Module 3 Practice Problems
20 Comprehensive Problems • Covers All Module 3 Topics
How to Use These Practice Problems
- Work through problems on paper before checking solutions
- Use the hints if you're stuck
- Problems are organized by topic and difficulty
- Show all your work - probability is about the process!
Problems 1-5: Basic Probability & Sample Spaces
Problem 1 Finding Sample Spaces Easy
List the sample space for flipping a coin three times.
Solution:
Sample Space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total outcomes: 2 × 2 × 2 = 8
Problem 2 Basic Probability Calculation Easy
A bag contains 6 red marbles, 4 blue marbles, and 5 green marbles. If you randomly select one marble, what is the probability it's blue?
Solution:
- Total marbles = 6 + 4 + 5 = 15
- Blue marbles = 4
- P(blue) = 4/15 ≈ 0.267
Answer: 4/15 or about 26.7%
Problem 3 Complement Rule Easy
If the probability of a student passing an exam is 0.82, what is the probability the student does NOT pass?
Solution:
P(not pass) = 1 − P(pass)
= 1 − 0.82
= 0.18
Answer: 0.18 or 18%
Problem 4 Rolling Dice Probability Medium
You roll two six-sided dice. What is the probability that the sum is 7?
Solution:
Ways to get sum of 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
Total possible outcomes = 6 × 6 = 36
P(sum = 7) = 6/36 = 1/6 ≈ 0.167
Answer: 1/6 or about 16.7%
Problem 5 Card Probability Medium
What is the probability of drawing a face card (Jack, Queen, or King) OR a heart from a standard 52-card deck?
Solution:
Face cards: 12 total (3 per suit × 4 suits)
Hearts: 13 total
Face cards that are hearts: 3 (J, Q, K)
Using inclusion: favorable = 12 + 13 − 3 = 22
P(face or heart) = 22/52 = 11/26 ≈ 0.423
Answer: 11/26 or about 42.3%
Problems 6-10: Counting Methods
Problem 6 Fundamental Counting Principle Easy
A restaurant offers 4 appetizers, 6 main courses, 3 side dishes, and 5 desserts. How many different four-course meals can you order?
Solution:
Using Fundamental Counting Principle:
Total meals = 4 × 6 × 3 × 5 = 360
Answer: 360 different meals
Problem 7 Factorial Calculation Easy
How many ways can 5 students be arranged in a single-file line?
Solution:
Number of arrangements = 5!
= 5 × 4 × 3 × 2 × 1
= 120
Answer: 120 ways
Problem 8 Permutations Medium
From a group of 10 finalists, how many ways can awards be given for 1st, 2nd, and 3rd place?
Solution:
Order matters, so use permutation:
P(10,3) = 10!/(10−3)! = 10!/7!
= 10 × 9 × 8
= 720
Answer: 720 ways
Problem 9 Combinations Medium
A committee of 4 people must be chosen from a group of 12. How many different committees are possible?
Solution:
Order doesn't matter, so use combination:
C(12,4) = 12!/(4! × 8!)
= (12 × 11 × 10 × 9)/(4 × 3 × 2 × 1)
= 11,880/24
= 495
Answer: 495 committees
Problem 10 Permutation vs Combination Hard
A password must contain 3 letters (from A-Z) followed by 2 digits (0-9). Letters and digits can repeat. How many passwords are possible?
Solution:
First letter: 26 choices
Second letter: 26 choices (can repeat)
Third letter: 26 choices
First digit: 10 choices
Second digit: 10 choices
Total = 26 × 26 × 26 × 10 × 10
= 17,576 × 100
= 1,757,600
Answer: 1,757,600 passwords
Problems 11-15: Conditional Probability & Independence
Problem 11 Conditional Probability Medium
You roll a fair die. Given that you rolled an even number, what is the probability you rolled a 4?
Solution:
Given: Even number, so outcomes are {2, 4, 6}
Of these 3 outcomes, one is a 4
P(4 | even) = 1/3 ≈ 0.333
Answer: 1/3 or about 33.3%
Problem 12 Two-Way Table Medium
A survey of 300 people asked about pet ownership:
| Dog | No Dog | Total | |
|---|---|---|---|
| Cat | 50 | 70 | 120 |
| No Cat | 80 | 100 | 180 |
| Total | 130 | 170 | 300 |
What is P(has dog | has cat)?
Solution:
Given: Person has a cat (120 people)
Of these 120, how many have a dog? 50
P(has dog | has cat) = 50/120 = 5/12 ≈ 0.417
Answer: 5/12 or about 41.7%
Problem 13 Multiplication Rule Medium
You draw 2 cards from a standard deck WITHOUT replacement. What is the probability both are aces?
Solution:
P(1st ace) = 4/52 = 1/13
P(2nd ace | 1st ace) = 3/51 = 1/17
P(both aces) = (1/13) × (1/17) = 1/221
≈ 0.0045
Answer: 1/221 or about 0.45%
Problem 14 Testing Independence Hard
If P(A) = 0.4, P(B) = 0.3, and P(A and B) = 0.12, are A and B independent?
Solution:
For independence: P(A and B) should equal P(A) × P(B)
P(A) × P(B) = 0.4 × 0.3 = 0.12
P(A and B) = 0.12 (given)
Since 0.12 = 0.12, the events ARE independent!
Answer: Yes, A and B are independent
Problem 15 Independent Events Medium
The probability of rain tomorrow is 0.3. The probability your team wins their game tomorrow is 0.6. Assuming these are independent, what's the probability it rains AND your team wins?
Solution:
Since independent:
P(rain and win) = P(rain) × P(win)
= 0.3 × 0.6
= 0.18
Answer: 0.18 or 18%
Problems 16-20: Probability Distributions
Problem 16 Valid Distribution? Easy
Is this a valid probability distribution?
| X | P(X) |
|---|---|
| 0 | 0.15 |
| 1 | 0.35 |
| 2 | 0.25 |
| 3 | 0.25 |
Solution:
Check 1: All probabilities between 0 and 1? Yes
Check 2: Sum = 0.15 + 0.35 + 0.25 + 0.25 = 1.00 Yes
Answer: YES, this is a valid distribution
Problem 17 Expected Value Medium
Calculate the expected value of X:
| X | P(X) |
|---|---|
| 2 | 0.3 |
| 4 | 0.5 |
| 6 | 0.2 |
Solution:
E(X) = (2)(0.3) + (4)(0.5) + (6)(0.2)
= 0.6 + 2.0 + 1.2
= 3.8
Answer: E(X) = 3.8
Problem 18 Game Expected Value Hard
A carnival game costs $5 to play. You roll a die: if you roll a 6, you win $20; otherwise you lose your $5. What is the expected value of your net winnings?
Solution:
Let X = net winnings
| Outcome | X | P(X) |
|---|---|---|
| Roll 6 | +$15 | 1/6 |
| Roll 1-5 | −$5 | 5/6 |
E(X) = (15)(1/6) + (−5)(5/6)
= 15/6 − 25/6
= −10/6 = −$1.67
Answer: E(X) = −$1.67
On average, you lose $1.67 per game - not favorable!
Problem 19 Creating a Distribution Hard
Let X = the number of heads when flipping 3 fair coins. Create the probability distribution for X.
Solution:
Sample space: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
| X (# heads) | Outcomes | P(X) |
|---|---|---|
| 0 | TTT | 1/8 = 0.125 |
| 1 | HTT, THT, TTH | 3/8 = 0.375 |
| 2 | HHT, HTH, THH | 3/8 = 0.375 |
| 3 | HHH | 1/8 = 0.125 |
| Total: | 1.000 | |
This is the probability distribution of X!
Problem 20 Variance Calculation Hard
For the distribution below, calculate the variance. (Given: E(X) = 2.0)
| X | P(X) |
|---|---|
| 1 | 0.4 |
| 2 | 0.3 |
| 3 | 0.3 |
Solution:
μ = 2.0 (given)
| X | P(X) | (x−μ)² | (x−μ)²·P(X) |
|---|---|---|---|
| 1 | 0.4 | (1−2)² = 1 | 1(0.4) = 0.4 |
| 2 | 0.3 | (2−2)² = 0 | 0(0.3) = 0.0 |
| 3 | 0.3 | (3−2)² = 1 | 1(0.3) = 0.3 |
| Var(X) = | 0.7 | ||
Answer: Variance = 0.7
Standard Deviation = √0.7 ≈ 0.837
Great Work!
You've completed all 20 practice problems! Make sure you understand each solution before moving on to the quiz.