Advaced ACT probability questions

ACT Probability Questions & Concepts

Probability is a key topic in ACT Math, and probability questions are becoming more common—especially among the most difficult problems on the exam. This section focuses on advanced ACT probability problems that require strong reasoning and problem-solving skills. While some of these concepts go beyond the standard high school curriculum, they closely reflect the types of ACT probability questions students will encounter on test day. These concepts are also covered in the ACT Math Book and available on AMAZON.

Probability Without Replacement ACT

One of the most important concepts in ACT Math probability is probability without replacement. In these problems, each selection affects the next because the total number of outcomes decreases after every draw. This makes these ACT probability problems more challenging and common in advanced-level questions.

For example, the probability that the first card drawn from a standard deck is an ace is $\dfrac{4}{52}$. If the first card is an ace, the probability that the second card is also an ace becomes $\dfrac{3}{51}$, since one ace and one total card have already been removed.

Consider a container with 7 red marbles and 7 green marbles. If 3 marbles are drawn without replacement, the probability that all 3 are red is:
$$ \dfrac{7}{14}\cdot\dfrac{6}{13}\cdot\dfrac{5}{12}=\dfrac{5}{52}. $$

Alternatively, this ACT probability question can be solved using permutations:
$$ \dfrac{_7P_3}{_{14}P_3}=\dfrac{7\cdot6\cdot5}{14\cdot13\cdot12}=\dfrac{5}{52}. $$

1. There are 15 green mints and 10 white mints in a bowl. If Portia takes two green mints out of the bowl and eats them and then randomly selects another mint to eat, what is the probability that the third mint is green? 2

$ \dfrac{7}{12} $
$ \dfrac{8}{23} $
$ \dfrac{13}{23} $
$ \dfrac{10}{23} $

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2. If 3 purple marbles and 2 orange marbles are in a bowl and 2 marbles are taken at random without replacement, what is the probability that both marbles selected will be purple? 4

$ \dfrac{3}{10} $
$ \dfrac{9}{25} $
$ \dfrac{3}{5}$
$ \dfrac{1}{4} $

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3. Pocket aces is the best hand in Texas Hold'em, but it can win or lose large amounts of chips. If Candice is dealt 2 cards from a standard deck, what is the probability that both cards are aces? 4

$ \dfrac{1}{200} $
$ \dfrac{1}{221}$
$ \dfrac{1}{26}$
$ \dfrac{1}{256} $

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4. There are 7 red and 3 blue marbles in a silver bowl on an ornate carved wooden table and Ingrid takes 2 marbles randomly without replacement. What is the probability that both marbles will be the same color? 4

$\dfrac{23}{30}$
$ \dfrac{8}{15} $
$ \dfrac{7}{15} $
$ \dfrac{3}{5} $

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5. If there are 5 red and 4 blue marbles in a bowl, and Sigfried draws 3 marbles without replacement, what is the probability that all the marbles drawn are red? 4

$\dfrac{11}{84}$
$\dfrac{9}{84}$
$ \dfrac{5}{42} $
$ \dfrac{1}{14} $

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6. Four balls numbered 1 to 4 are placed in a bin on an ornately carved octagonal table. Emma draws two balls without putting either ball back into the bin. What is the probability that the sum of the numbers Emma draws is exactly 5? 4

$ \dfrac{10}{31} $
$\dfrac{3}{10} $
$ \dfrac{1}{3} $
$\dfrac{3}{8} $

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7. There are 2 red marbles, 2 blue marbles, and 2 green marbles in a bowl. If 3 marbles are drawn at random without replacement, what is the probability all three marbles drawn will be of different colors? 5

$ \dfrac{1}{3} $
$ \dfrac{3}{8}$
$ \dfrac{2}{5} $
$ \dfrac{3}{7}$

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8. If there are 6 red marbles, 4 blue marbles, and 2 green marbles in a bowl and 3 marbles are selected at random, what is the probability to the nearest hundredth they will all be of different colors? 5

$ 0.22 $
$ 0.25 $
$ 0.3 $
$ 0.33 $

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Multiple Events

Independent Events and Unions

9. $A$ and $B$ are independent events and $P(A)=0.8$ and $P(B)=0.7$. What is $P(A\textup{ and }B)$? 3

$ 0.1 $
$ 0.28 $
$ 0.56 $
$ 0.8$

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10. $A$ and $B$ are independent events and $P(A)=0.8$ and $P(B)=0.7$. What is $P(A\textup{ or }B)$? 4

$0.8 $
$ 0.84 $
$ 0.92 $
$ 0.94 $

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11. The probability that the Tigers win the first game is 0.8 and the probability that they win the second game is 0.6. What is the probability that the Tigers win one game and lose the other if the games are independent events? 4

$0.25 $
$ 0.40 $
$0.44 $
$ 0.50 $

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12. If the probability that the Tigers win each game against the Red Sox is 0.7 and the games are independent events, what is the probability that the Tigers win all 3 games of a 3-game series? 3

$ 0.125$
$ 0.25 $
$ 0.3 $
$ 0.343 $

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13. There are 5 multiple choice answer choices to each ACT problem. If Cyrus had only 2 problems left blank at the end of Math ACT and randomly guessed both problems, what is the probability that he got exactly 1 of the 2 questions right? 4

$ \dfrac{6}{25} $
$ \dfrac{1}{3} $
$ \dfrac{2}{5} $
$ \dfrac{8}{25} $

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14. If the Giants have a 60% chance of winning each of two games, and the games are independent events, what is the probability that the Giants lose both games? 4

$ 16\% $
$ 18\% $
$ 20\% $
$ 25\% $

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Repeated Trials

To determine the probability that an event occurs at least once in a series of trials, subtract the probability of zero occurrences from 1. For example, if a product has a defect rate of 0.05, the probability that at least 1 item in a group of 10 is defective is: $$ 1 - 0.95^{10} \approx 0.401. $$ The probability that Sylvester obtains exactly 4 heads when flipping 8 coins is the ratio of the number of favorable outcomes to the total number of possible outcomes: $$ \dfrac{_8C_4}{2^8}=\dfrac{70}{256}=\dfrac{35}{128}. $$

15. As part of a game, Attila tosses 7 fair coins simultaneously. In order for Attila to win the game, all the coins must land heads face up. What is the probability that Attila wins? 4

$ \dfrac{1}{64}$
$ \dfrac{1}{128} $
$ \dfrac{1}{256} $
$ \dfrac{5}{512}$

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Dice Problems

When rolling two 6-sided dice, the number of ways to obtain a specific sum follows a clear pattern. There is 1 way to roll a 2 ($(1,1)$), so the probability is $\dfrac{1}{36}$. There are 2 ways to roll a 3 ($(1,2), (2,1)$), making the probability $\dfrac{2}{36}=\dfrac{1}{18}$. This pattern continues peaking at the sum of 7, which has 6 outcomes ($(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$) and a probability of $\dfrac{6}{36}=\dfrac{1}{6}$.The probability of rolling a sum of 8 is equal to that of a sum of 6. This symmetry also applies to the pairs $9\&5$, $10\&4$, $11\&3$, and finally $12\&2$. Similar questions may involve rolling two 10-sided dice.

	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

The table illustrates the outcomes for a sum of 5, which corresponds to a probability of $\dfrac {4} {36} = \dfrac {1} {9}$. Although the problem solutions do not explicitly show tables, constructing one is often the most reliable approach. High-scoring students are advised to practice this method.

16. If Crassius rolls a standard die and flips a coin, what is the probability of getting either heads or a 6? 3

$\dfrac{5}{12} $
$\dfrac{13}{24}$
$ \dfrac{7}{12} $
$ \dfrac{2}{3} $

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17. If Denise rolls two standard dice, what is the probability that the sum is exactly 9? 5

$\dfrac{1}{9} $
$\dfrac{5}{36} $
$ \dfrac{1}{6} $
$ \dfrac{1}{12} $

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18. If Alberto rolls two 6-sided dice and wins a prize when the sum is 8 or greater, what is the probability Alberto wins a prize?5

$ \dfrac{7}{18}$
$\dfrac{5}{12}$
$\dfrac{4}{9}$
$ \dfrac{1}{3} $

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Is the Product Odd or Positive?

For the product of a set of integers to be odd, every factor must be odd. Conversely, for the product of two numbers to be positive, both numbers must share the same sign (either both positive or both negative).Consider two sets of cards: one numbered 1--11 and another numbered 21--35. If Pedro randomly selects a card from each set, the probability that the product is odd depends on both selected numbers being odd. Since the selections are independent events, the probability of drawing two odd numbers is the product of the individual probabilities: $$ \dfrac{6}{11} \cdot \dfrac{8}{15} = \dfrac{48}{165} = \dfrac{16}{55}. $$

19. Gertrude randomly selects 3 numbers from slips of paper $\{1,2,3,4,5\}$. After selecting each slip, she puts it back with the others and thoroughly mixes up the slips. What is the probability that the product of all 3 numbers selected is odd?5

$ \dfrac{1}{8}$
$ \dfrac{27}{125}$
$ \dfrac{1}{4}$
$ \dfrac{1}{10}$

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20. Suppose Samantha randomly selects 2 different numbers from the set $\{1,2,3,4,5\}$ (that is, after choosing the first number, she selects from the four remaining numbers). What is the probability that the product of the selected numbers is odd? 5

$\dfrac{1}{2}$
$\dfrac{3}{5}$
$\dfrac{3}{10}$
$\dfrac{1}{4}$

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21. Mary Ann randomly selects one card from a set of cards numbered $\{1,2,3,4,5\}$ and another card from a pile of cards numbered $\{6,7,8,9,10\}$ . What is the probability that the product of the numbers on those two cards is odd? 5

$ \dfrac{2}{9} $
$ \dfrac{1}{3} $
$ \dfrac{17}{75}$
$ \dfrac{6}{25} $

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Is the Product Odd or Positive?

These problems assess logical reasoning and are rarely emphasized in standard textbooks.For example, consider a scenario where 10 people are divided into two groups of 5. What is the probability that Alice and Lisa are in the same group? Assume Alice is assigned to one group. There are 9 remaining spots for Lisa, 4 of which are in Alice's group. Thus, the probability is: $$ \dfrac{4}{9}. $$

22. Bob and Kevin will be among 10 people randomly seated at a circular table in Chemistry class. What is the probability that they will be seated next to each other? 5

$ \dfrac{2}{9} $
$ \dfrac{4}{9} $
$ \dfrac{1}{3} $
$ \dfrac{1}{4}$

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23. The coach splits 6 team members randomly into groups of 3. If Alvin and Bob are among the 6, what is the probability that they will be in the same group?5

$ \dfrac{2}{5} $
$ \dfrac{4}{9} $
$ \dfrac{3}{7} $
$ \dfrac{1}{3} $

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24. Jane and Julie are among 8 players in a tennis tournament in June at a park in Santa Monica. If the pairings for each match are random, what is the probability that they will be paired together in the first round? 5

$ \dfrac{1}{5}$
$ \dfrac{3}{14} $
$\dfrac{1}{4}$
$ \dfrac{1}{7} $

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Drawing One Card

The probability of either of two events occurring is the sum of their individual probabilities minus the probability that both occur. This subtraction is essential to avoid double counting the outcomes where both events happen simultaneously. Therefore, the probability of drawing an Ace or a Spade from a standard deck is: $$ \dfrac{13}{52}+\dfrac{4}{52}-\dfrac{1}{52}=\dfrac{16}{52}=\dfrac{4}{13}. $$ In contrast, for "and" conditions involving independent events (such as rank and suit), multiply the individual probabilities.

25. If Caligula draws a card from a standard deck, what is the probability that the card is a spade and a face card? 4

$\dfrac{11}{26}$
$\dfrac{1}{52}$
$\dfrac{3}{52}$
$\dfrac{1}{13}$

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26. If Benedict draws a card from a standard deck, what is the probability that the card is a spade or a face card? 4

$\dfrac{25}{52} $
$\dfrac{5}{13} $
$ \dfrac{23}{52} $
$ \dfrac{11}{26} $

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Answer key: 1.C, 2.A, 3.B, 4.B, 5.C, 6.C, 7.C, 8.A, 9.C, 10.D, 11.C, 12.D, 13.D, 14.A, 15.B, 16.C, 17.A, 18.B, 19.B, 20.C, 21.D, 22.A, 23.A, 24.D, 25.C, 26.D.

Solutions

1. (C) After eating two green mints, there are $15-2=13$ green mints remaining out of a new total of $25-2=23$ mints. Therefore, the probability is $\dfrac{13}{23}$.

2. (A) Initially, there are 5 marbles, 3 of which are purple. After drawing a purple marble, there are 4 marbles remaining, 2 of which are purple. Multiplying the probabilities yields: $$ \dfrac{3}{5}\cdot \dfrac{2}{4}=\dfrac{6}{20}=\dfrac{3}{10}. $$

3. (B) Initially, there are 4 Aces among 52 cards. After drawing the first Ace, there are 3 Aces remaining among 51 cards. Multiplying the probabilities yields: $$ \dfrac{4}{52}\cdot\dfrac{3}{51}=\dfrac{1}{221}. $$

4. (B) To find the probability that both marbles are the same color, sum the probability of selecting two red marbles and the probability of selecting two blue marbles: $$ \dfrac{7}{10}\cdot\dfrac{6}{9}+\dfrac{3}{10}\cdot\dfrac{2}{9}=\dfrac{42+6}{90}=\dfrac{48}{90}=\dfrac{8}{15}. $$

5. (C) Because the selections are without replacement, the number of red marbles and the total number of marbles decrease after each draw. Multiplying the sequential probabilities yields: $$ \dfrac{5}{9}\cdot\dfrac{4}{8}\cdot\dfrac{3}{7}=\dfrac{5}{42}. $$

6. (C) There are $4 \times 3 = 12$ total possible outcomes, as there are 4 choices for the first draw and 3 for the second. The four outcomes with a sum of 5 are $(1,4)$, $(2,3)$, $(3,2)$, and $(4,1)$. Therefore, the probability is: $$ \dfrac{4}{12}=\dfrac{1}{3}. $$

7. (C) There are $3!$ distinct arrangements for drawing three different colors (RBG, RGB, BRG, BGR, GRB, GBR). Multiplying the probability of one specific sequence by the number of arrangements yields: $$ 3!\cdot \left(\dfrac{2}{6}\cdot \dfrac{2}{5}\cdot \dfrac{2}{4}\right) = 6\cdot \dfrac{8}{120}=\dfrac{48}{120}=\dfrac{2}{5}. $$

8. (A) There are $3!$ distinct arrangements for selecting three marbles of different colors. Multiplying the probability of a single sequence by the number of arrangements yields: $$ 3! \cdot \left(\dfrac{6}{12}\cdot \dfrac{4}{11}\cdot \dfrac{2}{10}\right) = 6 \cdot \dfrac{48}{1320} \approx 0.22. $$

9. (C) Since events $A$ and $B$ are independent, the probability of both occurring is the product of their individual probabilities. Multiplying the values yields: $$ P(A \textup{ and } B) = P(A) \cdot P(B) = 0.8 \cdot 0.7 = 0.56. $$

10. (D) The probability of $A$ or $B$ occurring is the sum of their individual probabilities minus the probability that both occur. Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ yields: $$ 0.8 + 0.7 - (0.8 \cdot 0.7) = 1.5 - 0.56 = 0.94. $$

11. (C) To find the probability that the Tigers win exactly one game, sum the probability of winning the first and losing the second with the probability of losing the first and winning the second. This calculation yields: $$ (0.8 \cdot 0.4) + (0.2 \cdot 0.6) = 0.32 + 0.12 = 0.44. $$ Alternatively, subtract the probabilities of winning both games and losing both games from 1: $$ 1 - (0.8 \cdot 0.6) - (0.2 \cdot 0.4) = 1 - 0.48 - 0.08 = 0.44. $$

12. (D) Since the games are independent events, the probability of winning all three games is the product of the individual probabilities. Calculating the value yields: $$ 0.7^3 = 0.7 \cdot 0.7 \cdot 0.7 = 0.343. $$

13. (D) To obtain exactly one correct answer, one guess must be correct ($\dfrac{1}{5}$) and the other incorrect ($\dfrac{4}{5}$). Since this can occur in two orders (Correct-Incorrect or Incorrect-Correct), multiply the probability of a single sequence by 2: $$ 2\cdot\left(\dfrac{1}{5}\cdot\dfrac{4}{5}\right)=\dfrac{8}{25}. $$ Alternatively, subtract the probabilities of getting both questions correct and both questions incorrect from 1: $$ 1-\left(\dfrac{1}{5}\right)^2-\left(\dfrac{4}{5}\right)^2= 1-\dfrac{1}{25}-\dfrac{16}{25}=\dfrac{8}{25}. $$

14. (A) The probability of losing a single game is $1 - 0.60 = 0.40$. Since the games are independent, the probability of losing both is the product of the individual probabilities. Multiplying the values yields: $$ 0.40 \cdot 0.40 = 0.16 = 16\%. $$

15. (B) Since the coin tosses are independent events, the probability of obtaining heads on all 7 coins is the product of the individual probabilities. Calculating the value yields: $$ \left(\dfrac{1}{2}\right)^7 = \dfrac{1}{2^7} = \dfrac{1}{128}. $$

16. (C) Subtract the probability that neither event occurs (Tails and a non-6 roll) from 1. The calculation yields: $$ 1-\left(\dfrac{1}{2}\cdot\dfrac{5}{6}\right)=1-\dfrac{5}{12}=\dfrac{7}{12}. $$

17. (A) A sum of 9 corresponds to the outcomes $(3,6)$, $(4,5)$, $(5,4)$, and $(6,3)$. There are $6 \times 6 = 36$ total outcomes for two dice. Therefore, the probability is: $$ \dfrac{4}{36}=\dfrac{1}{9}. $$ Note: Constructing a table with rows and columns representing the values of each die can help visualize the sample space.

18. (B) The favorable outcomes for a sum of 8 or greater are: $$ \begin{gathered} (2,6) \\ (3,5), (3,6) \\ (4,4), (4,5), (4,6) \\ (5,3), (5,4), (5,5), (5,6) \\ (6,2), (6,3), (6,4), (6,5), (6,6) \end{gathered} $$ Summing these yields $1+2+3+4+5=15$ favorable outcomes out of 36 total outcomes. Therefore, the probability is: $$ \dfrac{15}{36}=\dfrac{5}{12}. $$

19. (B) For the product of three integers to be odd, all three integers must be odd. The set $\{1, 2, 3, 4, 5\}$ contains three odd numbers ($1, 3, 5$) out of five total. Since the selections are made with replacement, the events are independent. Therefore, the probability is: $$ \left(\dfrac{3}{5}\right)^3 = \dfrac{27}{125}. $$

20. (C) For the product of two integers to be odd, both integers must be odd. Initially, the set $\{1, 2, 3, 4, 5\}$ contains 3 odd numbers out of 5. After one odd number is selected, there are 2 odd numbers remaining out of 4. Multiplying the probabilities yields: $$ \dfrac{3}{5} \cdot \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}. $$

21. (D) For the product of two integers to be odd, both integers must be odd. The first set, $\{1, 2, 3, 4, 5\}$, contains three odd numbers ($1, 3, 5$) out of five. The second set, $\{6, 7, 8, 9, 10\}$, contains two odd numbers ($7, 9$) out of five. Multiplying the probabilities yields: $$ \dfrac{3}{5}\cdot\dfrac{2}{5}=\dfrac{6}{25}. $$

22. (A) Fix the position of one person (e.g., Bob). There are 9 remaining seats available for Kevin. Two of these seats are adjacent to Bob (one to his left and one to his right). Therefore, the probability is: $$ \dfrac{2}{9}. $$

23. (A) Fix Alvin's placement in a group. Since there are 6 people total, there are 5 remaining spots for Bob. Of these 5 spots, 2 are in Alvin's group and 3 are in the other group. Therefore, the probability that Bob lands in one of the 2 spots in Alvin's group is: $$ \dfrac{2}{5}. $$

24. (D) Consider Jane's perspective. She must be paired with one of the other 7 participants in the tournament. Since the pairings are random and Julie is one of these 7 players, the probability is: $$ \dfrac{1}{7}. $$

25. (C) The event of drawing a spade and the event of drawing a face card are based on independent attributes (suit and rank). The probability of drawing a spade is $\dfrac{1}{4}$ and the probability of drawing a face card is $\dfrac{12}{52} = \dfrac{3}{13}$. Multiplying these probabilities yields: $$ \dfrac{1}{4} \cdot \dfrac{3}{13} = \dfrac{3}{52}. $$

26. (D) There are 13 spades and 12 face cards in a standard deck. However, 3 cards are both spades and face cards (Jack, Queen, and King of Spades). To avoid double counting, subtract the intersection: $$ \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} = \dfrac{22}{52} = \dfrac{11}{26}. $$