Probability - ACT Math

1. Two fair six-sided dice are rolled. What is the probability that the sum is 7?
   Answer: $\frac{1}{6}$ The sample space for two dice has 36 equally likely outcomes. The favorable outcomes for sum = 7 are $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} = 6$ outcomes. $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$. This is the most likely sum when rolling two dice.
2. A deck of cards is shuffled. What is the probability of drawing a card that is either an Ace or a diamond?
   Answer: 4/13 A standard deck has 4 Aces and 13 diamonds, but the Ace of Diamonds creates overlap. Using inclusion-exclusion: P(Ace or diamond) = P(Ace) + P(diamond) - P(Ace and diamond) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13. The overlap of 1/52 must be subtracted to avoid double-counting the Ace of Diamonds. Choice C (1/4) equals 13/52, which ignores the Aces entirely.
3. If a die is rolled, what is the probability of rolling a 2 or a 5?
   Answer: 1/3 When rolling a die, the sample space consists of 6 equally likely outcomes: ${1, 2, 3, 4, 5, 6}$. The event "rolling a 2 or a 5" includes favorable outcomes ${2, 5}$, giving us 2 favorable outcomes. Since these are mutually exclusive events, $P(2 \text{ or } 5) = P(2) + P(5) = 1/6 + 1/6 = 2/6 = 1/3$. Choice A (1/6) represents the probability of rolling just one specific number.
4. A jar contains 10 candies: 4 red, 3 blue, and 3 green. If one candy is chosen at random, what is the probability that it is not blue?
   Answer: $\frac{7}{10}$ The total number of candies is 4 red + 3 blue + 3 green = 10. The favorable outcomes for "not blue" include all red and green candies: 4 + 3 = 7 candies. Using the probability rule P(not blue) = favorable/total = 7/10. Alternatively, P(not blue) = 1 - P(blue) = 1 - 3/10 = 7/10. Choice B (3/10) gives the probability of selecting blue instead of not blue.
5. In a deck of 52 cards, what is the probability of drawing an Ace or a King?
   Answer: $\frac{2}{13}$ In a standard deck, there are 4 Aces and 4 Kings, giving us 8 favorable cards total. The total number of cards is 52. Since Aces and Kings are mutually exclusive events (no card can be both), we use $P(\text{Ace or King}) = P(\text{Ace}) + P(\text{King}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$. Choice B ($\frac{1}{13}$) incorrectly counts only Aces or only Kings instead of both.
6. If two dice are rolled, what is the probability that the sum is 7?
   Answer: $\frac{1}{6}$ When rolling two dice, there are $6 \times 6 = 36$ total possible outcomes. The favorable outcomes that sum to 7 are: ${(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}$, giving us 6 favorable outcomes. Using $P(\text{sum} = 7) = \frac{\text{favorable}}{\text{total}} = \frac{6}{36} = \frac{1}{6}$. Choice C ($1/36$) incorrectly counts only one way to get a sum of 7 instead of six ways.
7. A dice is rolled twice. What is the probability of rolling a 3 and then a 4?
   Answer: $\frac{1}{36}$ When rolling a die twice, each roll is independent with outcomes from 1 to 6. The probability of rolling a 3 on the first roll is $\frac{1}{6}$, and the probability of rolling a 4 on the second roll is also $\frac{1}{6}$. Since these are independent events, $P(3 \text{ then } 4) = P(3) \times P(4) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$. Choice D ($\frac{1}{6}$) incorrectly uses the probability for just one roll instead of both.
8. What is the probability of drawing a spade or a club from a standard deck of cards?
   Answer: 1/2 A standard deck has 52 cards with 13 spades and 13 clubs, giving us 26 favorable outcomes total. Since drawing a spade and drawing a club are mutually exclusive events, P(spade or club) = P(spade) + P(club) = 13/52 + 13/52 = 26/52 = 1/2. Choice B (1/4) represents the probability of drawing cards from just one suit.
9. A fair coin is flipped and a fair six-sided die is rolled. What is the probability of getting heads and rolling an even number? (Assume the events are independent.)
   Answer: $\tfrac{1}{4}$ The sample space involves two independent events: flipping a coin (2 outcomes) and rolling a die (6 outcomes). The favorable outcomes are getting heads (probability $1/2$) AND rolling an even number (2, 4, or 6, so probability $3/6 = 1/2$). Since the events are independent, we multiply: $P(\text{heads and even}) = P(\text{heads}) \times P(\text{even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Choice B incorrectly uses 1/12, possibly by multiplying $1/2 \times 1/6$ instead of recognizing there are 3 even numbers on a die.
10. What is the probability of rolling an even number on a fair 6-sided die?
    Answer: $\frac{1}{2}$ On a fair 6-sided die, the sample space is $ \{1, 2, 3, 4, 5, 6\} $, giving us 6 total outcomes. The favorable outcomes for rolling an even number are $ \{2, 4, 6\} $, which gives us 3 favorable outcomes. Using $P(\text{even}) = \frac{\text{favorable}}{\text{total}} = \frac{3}{6} = \frac{1}{2}$. Choice B ($\frac{1}{3}$) incorrectly counts only 2 even numbers, while choice D ($\frac{2}{3}$) gives the probability of rolling an odd number.
11. If a number is randomly selected from 1 to 10, what is the probability of selecting a prime number?
    Answer: 2/5 Among numbers 1 to 10, the prime numbers are {2, 3, 5, 7}, giving us 4 favorable outcomes out of 10 total outcomes. Using the probability rule P(event) = favorable / total, we get P(prime) = 4/10 = 2/5. Choice C (1/5) incorrectly counts only 2 prime numbers, while choice D (3/10) suggests only 3 prime numbers were counted.
12. If an event occurs with probability $0.7$, what is $P(\text{not event})$?
    Answer: 0.3 The probability of an event and its complement must sum to 1. If P(event) = 0.7, then P(not event) = 1 - P(event) = 1 - 0.7 = 0.3. This is a fundamental rule of probability where complementary events are mutually exclusive and exhaustive. Choice B (0.7) incorrectly gives the probability of the original event instead of its complement.
13. A fair six-sided die is rolled once. What is the probability of rolling an even number or a number greater than 4? (Assume outcomes are mutually exclusive only when appropriate.)
    Answer: $\frac{2}{3}$ The sample space is {1, 2, 3, 4, 5, 6}. Even numbers are {2, 4, 6} and numbers greater than 4 are {5, 6}. Since 6 appears in both sets, favorable outcomes are {2, 4, 5, 6} = 4 outcomes. $P(\text{even or } >4) = \frac{4}{6} = \frac{2}{3}$. Choice D incorrectly treats these as mutually exclusive.
14. What is the probability of drawing a red card from a standard deck of 52 cards?
    Answer: $\frac{1}{2}$ A standard deck contains 52 cards with 26 red cards (13 hearts + 13 diamonds) and 26 black cards (13 clubs + 13 spades). The favorable outcomes for drawing a red card are the 26 red cards. Using $P(\text{red}) = \frac{\text{favorable}}{\text{total}} = \frac{26}{52} = \frac{1}{2}$. Choice B ($\frac{1}{4}$) incorrectly assumes only one red suit, while choice C ($\frac{3}{4}$) gives the probability of drawing either red or most black cards.
15. A fair die is rolled once. If the probability of rolling a number less than 5 is $p$, what is $P(\text{not less than 5})$ in terms of $p$?
    Answer: $1-p$ Numbers less than 5 on a die are {1, 2, 3, 4}, so p = 4/6 = 2/3. The complement event "not less than 5" means {5, 6}. Using the complement rule: P(not less than 5) = 1 - P(less than 5) = 1 - p. This fundamental probability relationship always holds for complementary events.
16. A box contains 5 red, 4 yellow, and 3 green balls. What is the probability of drawing a yellow or green ball?
    Answer: $7/12$ The box contains $5 + 4 + 3 = 12$ balls total. The event "drawing a yellow or green ball" includes $4 + 3 = 7$ favorable outcomes since these are mutually exclusive events. Using $P(event) = favorable / total$, we get $P(yellow or green) = 7/12$. Choice A ($5/12$) incorrectly counts only the red balls instead of yellow and green combined.
17. A box contains 4 red and 6 black balls. What is the probability of picking a red ball?
    Answer: 2/5 The box contains 4 + 6 = 10 balls total, with 4 red balls as favorable outcomes. Using the probability rule P(event) = favorable / total, we get P(red) = 4/10 = 2/5. Choice B (1/3) would be incorrect, and choice A (2/3) represents the probability of picking a black ball instead of red.
18. A coin is flipped twice. What is the probability of getting two heads?
    Answer: $\frac{1}{4}$ When flipping a coin twice, the sample space consists of 4 equally likely outcomes: {HH, HT, TH, TT}. The favorable outcome for getting two heads is just {HH}, which is 1 outcome. Since the coin flips are independent events, we can also calculate this as $P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Choice C (1/8) might come from incorrectly considering three flips instead of two.
19. If you flip a coin twice, what is the probability of getting two heads?
    Answer: 1/4 When flipping a coin twice, the sample space contains 4 equally likely outcomes: {HH, HT, TH, TT}. The event "getting two heads" has only 1 favorable outcome: {HH}. Applying the probability rule P(event) = favorable / total, we get P(two heads) = 1/4. Alternatively, since coin flips are independent, P(H and H) = P(H) × P(H) = (1/2)(1/2) = 1/4.
20. A bag contains 10 white and 15 black balls. What is the probability of selecting a white ball?
    Answer: 2/5 The bag contains 10 + 15 = 25 balls total, with 10 white balls as favorable outcomes. Using the probability rule P(event) = favorable / total, we get P(white) = 10/25 = 2/5. Choice B (1/2) would be incorrect since white balls don't make up half the total, and choice D (1/3) suggests a different ratio.

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