ACT Math: Probability
Probability questions on the ACT reward recognizing the right method (add or multiply, permutation or combination) more than heavy computation. This free, complete lesson covers basic probability and the complement rule, independent and dependent events, the fundamental counting principle, permutations and combinations, and expected value. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.
1. Basic Probability and the Complement Rule
P(event) = favorable outcomestotal outcomes
The complement rule: the probability an event does NOT happen equals 1 minus the probability it does.
Worked Example: A bag has 5 red, 3 blue, and 2 green marbles. Find P(red).
Total marbles: 5 + 3 + 2 = 10.
P(red) = 510 = 12
1. A bag has 4 red and 6 blue marbles. Find P(blue).
Explanation: Total marbles: 10. P(blue) = 610 = 35.
2. A standard six-sided die is rolled once. What is P(rolling a number greater than 4)?
Explanation: Numbers greater than 4: {5, 6}, so 2 favorable outcomes out of 6: 26 = 13.
3. The probability of rain tomorrow is 0.35. What is the probability it does NOT rain?
Explanation: 1 − 0.35 = 0.65.
4. A jar has 12 candies, and the probability of picking a red one is 13. How many red candies are in the jar?
Explanation: 12 × 13 = 4.
2. Independent and Dependent Events
Events are independent if one doesn't affect the other's probability (like separate coin flips, or drawing "with replacement"). Events are dependent if one changes the probability of the next (like drawing "without replacement"). Either way, to find P(A and B), multiply the two probabilities, recalculating the second probability if the events are dependent. Use addition instead when you want P(A or B) for outcomes that can't happen together.
Worked Example: A coin is flipped and a six-sided die is rolled. Find P(heads AND rolling a 4).
These are independent events.
P(heads) × P(4) = 12 × 16 = 112
1. Two six-sided dice are rolled. What is P(both show a 6)?
Explanation: These are independent events: 16 × 16 = 136.
2. A bag has 5 red and 3 blue marbles (8 total). One marble is drawn, replaced, and a second is drawn. Find P(both red).
Explanation: With replacement, these are independent: 58 × 58 = 2564.
3. The same bag (5 red, 3 blue, 8 total) has one marble drawn and NOT replaced, then a second is drawn. Find P(both red).
Explanation: Without replacement, these are dependent: 58 × 47 = 2056 = 514.
4. A deck of 52 cards has 4 aces. Two cards are drawn without replacement. What is P(first card is an ace AND second card is an ace)?
Explanation: 452 × 351 = 122652 = 1221.
5. Is rolling a die twice in a row an example of independent or dependent events?
Explanation: Each roll of the die has no effect on the outcome of any other roll, making them independent.
3. The Fundamental Counting Principle
When a process has multiple independent stages, multiply the number of choices at each stage to find the total number of possible outcomes.
Worked Example: A restaurant offers 4 appetizers, 6 entrees, and 3 desserts. How many different 3-course meals are possible?
4 × 6 × 3 = 72
1. A license plate has 3 letters followed by 4 digits. If letters and digits can repeat, how many different license plates are possible?
Explanation: 26 × 26 × 26 × 10 × 10 × 10 × 10 = 17,576 × 10,000 = 175,760,000.
2. A store has 5 shirt styles, 4 pants styles, and 2 shoe styles. How many different complete outfits (one of each) can be made?
Explanation: 5 × 4 × 2 = 40.
3. How many different 4-digit codes can be made using digits 0-9, if digits can repeat?
Explanation: 10 × 10 × 10 × 10 = 10,000.
4. How many different 4-digit codes can be made using digits 0-9, if NO digit can repeat?
Explanation: With no repeats, each choice reduces the options by one: 10 × 9 × 8 × 7 = 5,040.
4. Permutations and Combinations
A permutation counts arrangements where order matters (like assigning distinct roles). A combination counts selections where order does NOT matter (like choosing a group with no distinct roles).
Worked Example: In how many ways can 5 runners finish 1st, 2nd, and 3rd place?
Order matters here (this is a permutation):
5 × 4 × 3 = 60
1. A club has 8 members. In how many ways can a president, vice president, and secretary (3 distinct roles) be chosen?
Explanation: Since the roles are distinct, order matters: 8 × 7 × 6 = 336.
2. A club has 8 members. In how many ways can a 3-person committee (no distinct roles) be chosen?
Explanation: Since a committee has no distinct roles, order doesn't matter. Take the permutation (336) and divide by the number of ways to arrange 3 people (3! = 6): 336 ÷ 6 = 56.
3. Does choosing a team captain and co-captain from 10 players require a permutation or a combination?
Explanation: Captain and co-captain are distinct roles, so which player fills which role matters, this is a permutation.
4. How many ways can you choose 2 pizza toppings from a menu of 6 toppings, if order doesn't matter?
Explanation: The permutation would be 6 × 5 = 30. Since order doesn't matter, divide by 2! = 2: 30 ÷ 2 = 15.
5. Expected Value
Expected value is a weighted average of all possible outcomes, multiply each outcome's value by its probability, then add up the results.
Worked Example: A game costs $2 to play. You win $10 with probability 0.1, and win nothing otherwise. Find the expected net value of playing.
Expected winnings: 0.1(10) + 0.9(0) = 1.
Net expected value: 1 − 2 = −1 (an average loss of $1 per play)
1. A spinner has 4 equal sections: win $8 (1 section), win $2 (1 section), and win $0 (2 sections). What is the expected value of one spin?
Explanation: 14(8) + 14(2) + 24(0) = 2 + 0.5 + 0 = $2.50.
2. A raffle has 100 tickets. One ticket wins $200, and the rest win nothing. If a ticket costs $3, what is the expected net value of buying one ticket?
Explanation: Expected winnings: 1100(200) = $2. Net: 2 − 3 = −$1.
3. A die is rolled. You win $6 if it lands on 6, and lose $1 otherwise. What is the expected value?
Explanation: 16(6) + 56(−1) = 1 − 56 = 16.
4. An insurance policy pays $50,000 if a rare event occurs (probability 0.001), and $0 otherwise. What is the expected payout?
Explanation: 0.001 × 50,000 = $50.
Common Mistakes to Avoid
- Adding probabilities when you should multiply, or vice versa. Use multiplication for "and" (both events happening together); use addition for "or" between outcomes that can't both happen.
- Forgetting to adjust the denominator for dependent events. "Without replacement" means the total number of outcomes shrinks for the second event, recalculate the probability using the new total.
- Confusing permutations and combinations. If the roles or positions are distinguishable (order matters), use a permutation. If you're just selecting a group with no distinct roles (order doesn't matter), use a combination.
- Forgetting to subtract the cost when finding a net expected value. Expected value calculates the average payout; if there's a cost to play, subtract it separately at the end.
- Treating a "greater than" or "at least" condition as excluding the boundary value incorrectly. Carefully check whether the condition (like "more than 4" versus "4 or more") includes the boundary number itself before counting favorable outcomes.
Frequently Asked Questions
How can I quickly tell if I need a permutation or a combination?
Ask whether swapping two selected items would create a meaningfully different outcome. If yes (like assigning a president versus a vice president), order matters, use a permutation. If no (like just picking a group of people), order doesn't matter, use a combination.
Do I need to memorize the permutation and combination formulas for the ACT?
They're useful, but many ACT counting problems can be solved just as quickly with the fundamental counting principle or the "permutation, then divide by the redundant arrangements" approach shown in this lesson, without needing to plug into a formula.
What does a negative expected value actually mean?
It means that, on average over many repeated trials, you'd expect to lose money (or whatever quantity is being measured) rather than gain it, even though any single outcome could still be a win.
Where can I practice more problems like these?
The Probability quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.