Probability - SAT Math

1. A number is chosen at random from the integers 1 through 20, inclusive. What is the probability that the number is a multiple of 4 or a multiple of 5?
   Answer: $\frac{8}{20}$ We need P(multiple of 4 OR multiple of 5) from integers 1-20. Multiples of 4: {4, 8, 12, 16, 20} = 5 numbers. Multiples of 5: {5, 10, 15, 20} = 4 numbers. The number 20 appears in both sets (it's the LCM of 4 and 5). Using inclusion-exclusion: favorable outcomes = 5 + 4 - 1 = 8. Therefore $P = \frac{8}{20}$. We can verify by listing: {4, 5, 8, 10, 12, 15, 16, 20} = 8 numbers. The key is recognizing that 20 is counted in both sets and must not be double-counted.
2. A class has 30 students: 18 take Spanish, 12 take French, and 5 take both Spanish and French. If one student is chosen at random, what is the probability the student takes Spanish or French?
   Answer: $\frac{25}{30}$ This question asks for the probability that a randomly chosen student from a class of 30 takes Spanish or French, where 18 take Spanish, 12 take French, and 5 take both. The relevant outcomes are the students taking at least one language, which is $18 + 12 - 5 = 25$. The probability is $25/30$. This uses inclusion-exclusion to avoid double-counting the 5 who take both. A common error is adding without subtracting, giving $30/30$, which overcounts. Always subtract the intersection when finding the union in 'or' probabilities.
3. A jar contains 6 white and 4 black beads. Two beads are drawn at random with replacement (the first bead is returned before drawing the second). What is the probability that at least one of the two beads drawn is black?
   Answer: $\tfrac{16}{25}$ This question asks for the probability that at least one of two beads drawn with replacement is black, from a jar with 6 white and 4 black beads (total 10). The total outcomes for each draw are 10, and 'at least one black' is the complement of both white. The probability of both white is (6/10) × (6/10) = 36/100 = 9/25, so P(at least one black) = 1 - 9/25 = 16/25. This uses the complement because it's easier than adding P(first black and second white) + P(first white and second black) + P(both black). A common error is forgetting replacement, treating it as without and using 6/10 × 5/9 for both white. Another mistake is calculating only both black. When using complements, ensure the sample space is correctly identified for independent events.
4. A jar contains 6 red marbles, 5 blue marbles, and 4 green marbles. Two marbles are drawn at random without replacement. What is the probability that both marbles drawn are blue?
   Answer: $\frac{2}{21}$ This problem asks for the probability of drawing two blue marbles without replacement from a jar containing 6 red, 5 blue, and 4 green marbles (15 total). For the first draw, P(blue) = 5/15 = 1/3. After removing one blue marble, there are 4 blue marbles left out of 14 total marbles, so P(second blue | first blue) = 4/14 = 2/7. The probability of both events is (5/15) × (4/14) = (1/3) × (2/7) = 2/21. A common mistake is treating this as replacement, which would give (5/15)² = 25/225. When drawing without replacement, always adjust both the numerator and denominator for the second draw.
5. A bag contains 4 red balls and 6 yellow balls. Two balls are drawn at random without replacement. What is the probability that the two balls are different colors?
   Answer: $\frac{8}{15}$ This problem asks for the probability of drawing two balls of different colors without replacement from 4 red and 6 yellow balls. There are two ways: red then yellow, or yellow then red. P(red then yellow) = (4/10) × (6/9) = 24/90, and P(yellow then red) = (6/10) × (4/9) = 24/90. Total probability = 24/90 + 24/90 = 48/90 = 8/15. The key insight is considering both orderings and adjusting the denominator for the second draw without replacement.
6. A jar contains 5 lemon candies and 7 cherry candies. Two candies are drawn without replacement. What is the probability that both candies are cherry?
   Answer: $\tfrac{7}{12}\cdot\tfrac{6}{11}$ This question involves drawing two candies without replacement, asking for the probability that both are cherry. The jar has 5 lemon + 7 cherry = 12 total candies. For the first draw, $P(\text{cherry}) = \frac{7}{12}$. After removing one cherry candy, there are 6 cherry candies left out of 11 total, so $P(\text{second cherry}|\text{first cherry}) = \frac{6}{11}$. Therefore, $P(\text{both cherry}) = \frac{7}{12} \times \frac{6}{11}$. The key distinction is "without replacement," which means the second probability changes based on the first outcome. A common error is using $\frac{7}{12} \times \frac{7}{12}$, which would only be correct for drawing with replacement.
7. A box contains 3 defective bulbs and 9 nondefective bulbs. Two bulbs are selected at random without replacement. What is the probability that exactly one of the two bulbs is defective?
   Answer: $2\cdot\tfrac{3}{12}\cdot\tfrac{9}{11}$ This question asks for the probability of selecting exactly one defective bulb when drawing two bulbs without replacement from a box with 3 defective and 9 nondefective bulbs. Exactly one defective means either (defective first, nondefective second) OR (nondefective first, defective second). For the first case: P(D then N) = $\frac{3}{12} \times \frac{9}{11}$. For the second case: P(N then D) = $\frac{9}{12} \times \frac{3}{11}$. Adding these: P(exactly one defective) = $\frac{3}{12} \times \frac{9}{11} + \frac{9}{12} \times \frac{3}{11} = 2 \times \left( \frac{3}{12} \times \frac{9}{11} \right)$. The factor of 2 accounts for both possible orders. A common error is calculating only one order or forgetting that the probabilities change without replacement.
8. A quiz has 5 multiple-choice questions, each with 4 answer choices, and a student guesses randomly on every question. What is the probability that the student gets at least one question correct?
   Answer: $\tfrac{781}{1024}$ The question is asking for the probability that the student gets at least one question correct by guessing. The relevant outcomes are the possible guess results, but since independent, easier to compute complement. The probability of getting a question wrong is $\frac{3}{4}$, so $P(\text{all wrong}) = \left( \frac{3}{4} \right)^5 = \frac{243}{1024}$. Then $P(\text{at least one correct}) = 1 - \frac{243}{1024} = \frac{781}{1024}$. A common error is to calculate P(exactly one) or something instead of at least one. Another error is to use $\left( \frac{1}{4} \right)^5$ for all correct. Strategy: For 'at least one,' consider the complement 'none' which is often easier to calculate.
9. A jar has 6 black beads and 4 white beads. Two beads are drawn without replacement. What is the probability that both beads are black?
   Answer: $\frac{1}{3}$ This problem involves drawing without replacement, so the draws are not independent. Initially: 6 black, 4 white (10 total). P(first black) = $\frac{6}{10}$; After drawing one black bead: 5 black, 4 white (9 total remaining); P(second black|first black) = $\frac{5}{9}$. Therefore, P(both black) = $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$. The key error to avoid is treating the draws as independent and calculating $\frac{6}{10} \times \frac{6}{10}$. Without replacement means the first draw affects the second draw's probabilities.
10. A jar contains 5 black beads and 7 white beads. One bead is drawn at random and not replaced, then a second bead is drawn. What is the probability that the two beads are different colors?
    Answer: $\frac{35}{66}$ This problem asks for the probability of drawing two beads of different colors from a jar with 5 black and 7 white beads (12 total). There are two ways this can happen: black then white, or white then black. P(black then white) = (5/12) × (7/11) = 35/132, and P(white then black) = (7/12) × (5/11) = 35/132. The total probability is 35/132 + 35/132 = 70/132 = 35/66. A common mistake is calculating only one ordering instead of both. When finding probabilities of "different" outcomes, consider all possible orderings.
11. A teacher has 30 students. Of them, 18 submitted homework on time, and 12 did not. The teacher randomly selects 2 students without replacement. What is the probability that both selected students submitted homework on time?
    Answer: $\frac{51}{145}$ This problem involves selecting 2 students without replacement from 30 total, where 18 submitted homework on time. For the first selection, P(on time) = 18/30 = 3/5, and for the second selection (after removing one on-time student), P(on time) = 17/29. The probability both are on time is (18/30) × (17/29) = 306/870 = 51/145. The key is recognizing that without replacement changes both the numerator and denominator for the second draw. A common error is using 18/30 for both draws, which would apply only with replacement.
12. In a class of 30 students, 18 have a dog, 12 have a cat, and 5 have both a dog and a cat. What is the probability that a student has a dog or a cat?
    Answer: $\frac{5}{6}$ This question asks for the probability that a student has a dog OR a cat, which requires using the addition principle for overlapping events. $P(dog or cat) = P(dog) + P(cat) - P(both) = (18 + 12 - 5)/30 = 25/30 = 5/6$.
13. A card is drawn at random from a standard 52-card deck. What is the probability that the card is a heart or a face card (J, Q, or K)?
    Answer: $\frac{22}{52}$ This problem asks for $P(\text{heart OR face card})$ from a standard 52-card deck. Hearts: 13 cards; Face cards (J, Q, K in all suits): 12 cards; Hearts that are face cards: 3 cards (J♥, Q♥, K♥). Using $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$: $P = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52}$. The critical step is subtracting the 3 cards that are both hearts AND face cards to avoid double-counting. When dealing with playing cards, always visualize the overlap between categories.
14. A raffle has 10 tickets numbered 1–10 in a box. Two tickets are drawn without replacement. What is the probability that both ticket numbers are greater than 6?​
    Answer: $\tfrac{4}{10}\cdot\tfrac{3}{9}$ This question involves drawing two tickets without replacement from tickets numbered 1-10, asking for the probability that both numbers are greater than 6. Numbers greater than 6 are {7, 8, 9, 10}, so there are 4 favorable tickets initially. For the first draw, P(>6) = 4/10. After removing one ticket greater than 6, there are 3 such tickets left out of 9 total, so P(second >6|first >6) = 3/9. Therefore, P(both >6) = 4/10 × 3/9. The key is recognizing that without replacement, the second probability depends on the first outcome. A common error is using 4/10 × 4/10, which would only apply with replacement.
15. A bag contains 3 black socks and 7 white socks. Two socks are selected at random without replacement. What is the probability that at least one of the socks selected is black?​
    Answer: $\frac{8}{15}$ This problem asks for P(at least one black sock) when drawing 2 socks from 3 black and 7 white (10 total) without replacement. It's easier to use the complement: P(at least one black) = 1 - P(both white). P(both white) = (7/10) × (6/9) = 42/90 = 7/15. Therefore, P(at least one black) = 1 - 7/15 = 8/15. The complement approach is often simpler than calculating P(exactly one black) + P(both black). When you see "at least one," consider using the complement rule with "none."
16. A password is formed by randomly choosing one letter from {A, B, C, D} and then randomly choosing one digit from {1, 2, 3}. All choices are equally likely. What is the probability that the password contains the letter A and an even digit?
    Answer: $\frac{1}{12}$ We need P(letter A AND even digit) when forming a password. First, choose from {A, B, C, D}: $P(A) = \frac{1}{4}$. Then choose from {1, 2, 3}: only 2 is even, so $P(\text{even}) = \frac{1}{3}$. Since the choices are independent, $P(A \text{ and even}) = P(A) \times P(\text{even}) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$. We can verify by counting: there are $4 \times 3 = 12$ total passwords, and only one (A2) has both A and an even digit, giving $\frac{1}{12}$. The key insight is recognizing that letter and digit choices are independent events.
17. A bag contains 8 black tiles and 2 white tiles. One tile is drawn at random, replaced, and then a second tile is drawn. What is the probability that exactly one of the two tiles drawn is white?
    Answer: $\frac{8}{25}$ This problem involves drawing tiles with replacement, asking for the probability of getting exactly one white tile in two draws from a bag with 8 black and 2 white tiles. There are two ways this can happen: white then black, or black then white. P(white then black) = (2/10) × (8/10) = 16/100, and P(black then white) = (8/10) × (2/10) = 16/100. The total probability is 16/100 + 16/100 = 32/100 = 8/25. The key insight is that with replacement, each draw has the same probabilities, and we must consider both orderings.
18. A bag contains 5 black tiles and 3 white tiles. One tile is drawn at random, its color is recorded, and then it is replaced. A second tile is drawn. What is the probability that the two tiles drawn are the same color?
    Answer: $\frac{17}{32}$ This question asks for the probability that two tiles drawn with replacement from a bag with 5 black and 3 white tiles are the same color. The total tiles are 8, with each draw independent. The probability is $P(\text{both black}) + P(\text{both white}) = \left(\frac{5}{8}\right)^2 + \left(\frac{3}{8}\right)^2 = \frac{25}{64} + \frac{9}{64} = \frac{34}{64} = \frac{17}{32}$. This considers the two mutually exclusive ways to get the same color. A common error is treating it as without replacement, calculating $\left(\frac{5}{8}\right)\left(\frac{4}{7}\right) + \left(\frac{3}{8}\right)\left(\frac{2}{7}\right) = \frac{15}{32}$. Since replacement is specified, maintain the same sample space for each draw.
19. A box contains 12 tickets numbered 1 through 12. One ticket is selected at random. What is the probability that the number on the ticket is a multiple of 3 or is prime? Express your answer as a fraction in simplest form.
    Answer: $\frac{2}{3}$ This question asks for $P(\text{multiple of 3 OR prime})$ from tickets numbered 1-12. Multiples of 3 are: 3, 6, 9, 12 (4 numbers). Prime numbers from 1-12 are: 2, 3, 5, 7, 11 (5 numbers). Number 3 appears in both sets, so using inclusion-exclusion: favorable outcomes = $4 + 5 - 1 = 8$. Therefore, $P(\text{multiple of 3 OR prime}) = \frac{8}{12} = \frac{2}{3}$. Be careful not to double-count numbers that satisfy both conditions when dealing with OR probabilities.
20. A card is drawn at random from a set of 10 cards labeled 1 through 10. What is the probability that the number is a multiple of 3 or a multiple of 5?
    Answer: $\tfrac{1}{2}$ This asks for the probability that a number from 1-10 is a multiple of 3 OR a multiple of 5. Multiples of 3: $ \{3, 6, 9\} $ (3 numbers). Multiples of 5: $ \{5, 10\} $ (2 numbers). No overlap since 15 > 10. Using the addition rule: $P(\text{mult. of 3 OR mult. of 5}) = 3/10 + 2/10 = 5/10 = 1/2$. A common error would be looking for common multiples when the ranges don't allow for any. When the LCM of two numbers exceeds your range, there's no overlap to subtract.

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