ACT Math Probability: 25 Practice Problems with Step-by-Step Explanations
Probability questions on the ACT build on one simple foundation — probability equals favorable outcomes divided by total outcomes — but layer on more sophisticated rules as the problems get harder: combining probabilities with "and" versus "or," distinguishing independent events from events without replacement, reading two-way data tables, and calculating expected value. Recognizing which rule a problem is testing is usually the real challenge, more than the arithmetic itself.
For "and" problems (this event and that event both happening), multiply the probabilities — but only directly multiply if the events are independent. If one event changes the probability of the next (like drawing marbles without putting them back), you have to adjust each probability based on what was already removed. For "or" problems, add the probabilities, but if the two events can happen at the same time (like drawing a card that's both a 7 and a heart), you have to subtract the overlap so it isn't counted twice — this is the difference between mutually exclusive events (just add) and overlapping events (add, then subtract the overlap).
Two-way tables (like exam grades, diet outcomes, or pizza orders broken down by two categories) test whether you can correctly identify the right numerator and denominator for a given question — reading "what fraction of dogs who lost weight received diet B" requires dividing by the total who lost weight, not the grand total. Expected value problems ask for the average outcome over many trials, found by multiplying each possible outcome by its probability (or, more simply, multiplying a single probability by the number of trials for repeated identical experiments). Work through the 25 problems below, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Basic probability
- Problem 2 — Complement rule
- Problem 3 — Mutually exclusive "or"
- Problem 4 — Probability from a table
- Problem 5 — Expected value with replacement
- Problem 6 — Complement of a sum
- Problem 7 — Without replacement
- Problem 8 — Mutually exclusive "or"
- Problem 9 — Solving for an unknown count
- Problem 10 — Fraction from a table
- Problem 11 — Fraction from a table
- Problem 12 — Geometric probability
- Problem 13 — Independent repeated trials
- Problem 14 — Probability of "or"
- Problem 15 — Independent "and"
- Problem 16 — Probability from a two-way table
- Problem 17 — Mutually exclusive categories
- Problem 18 — Combining "or" and without replacement
- Problem 19 — Overlapping "or"
- Problem 20 — Independent "and" with a loss
- Problem 21 — Overlapping "or" with a deck of cards
- Problem 22 — Counting pairings
- Problem 23 — Probability of a product
- Problem 24 — Expected value
- Problem 25 — Weighted probability
Practice Problems
Andy has a big bag of marbles. If the bag has 30 blue marbles, 25 red marbles, and 15 green marbles, what is the probability that he will randomly select a red marble?
- A) 5/14
- B) 6/15
- C) 3/7
- D) 5/7
The total number of marbles is 30+25+15 = 70. Probability is favorable outcomes over total: 25/70, which simplifies to 5/14.
Jerry is in a rush and needs to grab one of the premade sandwiches at the market. Jerry's favorite is the Italian sub, but the sandwiches have no labels on them. If there are 15 turkey sandwiches, 12 veggie subs, and 10 Italian subs, what is the probability that he does NOT get an Italian sub?
- A) 10/27
- B) 15/27
- C) 10/37
- D) 27/37
The total number of sandwiches is 15+12+10 = 37. The probability of getting an Italian sub is 10/37.
Using the complement rule, the probability of NOT getting an Italian sub is 1 − 10/37 = 27/37.
A bag contains 13 flavored pieces of candy: 5 lemon, 2 strawberry, 3 orange and 3 grape. What is the probability that one piece of randomly selected candy is flavored strawberry or grape?
- A) 3/13
- B) 5/13
- C) 8/13
- D) 5/8
Since a piece of candy can't be both strawberry and grape at the same time, these two outcomes are mutually exclusive, so their probabilities can simply be added.
P(strawberry or grape) = (2+3)/13 = 5/13.
The table below shows the grades that 50 students received on the Chemistry exam. What is the probability that a randomly selected student's exam score is less than a B?
| Exam Grade | Number of Students |
|---|---|
| A | 11 |
| B | 23 |
| C | 12 |
| D | 3 |
| F | 1 |
- A) 0.24
- B) 0.32
- C) 0.46
- D) 0.52
"Less than a B" means C, D, or F: 12+3+1 = 16 students.
Probability = 16/50 = 0.32.
A box contains 1 red bead, 6 white beads, and 3 blue beads. Jorge will randomly remove one bead from the box, record its color, and place it back in the box. If Jorge repeats this experiment 150 times, what is the expected number of times Jorge will record a bead that is blue?
- A) 30
- B) 45
- C) 75
- D) 105
The probability of drawing blue on any single trial is 3/10 (there are 10 beads total, and replacing the bead each time keeps this probability constant).
Expected count = probability × number of trials = (3/10)(150) = 45.
A group of backpackers were asked to pick their favorite country in Europe. No backpacker was allowed to pick more than one country. 1/4 picked Italy, 1/5 picked Spain, 1/6 picked France, and 1/20 picked Portugal. If a random backpacker from the group is selected, what is the probability that he or she did not pick any of these 4 countries?
- A) 3/20
- B) 1/3
- C) 4/10
- D) 1/2
Since no backpacker picked more than one country, these four outcomes are mutually exclusive, so add their probabilities. Using a common denominator of 60: 1/4=15/60, 1/5=12/60, 1/6=10/60, 1/20=3/60.
Sum = (15+12+10+3)/60 = 40/60 = 2/3. Using the complement rule, the probability of picking none of these countries is 1 − 2/3 = 1/3.
A box contains 12 solid-colored marbles: 5 blue, 4 red, 1 yellow, and 2 pink. Which of the following expressions gives the probability of drawing, at random and without replacement, a pink marble on the 1st draw and a red marble on the 2nd draw, and a pink marble on the 3rd draw?
- A) (2/12)(4/11)(2/10)
- B) (2/12)(4/11)(1/10)
- C) (2/12)(1/11)(4/11)
- D) (2/12)(4/12)(2/12)
Without replacement, each draw changes the pool size and the count of the color just drawn. Draw 1 (pink): 2/12. Draw 2 (red, pink not affected): 4/11 (11 marbles remain).
Draw 3 (pink): since one pink was already removed on draw 1, only 1 pink remains out of 10 total marbles left: 1/10.
A pencil will be randomly selected from a bag. The probability of selecting a white pencil is 3/13. The probability of selecting a blue pencil is 2/13. What is the probability of selecting a white or blue pencil?
- A) 1/13
- B) 5/13
- C) 5/26
- D) 6/26
A pencil can't be both white and blue at once, so these outcomes are mutually exclusive.
P(white or blue) = 3/13 + 2/13 = 5/13.
A bowl contains 12 red marbles, 6 blue marbles and an unknown number of black marbles. The probability of choosing a blue marble out of the bowl is 1/7. How many black marbles are in the bowl?
- A) 6
- B) 12
- C) 18
- D) 24
Since P(blue) = 6/total = 1/7, the total number of marbles is 6×7 = 42.
Black marbles = total − red − blue = 42 − 12 − 6 = 24.
Use the information below for questions 10 and 11. The table below shows the results of two diets on the weight of dogs.
| Diet | Gained | Lost | Total |
|---|---|---|---|
| A | 50 | 100 | 150 |
| B | 75 | 75 | 150 |
Based on the results in the table, what fraction of the dogs who lost weight received diet B?
- A) 75/175
- B) 75/150
- C) 100/175
- D) 175/300
The total number of dogs who lost weight (across both diets) is 100+75 = 175 — this is the correct denominator, not the grand total of 300.
Of those, 75 received diet B: 75/175.
Based on the results in the same table, what fraction of the dogs received diet A and gained weight?
- A) 1/2
- B) 1/3
- C) 1/6
- D) 5/12
This question asks about the whole group of dogs (150+150=300), not just one diet or one outcome.
Diet A and gained weight: 50 dogs. Fraction: 50/300 = 1/6.
Three squares with the same center have side lengths of 1, 2, and 3 respectively. What is the probability that a point randomly chosen in the interior of the largest square is also in the interior of the smallest square?
- A) 1/9
- B) 2/9
- C) 1/3
- D) 4/9
This is a geometric probability problem: probability equals favorable area over total area. The smallest square has area 12=1, and the largest has area 32=9.
Probability = 1/9.
Julia runs out of time on her test and needs to guess on the last 4 questions. Each question has 5 answers. If Julia answers each one randomly, what is the probability she answers all 4 questions correctly?
- A) 1/625
- B) 4/625
- C) 1/125
- D) 1/25
The probability of guessing any one question correctly is 1/5. Since each guess is independent, multiply the probability by itself for each of the 4 questions.
(1/5)4 = 1/625.
The probability that each of two independent events will occur is given in the table below. What is the probability that Event X or Event Y occurs?
| Event | Probability |
|---|---|
| X | 0.3 |
| Y | 0.5 |
- A) 0.15
- B) 0.4
- C) 0.7
- D) 0.8
Since the two events don't overlap in this scenario, the probability of X or Y occurring is found by adding their individual probabilities.
P(X or Y) = 0.3 + 0.5 = 0.8.
At the regional baseball tournament, two games are played simultaneously. In the first game, team A has a 35% chance of winning. In the second game, team C has a 60% chance of winning. What is the probability that both team A and team C win? (Note: Neither game can result in a tie.)
- A) 14%
- B) 21%
- C) 26%
- D) 47.5%
Since the two games are played simultaneously and don't affect each other, they're independent events. Multiply their probabilities.
0.35 × 0.60 = 0.21 = 21%.
Use the information below for questions 16–18. The table below shows the pizza orders at Pete's Pizzeria from 1–2pm on a Saturday afternoon.
| Thin | Deep Dish | Gluten Free | ||
|---|---|---|---|---|
| Topping | Cheese | 12 | 19 | 7 |
| Pepperoni | 8 | 25 | 9 |
What is the probability that a pizza ordered was a thin crust pizza?
- A) 1/5
- B) 1/4
- C) 2/5
- D) 19/40
Total pizzas ordered: 12+19+7+8+25+9 = 80. Thin crust pizzas: 12 (cheese) + 8 (pepperoni) = 20.
Probability = 20/80 = 1/4.
Based on the same pizza table, what is the probability that a pizza ordered was thin crust or gluten free?
- A) 19/28
- B) 21/40
- C) 19/40
- D) 9/20
A pizza can't be both thin crust and gluten free at the same time (they're different crust categories), so these are mutually exclusive. Thin crust: 20. Gluten free: 7+9=16.
Probability = (20+16)/80 = 36/80 = 9/20.
Max and Amelia are picking up 2 pizzas from Pete's Pizzeria for lunch. If 2 pizzas are randomly selected from the pizzas ordered from 1–2pm at Pete's Pizzeria, what is the probability that Max and Amelia get 2 deep dish pepperoni pizzas or 2 thin crust cheese pizzas?
- A) (25/80)(25/80) + (12/80)(12/80)
- B) (25/80)(12/79) + (24/80)(11/80)
- C) (25/80)(24/79) + (12/80)(11/79)
- D) (25/44)(12/20)
There are 25 deep dish pepperoni pizzas and 12 thin crust cheese pizzas, out of 80 total. Since pizzas are selected without replacement, each probability requires an adjusted second draw.
P(2 deep dish pepperoni) = (25/80)(24/79). P(2 thin crust cheese) = (12/80)(11/79). Since these two outcomes can't both happen, add them.
A box of donated toys has ten toys left. 3 are blue stuffed animals, 3 are colored balls, of which 2 are red and 1 is blue, and 4 are frisbees, of which 3 are blue and 1 is red. If the probability of drawing each object is the same, what is the probability that the next toy drawn from the box is red or a ball?
- A) 1/5
- B) 3/10
- C) 2/5
- D) 1/2
Red toys total: 2 (red balls) + 1 (red frisbee) = 3. Balls total: 3. These overlap: the 2 red balls are counted in both groups, so subtract that overlap.
P(red or ball) = 3/10 + 3/10 − 2/10 = 4/10 = 2/5.
A San Diego high school football team will play three games in a tournament to get to the final. The local newspaper gives them an 80% chance of winning the first game, a 60% chance of winning the second game, and a 20% chance of winning the third game. What is the probability that the team will win the first two games and lose the third game?
- A) 6.4%
- B) 9.6%
- C) 25.6%
- D) 38.4%
The probability of losing game 3 is the complement of winning: 1 − 0.20 = 0.80.
Multiply the three independent probabilities: 0.80 × 0.60 × 0.80 = 0.384 = 38.4%.
There are 52 cards in a deck of cards. There are 4 suits, spades, hearts, diamonds, and clubs, and each suit has 13 cards numbered 1 through 13. What is the probability that a randomly selected card is numbered 7 or a heart?
- A) 4/52
- B) 10/52
- C) 13/52
- D) 16/52
There are 4 sevens and 13 hearts, but one card, the 7 of hearts, is counted in both groups, so it needs to be subtracted once to avoid double-counting.
4 + 13 − 1 = 16, giving a probability of 16/52.
Billy, Randal, Mike, and Shaun are at the waterpark. They are randomly assigned into pairs at the top of a two-person waterslide. What is the probability that Billy and Shaun will be paired together?
- A) 1/12
- B) 1/6
- C) 1/4
- D) 1/3
Four people can be split into 2 unordered pairs in exactly 3 different ways total. Of those 3 arrangements, exactly 1 has Billy and Shaun paired together (with Randal and Mike automatically forming the other pair).
Probability = 1/3.
A slot machine has two buttons you press. The first button will randomly give you a number from 1 to 5. The second button will randomly give you a number from 1 to 3. What is the probability that the product of the two buttons is even?
- A) 2/3
- B) 3/5
- C) 8/15
- D) 7/15
A product is odd only when both numbers are odd, so it's easier to find P(both odd) and take the complement. Button 1 has odd values {1,3,5}: probability 3/5. Button 2 has odd values {1,3}: probability 2/3.
P(both odd) = (3/5)(2/3) = 6/15 = 2/5. P(product even) = 1 − 2/5 = 3/5.
At casino night, James will play a game in which he rolls three six-sided dice numbered 1 through 6. He will be awarded 4 points for each die that lands on an odd number. Let the random variable x represent the total number of points awarded on any roll of the dice. What is the expected value of x?
- A) 1
- B) 2
- C) 4
- D) 6
Each die has a 1/2 chance of landing odd (3 odd faces out of 6). Over 3 dice, the expected number of odd results is 3 × 1/2 = 1.5.
Since each odd result is worth 4 points, the expected total points is 4 × 1.5 = 6.
There is a 70% chance that it snows more than 4 inches today. If it snows more than 4 or more inches, there is a 90% chance that school will be cancelled tomorrow. If it snows less than 4 inches, there is a 30% school will be cancelled. What is the probability that school will be cancelled tomorrow?
- Answer choices weren't visible in the source image for this problem.
This requires weighting each cancellation scenario by how likely it is to occur. P(cancelled) = P(more than 4 in.) × P(cancelled | more than 4 in.) + P(less than 4 in.) × P(cancelled | less than 4 in.).
= (0.70)(0.90) + (0.30)(0.30) = 0.63 + 0.09 = 0.72, or 72%.
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