Start Practicing →
Progress
0 / 25 answered
ACT Math · Practice Questions
25 ACT Probability Questions
You Must Master Before Test Day
These 25 questions cover every probability topic the ACT tests — basic probability, independent and dependent events, the OR rule, conditional probability, counting, and expected value. Work through them all, check your answers, and read the explanations.
Basic Probability
Independent/Dependent
OR Probability
Conditional
Counting
Expected Value
Category 1
Basic Probability & Complement
Questions 1–5
1
Basic Probability
A bag has 4 red, 6 blue, and 2 green marbles. What is the probability of picking a blue marble?
A 1/2
B 1/3
C 2/3
D 1/6
Solution
Total marbles = 4 + 6 + 2 = 12
P(blue) = 6 / 12 = 1/2
⚠️ P(event) = favorable outcomes / total outcomes. Always find the total first — it's easy to forget one category when adding.
2
Basic Probability
A fair six-sided die is rolled. What is the probability of rolling a number greater than 4?
A 1/6
B 1/3
C 1/2
D 2/3
Solution
Numbers greater than 4 on a die: {5, 6} = 2 outcomes
P = 2/6 = 1/3
💡 List the favorable outcomes explicitly before dividing — don't guess at the count.
3
Basic Probability
A card is drawn from a standard 52-card deck. What is the probability it is a heart?
A 1/52
B 1/13
C 1/4
D 4/13
Solution
There are 13 hearts in 52 cards
P = 13/52 = 1/4
💡 A standard deck has 4 suits of 13 cards each: hearts, diamonds, clubs, spades.
4
Complement Rule
The probability it will rain on Saturday is 0.35. What is the probability it will NOT rain?
A 0.35
B 0.55
C 0.65
D 0.70
Solution
P(not A) = 1 − P(A) = 1 − 0.35 = 0.65
⚠️ Complement rule: P(not A) = 1 − P(A). An event and its complement always sum to 1.
5
Complement Rule
In a class of 30 students, 18 play sports. What is the probability a randomly chosen student does NOT play sports?
A 3/5
B 2/5
C 1/5
D 4/5
Solution
Students not playing sports = 30 − 18 = 12
P = 12/30 = 2/5
💡 Or use the complement: P(sports) = 18/30 = 3/5, so P(not sports) = 1 − 3/5 = 2/5.
Category 2
Independent & Dependent Events
Questions 6–10
6
Independent Events
A coin is flipped and a die is rolled. What is the probability of getting heads AND a 3?
A 1/6
B 1/12
C 1/3
D 1/8
Solution
P(heads) = 1/2
P(rolling a 3) = 1/6
Independent events: P = 1/2 × 1/6 = 1/12
⚠️ For independent events: P(A and B) = P(A) × P(B). Multiply the probabilities, never add them.
7
Independent Events
Two fair coins are flipped. What is the probability both land on tails?
A 1/2
B 1/4
C 1/3
D 3/4
Solution
P(T) = 1/2 for each flip
Independent: P(TT) = 1/2 × 1/2 = 1/4
💡 List the sample space: {HH, HT, TH, TT}. Only 1 of 4 equally likely outcomes is TT.
8
Independent Events
A bag has 5 red and 5 blue balls. One ball is drawn, replaced, then another is drawn. What is P(both red)?
A 1/4
B 2/5
C 1/5
D 4/9
Solution
With replacement, draws are independent
P(red) = 5/10 = 1/2 each time
P(both red) = 1/2 × 1/2 = 1/4
⚠️ "With replacement" = independent events. "Without replacement" = dependent events. This phrase decides your entire setup.
9
Dependent Events
A bag has 4 red and 6 blue balls. One red ball is drawn and NOT replaced. What is P(drawing a red ball next)?
A 4/10
B 3/9
C 3/10
D 4/9
Solution
After removing 1 red ball: 3 red remain, 9 total remain
P = 3/9 = 1/3
💡 Without replacement: both the numerator AND the denominator change after each draw.
10
Dependent Events
From a deck of 52 cards, 2 cards are drawn WITHOUT replacement. What is P(both are aces)?
A 1/169
B 1/221
C 4/52
D 1/52
Solution
P(1st ace) = 4/52
P(2nd ace | 1st was ace) = 3/51
P = 4/52 × 3/51 = 12/2652 = 1/221
⚠️ Without replacement: multiply 4/52 × 3/51 — both the ace count and the total count drop by 1 after the first draw.
Category 3
OR Probability (Addition Rule)
Questions 11–15
11
Mutually Exclusive
P(A) = 0.4 and P(B) = 0.3, and A and B are mutually exclusive. What is P(A or B)?
A 0.12
B 0.58
C 0.70
D 0.10
Solution
Mutually exclusive: P(A or B) = P(A) + P(B)
= 0.4 + 0.3 = 0.7
💡 Mutually exclusive means A and B cannot both happen, so there's no overlap to subtract.
12
Addition Rule
In a class, 40% like math, 50% like science, and 20% like both. What percent like math OR science?
A 90%
B 70%
C 60%
D 50%
Solution
P(M or S) = P(M) + P(S) − P(M and S)
= 0.40 + 0.50 − 0.20 = 0.70 = 70%
⚠️ Addition Rule: P(A or B) = P(A) + P(B) − P(A and B). Always subtract the overlap to avoid double-counting.
13
Addition Rule
A card is drawn from a 52-card deck. What is P(the card is a king OR a heart)?
A 17/52
B 16/52
C 4/13
D 1/4
Solution
P(king) = 4/52, P(heart) = 13/52, P(king of hearts) = 1/52
P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
⚠️ The king of hearts belongs to BOTH groups, so it's counted twice unless you subtract it once.
14
Addition Rule
P(A) = 0.5, P(B) = 0.4, and P(A and B) = 0.2. What is P(A or B)?
A 0.9
B 0.7
C 0.6
D 1.1
Solution
P(A or B) = P(A) + P(B) − P(A and B)
= 0.5 + 0.4 − 0.2 = 0.7
💡 Never end up with a probability greater than 1.0 — that signals you forgot to subtract the intersection.
15
Addition Rule
A number is chosen at random from 1–20. What is P(it is even OR divisible by 3)?
A 13/20
B 10/20
C 6/20
D 16/20
Solution
Even: {2,4,6,8,10,12,14,16,18,20} = 10 numbers
Div by 3: {3,6,9,12,15,18} = 6 numbers
Both (div by 6): {6,12,18} = 3 numbers
P = (10 + 6 − 3)/20 = 13/20
💡 List both sets explicitly, then find their intersection before applying the addition rule.
Category 4
Conditional Probability
Questions 16–19
16
Conditional Probability
In a class of 40, 25 are girls and 15 are boys. 10 girls and 5 boys play chess. Given a student plays chess, what is P(the student is a girl)?
A 1/4
B 2/3
C 5/8
D 10/25
Solution
Chess players total: 10 + 5 = 15
Girls who play chess: 10
P(girl | plays chess) = 10/15 = 2/3
⚠️ Conditional probability restricts your sample space to only those who meet the given condition — here, only chess players.
17
Conditional Probability
P(A and B) = 0.12 and P(B) = 0.40. What is P(A | B)?
A 0.48
B 0.30
C 0.12
D 0.052
Solution
P(A | B) = P(A and B) / P(B)
= 0.12 / 0.40 = 0.30
💡 Formula: P(A|B) = P(A and B) / P(B). Read as "probability of A GIVEN that B has occurred."
18
Conditional Probability
A survey found 60% own a car and 20% own both a car and a bike. What is P(owns a bike | owns a car)?
A 0.20
B 0.33
C 0.40
D 0.12
Solution
P(bike | car) = P(bike and car) / P(car)
= 0.20 / 0.60 = 1/3 ≈ 0.33
⚠️ The denominator in conditional probability is always the probability of the GIVEN event — here, owning a car.
19
Conditional Probability
A die is rolled. Given the result is even, what is P(the result is greater than 2)?
A 1/3
B 1/2
C 2/3
D 1/6
Solution
Even numbers: {2, 4, 6} — this is the new sample space
Greater than 2 within that set: {4, 6} = 2 outcomes
P = 2/3
💡 The new sample space is restricted to {2,4,6}. Count favorable outcomes only from that restricted set.
Category 5
Counting & Combinations
Questions 20–22
20
Permutations
How many different 3-letter arrangements can be made from the letters A, B, C, D if no letter repeats?
A 12
B 24
C 64
D 6
Solution
This is a permutation (order matters)
4 choices for 1st × 3 for 2nd × 2 for 3rd
= 4 × 3 × 2 = 24
⚠️ Order matters here (ABC ≠ CBA), so use permutations: nPr = n! / (n−r)!
21
Combinations
A committee of 3 is chosen from 7 people. How many different committees are possible?
A 21
B 35
C 42
D 210
Solution
Order does not matter for a committee
C(7,3) = 7! / (3! × 4!) = (7×6×5) / (3×2×1) = 210/6 = 35
⚠️ Order doesn't matter for committees/groups → use combinations C(n,r). Order DOES matter for arrangements → use permutations.
22
Counting Principle
A password requires 2 letters (A–Z, 26 options each) followed by 2 digits (0–9). How many passwords are possible if repetition is allowed?
A 6,760
B 67,600
C 58,500
D 6,500
Solution
26 × 26 × 10 × 10 = 676 × 100 = 67,600
💡 With repetition allowed, each slot is independent of the others. Multiply the number of options for each position.
Category 6
Expected Value
Questions 23–25
23
Expected Value
A game pays $10 if you roll a 6 and $0 otherwise. The ticket costs $2. What is the expected net gain per play?
A −$0.33
B $0.33
C −$1.67
D $1.67
Solution
E(payout) = (1/6)(10) + (5/6)(0) = 10/6 ≈ $1.67
Net = $1.67 − $2.00 (ticket cost) = −$0.33
⚠️ Expected value = sum of (outcome × probability). Always subtract any cost to get the net expected gain.
24
Expected Value
A spinner has 3 sections: $5 (probability 1/2), $10 (probability 1/3), $0 (probability 1/6). What is the expected payout?
A $5.00
B $5.83
C $6.00
D $4.17
Solution
E = 5(1/2) + 10(1/3) + 0(1/6)
= 2.50 + 3.33 + 0 = $5.83
💡 Expected value is a weighted average. Multiply each outcome by its probability, then add all the products.
25
Expected Value
A raffle sells 200 tickets at $5 each. One prize of $400 is given. What is the expected net gain for a ticket buyer?
A −$3.00
B −$2.00
C $2.00
D −$4.00
Solution
P(win) = 1/200
E(payout) = (1/200)(400) + (199/200)(0) = $2.00
Net = $2.00 − $5.00 (ticket cost) = −$3.00
⚠️ Expected net = expected payout minus the ticket cost. Most raffles have negative expected value for buyers — that's how they raise money.
Free · 2,000+ Questions · No Account Needed
Want More ACT Probability Practice?
The ACT Math QBank has 2,000+ free questions covering every probability topic in this set — plus algebra, geometry, trig, and more.
Go to ACT Math QBank →