PSAT Math Quiz: Probability

What this quiz covers

This quiz focuses on Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{1}{8}$
2. $\frac{3}{8}$
3. $\frac{7}{8}$ (correct answer)
4. $\frac{1}{2}$

Explanation: This asks for the probability of getting at least one head in 3 coin flips. The complement of "at least one head" is "no heads" (all tails). With a fair coin, P(tails on one flip) = 1/2, so P(all tails in 3 flips) = (1/2)³ = 1/8. Therefore, P(at least one head) = 1 - 1/8 = 7/8. The complement approach is often easier than counting all favorable outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH). For "at least one" problems, consider using the complement rule: P(at least one) = 1 - P(none).

Question 2

1. $\tfrac{2}{5}$
2. $\tfrac{4}{9}$ (correct answer)
3. $\tfrac{1}{3}$
4. $\tfrac{4}{10}$

Explanation: The question is asking for the conditional probability that the second book chosen is a math book, given that the first was a history book, without replacement from 4 math and 6 history. The relevant outcomes are after the first history is chosen, the remaining books: 4 math, 5 history, total 9. The probability is $\tfrac{4}{9}$, since given first is history, second is math with 4 left out of 9. Note this is conditional, so we don't need the probability of the condition. A key error is to think the first affects differently, but given it happened, just update the counts. Another error is to use the original proportions without updating. A test-taking strategy is to simulate the condition and recalculate the probabilities based on the updated sample space.

Question 3

1. $\frac{1}{3}$
2. $\frac{5}{11}$ (correct answer)
3. $\frac{5}{15}$
4. $\frac{5}{9}$

Explanation: This is a conditional probability question asking for P(blue | not green). We have 15 marbles total: 6 red, 5 blue, and 4 green, so 11 marbles are not green (6 red + 5 blue). Given that the marble is not green, our sample space reduces to these 11 marbles, of which 5 are blue. Therefore, P(blue | not green) = 5/11. A common mistake is calculating P(blue) = 5/15 without considering the condition. For conditional probability, always identify the reduced sample space first, then count favorable outcomes within that space.

Question 4

1. $\frac{1}{3}$ (correct answer)
2. $\frac{1}{4}$
3. $\frac{1}{9}$
4. $\frac{2}{3}$

Explanation: This problem asks for the probability that a 3-character code starts and ends with the same letter. The first character has 3 choices (A, B, or C), the second has 4 choices (1, 2, 3, or 4), and for the code to start and end with the same letter, the third character must match the first. So there are $3 \times 4 \times 1 = 12$ favorable codes out of $3 \times 4 \times 3 = 36$ total possible codes. Therefore, $P(\text{same first and last letter}) = 12/36 = 1/3$. A common error is treating the third character as independent when it's constrained by the condition. When events are dependent, count carefully.

Question 5

1. $\frac{11}{26}$ (correct answer)
2. $\frac{3}{13}$
3. $\frac{1}{4}$
4. $\frac{5}{13}$

Explanation: This question asks for the probability that a randomly drawn card from a standard 52-card deck is a heart or a face card (jack, queen, or king). The relevant outcomes are the 13 hearts and the 12 face cards, but subtracting the 3 heart face cards to avoid double-counting, giving $13 + 12 - 3 = 22$ favorable out of 52. The probability is $P(\text{heart or face}) = \frac{\text{number of hearts} + \text{number of face} - \text{number of heart face}}{52} = \frac{22}{52} = \frac{11}{26}$, using the inclusion-exclusion principle. Simplifying $22/52$ by dividing numerator and denominator by 2 yields $11/26$. This accounts for the overlap between the two categories. A key error is forgetting to subtract the intersection, resulting in $25/52$ instead. When dealing with 'or' in probability, always apply inclusion-exclusion to handle overlaps.

Question 6

1. $\tfrac{1}{9}$
2. $\tfrac{2}{21}$ (correct answer)
3. $\tfrac{5}{15}$
4. $\tfrac{25}{225}$

Explanation: The question is asking for the probability that both marbles drawn are blue, without replacement from a jar with 6 red, 5 blue, and 4 green marbles. The relevant outcomes are the possible ways to draw two marbles from 15, but since order matters in sequential draw, we use sequential probabilities. The calculation is $P(\text{first blue}) = \frac{5}{15} = \frac{1}{3}$, then $P(\text{second blue given first blue}) = \frac{4}{14} = \frac{2}{7}$, so $(\frac{1}{3}) \times (\frac{2}{7}) = \frac{2}{21}$. Alternatively, using combinations, number of ways to choose 2 blue from 5 is $C(5,2) = 10$, total ways to choose 2 from 15 = $C(15,2) = 105$, so $\frac{10}{105} = \frac{2}{21}$. A key error is to assume replacement, giving $(\frac{5}{15})^2 = \frac{1}{9}$. Another error is to add probabilities instead of multiplying for "both". A test-taking strategy is to decide if order matters or not; here, since without replacement, either sequential or combinations work.

Question 7

1. $\frac{9}{22}$
2. $\frac{5}{11}$ (correct answer)
3. $\frac{1}{4}$
4. $\frac{3}{11}$

Explanation: This question asks for the probability that at least one of two batteries selected without replacement is defective, from a box with 3 defective and 9 non-defective. The total batteries are 12, and 'at least one defective' includes outcomes with one or two defective. It is easier to calculate as $1 - P(\text{no defective}) = 1 - \left[ \frac{9}{12} \times \frac{8}{11} \right]$. First, $P(\text{no defective}) = \frac{9}{12} \times \frac{8}{11} = \frac{3}{4} \times \frac{8}{11} = \frac{24}{44} = \frac{6}{11}$. Thus, $P(\text{at least one}) = 1 - \frac{6}{11} = \frac{5}{11}$. Using combinations confirms: $P = \frac{ C(3,1) C(9,1) + C(3,2) C(9,0) }{ C(12,2) } = \frac{ (3 \times 9 + 3 \times 1) }{ 66 } = \frac{ (27 + 3) }{ 66 } = \frac{30}{66} = \frac{5}{11}$. A key error is calculating P(exactly one) only, missing the two defective case. For 'at least' probabilities, complementing with 'none' is an efficient strategy.

Question 8

1. $\tfrac{61}{120}$
2. $\tfrac{23}{30}$ (correct answer)
3. $\tfrac{11}{20}$
4. $\tfrac{3}{20}$

Explanation: This problem requires finding P(library card OR volunteers) using the addition rule for overlapping events. We have 72 members with library cards, 50 who volunteer, and 30 who do both, out of 120 total members. Using P(L or V) = P(L) + P(V) - P(L and V), we get 72/120 + 50/120 - 30/120 = 92/120. Simplifying by dividing both numerator and denominator by 4 gives 23/30. The key insight is recognizing that 30 members are counted in both groups, so we must subtract them once to avoid double-counting. A common error would be adding 72 + 50 = 122 without subtracting the overlap, which would incorrectly suggest more than 100% probability.

Question 9

1. $\tfrac{1}{3}$
2. $\tfrac{5}{11}$ (correct answer)
3. $\tfrac{5}{15}$
4. $\tfrac{5}{9}$

Explanation: This is a conditional probability problem asking for P(green | not blue). The condition "not blue" creates a reduced sample space of only red and green marbles: 6 red + 5 green = 11 marbles total (excluding the 4 blue ones). Among these 11 non-blue marbles, 5 are green, so P(green | not blue) = 5/11. This fraction is already in simplest form since gcd(5,11) = 1. A common mistake is using the original total of 15 marbles as the denominator, giving 5/15 = 1/3, but this ignores the given condition. When solving conditional probability, always identify the new sample space created by the condition before calculating.

Question 10

1. $\dfrac{1}{3}$
2. $\dfrac{1}{2}$
3. $\dfrac{5}{6}$ (correct answer)
4. $1$

Explanation: When you encounter probability questions involving "at least one" outcome, you have two main approaches: calculate the probability directly by considering all favorable cases, or use the complement rule (find the probability of the opposite event and subtract from 1). The complement approach is often faster. Let's use the complement rule here. The complement of "at least one defective bulb" is "no defective bulbs" (both bulbs are working). Since there are 2 working bulbs out of 4 total, the probability that the first bulb selected is working is $\frac{2}{4} = \frac{1}{2}$. After removing one working bulb, there's 1 working bulb left out of 3 remaining, so the probability the second is also working is $\frac{1}{3}$. The probability both are working is $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$. Therefore, the probability of at least one defective bulb is $1 - \frac{1}{6} = \frac{5}{6}$. The answer is C. Choice A ($\frac{1}{3}$) incorrectly calculates the probability of selecting exactly one defective bulb without considering the "exactly two defective" case. Choice B ($\frac{1}{2}$) represents the probability of the first bulb being defective, ignoring the second selection entirely. Choice D ($1$) would only be correct if all bulbs were defective, but we have 2 working bulbs. Strategy tip: For "at least one" probability questions, always consider using the complement rule: P(at least one) = 1 - P(none). It's usually simpler than adding multiple probability cases.

Question 11

1. $0.5$
2. $0.6$
3. $0.7$ (correct answer)
4. $0.8$

Explanation: When you encounter probability questions involving "or" statements, you're dealing with the union of two sets, which requires careful attention to overlap between the groups. To find the probability that a respondent owns a smartphone OR a tablet, you need to use the addition rule for probability: P(A or B) = P(A) + P(B) - P(A and B). The key insight is that simply adding the individual probabilities would double-count people who own both devices. Here's the calculation: P(smartphone or tablet) = P(smartphone) + P(tablet) - P(both) = 0.60 + 0.30 - 0.20 = 0.70. You subtract the overlap because those people are already counted in both the smartphone group and the tablet group. Looking at the wrong answers: Choice (A) 0.5 likely comes from incorrectly thinking this equals the larger probability minus the smaller one (0.60 - 0.30 = 0.30, but that's not even 0.5, so this represents some other calculation error). Choice (B) 0.6 is simply the probability of owning just a smartphone, ignoring the tablet owners entirely. Choice (D) 0.8 results from incorrectly adding all three percentages (0.60 + 0.30 + 0.20 = 1.10, but capped at probability ranges), or from adding the two main probabilities without subtracting the overlap (0.60 + 0.30 = 0.90, though this doesn't quite give 0.8 either). Remember: whenever you see "or" in probability, add the individual probabilities but always subtract any overlap to avoid double-counting.

Question 12

1. $\dfrac{1}{2}$
2. $\dfrac{4}{9}$
3. $\dfrac{5}{9}$ (correct answer)
4. $\dfrac{2}{3}$

Explanation: When you see probability questions involving drawing items "without replacement," you're dealing with dependent events where each draw affects the next. The key is to carefully track how the sample space changes. To find the probability of drawing different colors, you can use two approaches. Let's use the direct method: calculate the probability of one white and one black ball in any order. Case 1: White first, then black

• P(white first) = $\frac{5}{10}$
• P(black second | white first) = $\frac{5}{9}$ (5 black balls remain out of 9 total)
• P(white then black) = $\frac{5}{10} \times \frac{5}{9} = \frac{25}{90}$

• P(black first) = $\frac{5}{10}$
• P(white second | black first) = $\frac{5}{9}$
• P(black then white) = $\frac{5}{10} \times \frac{5}{9} = \frac{25}{90}$

Question 13

1. $\dfrac{3}{20}$
2. $\dfrac{1}{8}$
3. $\dfrac{9}{56}$
4. $\dfrac{3}{28}$ (correct answer)

Explanation: This is a probability question involving combinations without replacement. When you see "at random" with multiple selections where order doesn't matter, you're typically dealing with combinations rather than permutations. To find the probability that both positions go to seniors, you need the ratio of favorable outcomes to total possible outcomes. There are 3 seniors out of 8 total staff members. For the favorable outcomes: You're choosing 2 seniors from the 3 available seniors, which is $\binom{3}{2} = 3$ ways. For the total possible outcomes: You're choosing any 2 people from all 8 staff members, which is $\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28$ ways. Therefore, the probability is $\frac{3}{28}$, which is answer choice D. Let's examine why the other answers are wrong. Choice A ($\frac{3}{20}$) likely comes from incorrectly calculating the total outcomes or confusing this with a different probability setup. Choice B ($\frac{1}{8}$) might result from thinking about the probability of selecting one specific senior, rather than any two seniors. Choice C ($\frac{9}{56}$) appears to come from multiplying $\frac{3}{8} \times \frac{2}{7}$ and then doubling it, which incorrectly treats the positions as distinguishable when they're not. Study tip: For combination probability problems, always identify what you're selecting from what pool, then use $\frac{\text{favorable combinations}}{\text{total combinations}}$. Don't get trapped by treating identical outcomes as different.

Question 14

1. $\dfrac{1}{10}$
2. $\dfrac{11}{50}$ (correct answer)
3. $\dfrac{2}{9}$
4. $\dfrac{1}{4}$

Explanation: When you encounter probability questions involving "or" statements, you need to identify whether the events can overlap and apply the appropriate formula. Here, you're looking for codes that are either "E7" OR begin with "O". First, find the total number of possible codes: 5 vowels × 10 digits = 50 total codes. Next, identify the favorable outcomes. There's exactly 1 way to get "E7". For codes beginning with "O", the second character can be any of the 10 digits (0-9), giving us 10 codes: O0, O1, O2, ..., O9. Since "E7" and "begins with O" cannot happen simultaneously (a code can't start with both E and O), these are mutually exclusive events. Therefore, you simply add: 1 + 10 = 11 favorable outcomes. The probability is $\frac{11}{50}$, which is answer choice B. Answer choice A ($\frac{1}{10}$) only accounts for the probability of getting "E7" specifically, ignoring the "begins with O" condition. Answer choice C ($\frac{2}{9}$) incorrectly calculates the total sample space, possibly confusing the number of vowels with total possibilities. Answer choice D ($\frac{1}{4}$) might result from incorrectly thinking there are only 4 total outcomes or misapplying probability rules. Strategy tip: For "or" probability problems, always check if the events can overlap. If they're mutually exclusive (can't happen together), add the favorable outcomes. If they can overlap, use the formula P(A or B) = P(A) + P(B) - P(A and B).

Question 15

1. $\dfrac{2}{7}$
2. $\dfrac{3}{14}$
3. $\dfrac{9}{35}$ (correct answer)
4. $\dfrac{6}{35}$

Explanation: This is a probability question involving dependent events, where the outcome of one selection affects the probability of the next. When you see problems asking for specific positions in arrangements, think about calculating the probability of each event happening in sequence. To find the probability that the first student is a girl AND the last student is a boy, you multiply the individual probabilities. For the first position, there are 6 girls out of 15 total students, so $P(\text{first is girl}) = \frac{6}{15}$. Once a girl is placed first, 14 students remain for the remaining positions. For the last position, there are still 9 boys among these 14 remaining students, so $P(\text{last is boy | first is girl}) = \frac{9}{14}$. The combined probability is: $\frac{6}{15} \times \frac{9}{14} = \frac{54}{210} = \frac{9}{35}$, which is answer choice C. Let's examine why the other answers are incorrect. Choice A ($\frac{2}{7}$) equals $\frac{6}{21}$, which might result from incorrectly using 21 as a denominator. Choice B ($\frac{3}{14}$) could come from averaging $\frac{6}{15}$ and $\frac{9}{14}$ instead of multiplying them. Choice D ($\frac{6}{35}$) might result from using 6 (number of girls) in the numerator but forgetting to account for the 9 boys in the calculation. Remember: for "and" probability problems with dependent events, multiply the individual probabilities and adjust the totals as selections are made. Always reduce your fraction to lowest terms to match the answer choices.

Question 16

1. $\frac{1}{5}$
2. $\frac{2}{5}$
3. $\frac{4}{5}$ (correct answer)
4. $\frac{3}{5}$

Explanation: This question asks for the probability that a randomly selected candy is not lemon from a bag with 5 chocolate, 3 strawberry, and 2 lemon candies. The relevant outcomes are the non-lemon candies, which are 5 chocolate + 3 strawberry = $8$ favorable out of a total of 10 candies. The probability is $P(\text{not lemon}) = \frac{\text{number of non-lemon}}{\text{total}} = \frac{8}{10} = \frac{4}{5}$. This can also be calculated as $1 - P(\text{lemon}) = 1 - \frac{2}{10} = \frac{8}{10} = \frac{4}{5}$, confirming the result. The calculation assumes equal likelihood for each candy. A key error is miscounting the favorable outcomes, such as including lemon by mistake. When finding probabilities of 'not' an event, subtracting from 1 is often an efficient strategy.

Question 17

1. $\frac{1}{10}$
2. $\frac{1}{15}$ (correct answer)
3. $\frac{2}{15}$
4. $\frac{3}{10}$

Explanation: This problem asks for the probability that both marbles drawn are blue, without replacement. Initially there are 10 marbles total (5 red + 3 blue + 2 green), with 3 blue marbles. For the first draw, P(blue) = 3/10. After removing one blue marble, there are 2 blue marbles left out of 9 total marbles, so P(second blue | first blue) = 2/9. The probability of both events is (3/10) × (2/9) = 6/90 = 1/15. A common error is using 3/10 for both draws, forgetting that the first draw changes the sample space. Always adjust the counts for "without replacement" problems.

Question 18

1. $\frac{256}{625}$
2. $\frac{369}{625}$ (correct answer)
3. $\frac{1}{5}$
4. $\frac{257}{625}$

Explanation: This problem asks for the probability of getting at least one question correct when guessing randomly. With 5 choices per question, the probability of guessing correctly is $\frac{1}{5}$, and incorrectly is $\frac{4}{5}$. It's easier to use the complement: P(at least one correct) = 1 - P(all incorrect). The probability of getting all 4 questions wrong is $(\frac{4}{5})^4 = \frac{256}{625}$. Therefore, P(at least one correct) = 1 - \frac{256}{625} = \frac{369}{625}. A common error is trying to add individual probabilities directly. For "at least one" problems, the complement approach is often simpler.

Question 19

1. $\dfrac{1}{1024}$
2. $\dfrac{243}{1024}$
3. $\dfrac{781}{1024}$ (correct answer)
4. $\dfrac{5}{4}$

Explanation: The question asks for the probability that a student guessing on a 5-question quiz with 4 choices per question gets at least one correct. Each question has a 1/4 chance of being correct and 3/4 chance of being incorrect, following a binomial distribution. The probability of at least one correct is the complement of getting none correct: 1 - (3/4)^5. Calculate (3/4)^5 = 243/1024, so 1 - 243/1024 = 781/1024. A common error is miscalculating the power or confusing it with exactly one correct. For 'at least one' problems, computing the complement of 'none' simplifies the calculation by avoiding summing multiple terms.

Question 20

1. $\frac{5}{8}$ (correct answer)
2. $\frac{3}{8}$
3. $\frac{6}{8}$
4. $\frac{4}{8}$

Explanation: This question asks for the probability that a spinner result is either a multiple of 3 or an even number from sections 1-8. The multiples of 3 are {3, 6}, and the even numbers are {2, 4, 6, 8}. The union of these sets is {2, 3, 4, 6, 8}, which contains 5 numbers (note that 6 appears in both sets but is counted only once). Therefore, P(multiple of 3 OR even) = 5/8. A common error is adding 2 + 4 = 6 favorable outcomes by counting 6 twice. When finding probabilities involving 'or', always identify the overlap to avoid double-counting.