318, 0.4236548].

2. Step 2: Calculate mean and round

   Sum = 2.83530423, mean = 0.567060846, round(mean_val, 2) = 0.57.

3. Final Answer:

   0.57 -> Option B

4. Quick Check:

   Mean of samples = 0.57 [OK]

Hint: Seed fixes random values; mean rounds to 0.57 [OK]

Common Mistakes:

• Not setting seed causing different outputs
• Rounding incorrectly
• Calculating sum or mean wrongly

4. Identify the error in this Monte Carlo simulation code:

import numpy as np
samples = np.random.rand(1000)
result = sum(samples) / 100
print(result)

medium

A. Using np.random.rand() instead of np.random.randint()

B. Not setting random seed

C. Missing np.mean() function

D. Dividing by 100 instead of number of samples

Solution

1. Step 1: Check calculation of average

   sum(samples) sums 1000 values, but dividing by 100 underestimates average.

2. Step 2: Correct divisor

   To get average, divide sum by 1000 (number of samples), not 100.

3. Final Answer:

   Dividing by 100 instead of number of samples -> Option D

4. Quick Check:

   Average = sum / number of samples [OK]

Hint: Divide sum by sample count, not fixed number [OK]

Common Mistakes:

• Dividing by wrong number
• Confusing random functions
• Ignoring seed importance

5. You want to estimate the probability that a random point inside a 1x1 square falls within a circle of radius 0.5 centered at (0.5, 0.5) using Monte Carlo simulation. Which numpy code correctly estimates this probability?

hard

A. points = np.random.rand(10000, 2)
dist = np.sqrt((points[:,0]-0.5)**2 + (points[:,1]-0.5)**2)
prob = np.sum(dist <= 0.5) / 10000
print(prob)

B. points = np.random.rand(10000)
dist = np.sqrt((points-0.5)**2)
prob = np.sum(dist <= 0.5) / 10000
print(prob)

C. points = np.random.rand(2, 10000)
dist = np.sqrt((points[:,0]-0.5)**2 + (points[:,1]-0.5)**2)
prob = np.sum(dist <= 0.5) / 10000
print(prob)

D. points = np.random.randint(0, 1, (10000, 2))
dist = np.sqrt((points[:,0]-0.5)**2 + (points[:,1]-0.5)**2)
prob = np.sum(dist <= 0.5) / 10000
print(prob)

Solution

1. Step 1: Generate 2D points inside 1x1 square

   np.random.rand(10000, 2) creates 10000 points with x,y in [0,1].

2. Step 2: Calculate distance from center (0.5,0.5)

   Distance formula sqrt((x-0.5)^2 + (y-0.5)^2) applied to each point.

3. Step 3: Count points inside circle and estimate probability

   Count points with distance ≤ 0.5, divide by total points for probability.

4. Final Answer:

   np.random.rand(10000, 2) with column indexing for coordinates -> Option A

5. Quick Check:

   2D points + distance check = probability [OK]

Hint: Use 2D random points and distance formula [OK]

Common Mistakes:

• Using 1D points instead of 2D
• Wrong shape for points array
• Using randint instead of rand for floats