Probability: ACT Math Study Guide

The probability of an event is the number of favorable outcomes divided by the total number of equally likely outcomes in the sample space. For independent events, the probability that both occur is the product of their individual probabilities: P(A and B) = P(A) × P(B). For mutually exclusive events, which cannot both happen at once, the probability that either occurs is the sum: P(A or B) = P(A) + P(B).

Combinations and permutations count selections differently based on whether order matters. A combination counts groups where order doesn't matter, such as choosing a committee. A permutation counts arrangements where order does matter, such as filling distinct offices. Using the wrong one is the most common error on this topic.

The complement of an event A is everything that is not A; P(not A) = 1 − P(A). Expected value multiplies each possible outcome by its probability, then adds the results, giving the long-run average payout of a probabilistic game or scenario.

Where this appears as a model: a described collection of items. A bag containing 5 red marbles and 3 blue marbles gives a 3/8 probability of drawing blue, found once the collection is translated into a count of favorable and total outcomes.

▸ Combinations versus permutations

6C3 = (6×5×4)/(3×2×1) = 20

▸ Conditional probability from a two-way table

A conditional probability restricted to one row or one column of a two-way table divides by that row's or column's total, not the grand total. Reading which restriction applies is the actual skill; the division itself is straightforward once the right numbers are identified.

▸ General addition and multiplication rules

The addition rule above applies to mutually exclusive events; a more general version works whether or not two events overlap: P(A or B) = P(A) + P(B) − P(A and B). The overlap is subtracted once because it was already counted inside both P(A) and P(B). When A and B cannot happen together, P(A and B) = 0, and the general rule reduces to the simple addition already introduced.

A similar broadening applies to the multiplication rule. For events that occur in sequence, P(A and B) = P(A)·P(B|A), where P(B|A) is the probability of B given that A already happened. If the events are independent, P(B|A) = P(B), and the rule reduces to the simple product already introduced. The first question to ask is not “multiply or add?” It's whether the two events overlap, and whether the second event's probability depends on the first.

Worked Example In a group of students, P(plays soccer) = 0.5, P(plays tennis) = 0.3, and P(plays both) = 0.15. Find P(plays soccer or tennis).

The two events overlap, so use the general addition rule.

P(soccer or tennis) = 0.5 + 0.3 − 0.15 = 0.65

▸ Practice

Question 1

Events A and B are independent, with P(A) = 0.4 and P(B) = 0.7. Find P(A and B).

A. 1.1
B. 0.30
C. 0.28
D. 0.11

Question 2

Two fair six-sided dice are rolled. Find the probability that the sum is 7.

F. 1/6
G. 1/12
H. 1/9
J. 7/36

Question 3

How many ways can a committee of 2 be chosen from 5 people, where order does not matter?

A. 10
B. 20
C. 5
D. 25

Question 4

How many ways can 2 distinct offices (president and vice president) be filled from 5 candidates, where order matters?

F. 10
G. 25
H. 20
J. 5

Question 5

A raffle pays $100 with probability 0.1, pays $20 with probability 0.4, and costs $10 (a loss) with probability 0.5. Find the expected value.

A. $11
B. $4
C. $13
D. $110

Question 6 refers to the approval table below.

	Approve	Disapprove	Total
Under 30	40	10	50
Over 30	25	25	50
Total	65	35	100

Question 6

Using the Under 30/Over 30 approval table from the previous section, find P(Approve given Under 30).

F. 0.65
G. 0.40
H. 0.80
J. 0.50

Question 7

A fair six-sided die is rolled once. Find the probability of rolling a multiple of 3.

A. 1/2
B. 1/3
C. 1/6
D. 2/3

Question 8

Events A and B are independent, with P(A) = 0.6 and P(B) = 0.75. Find P(A and B).

F. 1.35
G. 0.45
H. 0.15
J. 0.6

Question 9 refers to the table below.

	Yes	No	Total
Group A	14	11	25
Group B	17	8	25
Total	31	19	50

Question 9

Using the table above, find the probability that a randomly chosen member of Group B answered “No.”

A. 0.32
B. 0.16
C. 0.68
D. 0.5

Question 10

A raffle pays $60 with probability 0.25, pays $15 with probability 0.4, and costs $10 (a loss) with probability 0.35. Find the expected value.

F. 15
G. 17.5
H. 20
J. 12.5

Question 11

How many ways can a committee of 3 be chosen from 8 people?

A. 336
B. 24
C. 168
D. 56

Question 12

How many ways can 4 distinct offices be filled from 7 candidates, where order matters?

F. 35
G. 210
H. 420
J. 840

Question 13

Events A and B are mutually exclusive, with P(A) = 0.3 and P(B) = 0.45. Find P(A or B).

A. 0.135
B. 0.75
C. 1.35
D. 0.15

Question 14

The probability of event A is 0.35. Find the probability that A does not occur.

F. 0.35
G. 1.35
H. 0.5
J. 0.65

Question 15

A fair six-sided die is rolled once. Find the probability of rolling a 5 or a 6.

A. 1/6
B. 1/3
C. 1/2
D. 2/3

Question 16

Events A and B are independent, with P(A) = 0.5 and P(B) = 0.6. Find P(A and B).

F. 0.3
G. 1.1
H. 0.83
J. 0.1

Question 17

How many ways can a committee of 2 be chosen from 7 people?

A. 42
B. 5
C. 14
D. 21

Question 18

How many ways can 2 distinct offices be filled from 5 candidates, where order matters?

F. 10
G. 20
H. 3
J. 25

Question 19

A game pays $30 with probability 0.2, pays $10 with probability 0.3, and costs $5 (a loss) with probability 0.5. Find the expected value.

A. $6.50
B. $7.50
C. $5.50
D. $8.50

Question 20

The probability of event A is 0.28. Find the probability that A does not occur.

F. 0.72
G. 0.28
H. 1.28
J. 0.5

▸ Answers and Explanations

Show the answers and explanations

1: C. P(A and B) = 0.4 × 0.7 = 0.28. Choice A adds the two probabilities instead of multiplying them, a common independent-events error.

2: F. Six of the 36 equally likely outcomes sum to 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1): 6/36 = 1/6. Choice G undercounts the favorable outcomes, finding only half of the six combinations that sum to 7.

3: A. Order does not matter, so this is a combination: 5C2 = 10. Choice B treats the selection as if order mattered, computing a permutation instead of a combination.

4: H. Order matters (the two offices are distinct), so this is a permutation: 5P2 = 20. Choice F treats the selection as if order didn't matter, computing a combination instead of a permutation.

5: C. Expected value = (100)(0.1) + (20)(0.4) + (−10)(0.5) = 10 + 8 − 5 = 13. Choice A comes from an arithmetic slip while combining the three weighted outcomes.

6: H. Restricting to the Under 30 row: 40 approve out of 50 total, 40/50 = 0.80. Choice G divides the 40 approvals by the full grand total instead of the Under 30 row total, halving the correct value.

7: B. The multiples of 3 on a die are 3 and 6, two of six equally likely outcomes: 2/6=1/3. Choice A overcounts the favorable outcomes, including one value that is not actually a multiple of 3.

8: G. 0.6×0.75=0.45. Choice F adds the two probabilities instead of multiplying them, a common independent-events error.

9: A. Restricting to Group B: 8 No out of 25 total, 8/25=0.32. Choice B reports half the correct proportion, likely from misreading the No count for Group B.

10: G. (60)(0.25)+(15)(0.4)+(−10)(0.35)=15+6−3.5=17.5. Choice F comes from an arithmetic slip while combining the three weighted outcomes.

11: D. Order does not matter, so this is a combination: 8C3=56. Choice A treats the selection as if order mattered, computing a permutation instead of a combination.

12: J. Order matters (distinct offices), so this is a permutation: 7P4=840. Choice F treats the selection as if order didn't matter, computing a combination instead of a permutation.

13: B. Mutually exclusive events cannot occur together, so P(A or B)=P(A)+P(B)=0.3+0.45=0.75. Choice A multiplies the two probabilities instead of adding them; mutually exclusive events use the addition rule, not the independence rule.

14: J. The complement rule: P(not A)=1−P(A)=1−0.35=0.65. Choice F reports P(A) itself instead of computing its complement.

15: B. 5 and 6 are two of six equally likely outcomes: 2/6=1/3. Choice A counts only one favorable outcome instead of both.

16: F. 0.5×0.6=0.3. Choice G adds the two probabilities instead of multiplying them, a common independent-events error.

17: D. Order does not matter, so this is a combination: 7C2=21. Choice A treats the selection as if order mattered, computing a permutation instead of a combination.

18: G. Order matters (distinct offices), so this is a permutation: 5P2=20. Choice F treats the selection as if order didn't matter, computing a combination instead of a permutation.

19: A. (30)(0.2)+(10)(0.3)+(−5)(0.5)=6+3−2.5=6.5. Choice B comes from an arithmetic slip while combining the three weighted outcomes.

20: F. The complement rule: P(not A)=1−P(A)=1−0.28=0.72. Choice G reports P(A) itself instead of computing its complement.