Probability - PSAT Math

1. A card is drawn at random from a standard 52-card deck. What is the probability the card is a heart or a face card (jack, queen, or king)?
   Answer: $\frac{11}{26}$ 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.
2. A box contains 10 cards numbered 1 through 10. One card is selected at random. What is the probability that the number on the card is not a multiple of 3?
   Answer: $\frac{7}{10}$ The question asks for the probability that the number on a randomly selected card from 1 to 10 is not a multiple of 3. The sample space is 10 cards, with multiples of 3 being 3,6,9 (3 outcomes). The favorable outcomes are the remaining 7 cards (1,2,4,5,7,8,10). Thus, P(not multiple of 3) = 7/10, or equivalently 1 - 3/10. A common error is miscounting multiples, like including 10 or forgetting 9. Carefully list the sample space and count favorable outcomes to avoid such mistakes.
3. A computer generates a 3-character code. The first character is a letter chosen uniformly from A, B, or C. The second character is a digit chosen uniformly from 1, 2, 3, or 4. The third character is a letter chosen uniformly from A, B, or C. What is the probability the code starts and ends with the same letter?
   Answer: $\frac{1}{3}$ 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.
4. A school newspaper must fill the positions of editor and photographer from its 8-member staff, 3 of whom are seniors. If the two positions are assigned at random and no student may hold more than one position, what is the probability that both positions are filled by seniors?
   Answer: $\dfrac{3}{28}$ 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.
5. An integer is chosen at random from $1$ through $12$. What is the probability that the integer is a prime number or a multiple of 4?
   Answer: $\dfrac{2}{3}$ When you encounter probability questions involving "or," you're dealing with the union of two sets. The key is identifying each group separately, then checking for any overlap to avoid double-counting. First, let's identify the integers from 1 to 12 that are prime numbers: 2, 3, 5, 7, and 11. That's 5 prime numbers. Next, find the multiples of 4: 4, 8, and 12. That's 3 multiples of 4. Now here's the crucial step: check if any numbers appear in both categories. Since prime numbers have exactly two factors (1 and themselves), and multiples of 4 have 4 as a factor, no number can be both prime and a multiple of 4. This means there's no overlap. The total count of favorable outcomes is $5 + 3 = 8$ integers out of 12 possible outcomes, giving us $\frac{8}{12} = \frac{2}{3}$. Choice A ($\frac{1}{2}$) likely comes from only counting one category or making an error in the setup. Choice C ($\frac{3}{4}$) might result from incorrectly including composite numbers or miscounting. Choice D ($\frac{5}{6}$) could arise from including too many numbers, perhaps confusing factors with multiples. The correct answer is B. Strategy tip: For "or" probability problems, always list out the elements in each category when dealing with small sets like this. It prevents counting errors and helps you spot overlaps. Remember that "or" in probability means addition, but watch for double-counting when sets intersect.
6. A fair coin is flipped 3 times. What is the probability of getting at least one head?
   Answer: $\frac{7}{8}$ 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).
7. A standard 52-card deck is shuffled. What is the probability that the first card dealt is a heart or a king?
   Answer: $\dfrac{4}{13}$ When you encounter probability questions involving "or," you need to carefully apply the addition rule while watching out for overlap between events. To find the probability of drawing a heart or a king, start by identifying each event separately. There are 13 hearts in a 52-card deck, so P(heart) = $\frac{13}{52}$. There are 4 kings in the deck, so P(king) = $\frac{4}{52}$. However, you can't simply add these probabilities because there's overlap—the king of hearts is counted in both events. Using the addition rule: P(heart or king) = P(heart) + P(king) - P(heart and king). Since there's exactly one king of hearts, P(heart and king) = $\frac{1}{52}$. Therefore: P(heart or king) = $\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$, which is choice B. Choice A ($\frac{1}{4}$) represents only the probability of drawing a heart, ignoring the kings entirely. Choice C ($\frac{1}{3}$) doesn't correspond to any logical calculation for this scenario. Choice D ($\frac{5}{13}$) is what you'd get if you incorrectly added P(heart) + P(king) without subtracting the overlap: $\frac{13+4}{52} = \frac{17}{52}$. Remember: whenever you see "or" in probability, check whether the events overlap. If they do, subtract the intersection to avoid double-counting. This is one of the most common traps in PSAT probability questions.
8. A multiple-choice quiz has 5 questions, each with 4 answer choices, and a student guesses randomly on every question. What is the probability that the student answers exactly 3 questions correctly?
   Answer: $\tfrac{45}{512}$ The question is asking for the probability that a student who guesses randomly on a 5-question quiz with 4 choices each gets exactly 3 correct. The relevant outcomes are the binomial trials, with success probability $p=1/4$ for each question, $n=5$, $k=3$. The calculation uses the binomial probability formula $P(k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}$. So, $C(5,3) =10$, $(1/4)^3 =1/64$, $(3/4)^2 =9/16$, then $10 \times (1/64) \times (9/16) = 90 / 1024 = 45/512$. A key error is to use $p=1/2$ instead of $1/4$. Another error is forgetting the binomial coefficient and just doing $p^k \cdot (1-p)^{n-k}$. A test-taking strategy is to recall the binomial formula for "exactly k" in independent trials.
9. A box contains 5 white balls and 5 black balls that are indistinguishable except for color. Two balls are drawn without replacement. What is the probability that the two balls are of different colors?
   Answer: $\dfrac{5}{9}$ 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) = $\frac{5}{10} \times \frac{5}{9} = \frac{25}{90}$ Case 2: Black first, then white - P(black first) = $\frac{5}{10}$ - P(white second | black first) = $\frac{5}{9}$ - P(black then white) = $\frac{5}{10} \times \frac{5}{9} = \frac{25}{90}$ Total probability = $\frac{25}{90} + \frac{25}{90} = \frac{50}{90} = \frac{5}{9}$ Answer A ($\frac{1}{2}$) incorrectly treats this as replacement or assumes equal likelihood without proper calculation. Answer B ($\frac{4}{9}$) likely comes from miscalculating the second draw probabilities. Answer D ($\frac{2}{3}$) might result from confusing this with a different probability setup. Strategy tip: For "without replacement" problems, always update your denominator after each draw, and remember that "different" or "at least one" scenarios often involve adding multiple cases together.
10. A jar contains 5 black beads and 7 white beads. One bead is chosen at random. What is the probability that the bead is not black?
    Answer: $\frac{7}{12}$ This question asks for the probability that a randomly chosen bead is NOT black. The jar contains 5 black beads and 7 white beads, for a total of 12 beads. The favorable outcomes are the white beads (not black), of which there are 7. Therefore, P(not black) = 7/12. A common error would be calculating P(black) = 5/12 instead of P(not black). When a question asks for the probability of NOT an event, count all outcomes except those in the specified event.
11. A spinner has 8 equal sections labeled 1 through 8. It is spun once. What is the probability that the result is a multiple of 3 or an even number?
    Answer: $\frac{5}{8}$ This asks for the probability that a spinner result is a multiple of 3 OR an even number. The multiples of 3 from 1-8 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 outcomes. Therefore, P(multiple of 3 OR even) = 5/8. Note that 6 appears in both sets but is counted only once in the union. A common error is adding 2 + 4 = 6 favorable outcomes without recognizing the overlap. For OR probabilities with overlapping events, use the inclusion-exclusion principle or count the union directly.
12. A spinner has 8 equal sections labeled 1 through 8. The spinner is spun once. Event $A$ is landing on an even number. Event $B$ is landing on a number greater than 5. What is $P(A \cap B)$?
    Answer: $\frac{1}{4}$ This question asks for the probability of the intersection of event $A$ (landing on an even number) and event $B$ (landing on a number greater than 5) when spinning a spinner with 8 equal sections labeled 1 through 8. The relevant outcomes for $A \cap B$ are the numbers that are both even and greater than 5, which are 6 and 8, so there are 2 favorable outcomes out of the total 8 possible. The probability is $P(A \cap B)$ = number of favorable outcomes / total outcomes = $2/8 = 1/4$. This can also be thought of as the overlap between even numbers (2,4,6,8) and numbers >5 (6,7,8). The calculation is straightforward since all sections are equally likely. A key error is confusing intersection with union, which would give $P(A \text{ or } B) = 5/8$ instead. When questions ask for $P(A \cap B)$, focus on outcomes satisfying both conditions simultaneously.
13. A box contains 3 batteries that are defective and 9 batteries that are not defective. Two batteries are selected at random without replacement. What is the probability that at least one of the two selected batteries is defective?
    Answer: $\frac{5}{11}$ 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.
14. A fair coin is flipped twice. What is the probability of obtaining exactly one head?
    Answer: $\dfrac{1}{2}$ When you encounter probability questions involving multiple coin flips, start by identifying all possible outcomes and then count the favorable ones. This systematic approach prevents missing cases or double-counting. For two coin flips, there are four equally likely outcomes: HH, HT, TH, and TT. Each has a probability of $\frac{1}{4}$ since each flip has a $\frac{1}{2}$ chance of heads or tails, and $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. To get exactly one head, you need either HT or TH. That's 2 favorable outcomes out of 4 total outcomes, giving us $\frac{2}{4} = \frac{1}{2}$. This makes C correct. Choice A ($\frac{1}{4}$) represents the probability of any single specific outcome, like getting exactly HH or exactly TT. Students often pick this by confusing the probability of one particular sequence with the probability of a category of sequences. Choice B ($\frac{1}{3}$) doesn't correspond to any logical counting of outcomes in this scenario. This might tempt students who incorrectly think there are only three possibilities: zero heads, one head, or two heads, then assume equal probability. Choice D ($\frac{3}{4}$) represents getting at least one head (HH, HT, or TH), which is the complement of getting no heads (TT only). Strategy tip: Always list all possible outcomes first, then count those that satisfy your condition. For coin problems, remember that order matters—HT and TH are different outcomes even though both have exactly one head.
15. A spinner has 8 equal sections numbered 1 through 8. The spinner is spun once. What is the probability that the result is a multiple of 3 or an even number?
    Answer: $\frac{5}{8}$ 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.
16. A spinner is divided into 8 equal sections labeled A, A, B, B, B, C, C, and D. The spinner is spun once. What is the probability of landing on a letter that is not B?
    Answer: $\frac{5}{8}$ The question asks for the probability of landing on a letter that is not B on a spinner with 8 equal sections: 2 A, 3 B, 2 C, 1 D. The sample space is 8 sections, with 3 B sections. The favorable outcomes are the 5 non-B sections (2 A + 2 C + 1 D). Thus, P(not B) = $\frac{5}{8}$, or 1 - $\frac{3}{8}$. A common error is treating letters as equally likely instead of sections, like assuming $\frac{3}{4}$ letters are not B. Emphasize counting actual outcomes in the sample space, not just unique labels.
17. A bag contains 6 black socks and 4 white socks. Two socks are selected at random without replacement. What is the probability that the two socks are different colors?
    Answer: $\frac{8}{15}$ This problem asks for the probability of selecting two socks of different colors. There are two favorable scenarios: black then white, or white then black. P(black first, white second) = (6/10) × (4/9) = 24/90. P(white first, black second) = (4/10) × (6/9) = 24/90. Total probability = 24/90 + 24/90 = 48/90 = 8/15. A common error is forgetting one of the orderings or not adjusting for the second draw. When order matters for the outcome, consider all possible sequences.
18. A jar contains 5 red marbles, 3 blue marbles, and 2 green marbles. One marble is drawn at random, not replaced, and then a second marble is drawn. What is the probability that both marbles drawn are blue?
    Answer: $\frac{1}{15}$ 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.
19. A multiple-choice quiz has 4 questions, each with 5 answer choices and exactly one correct answer. A student guesses randomly on every question. What is the probability the student gets at least one question correct?
    Answer: $\frac{369}{625}$ 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.
20. A club has 120 members. Of these, 72 have a library card, 50 volunteer monthly, and 30 do both. If one member is chosen at random, what is the probability the member has a library card or volunteers monthly?
    Answer: $\tfrac{23}{30}$ 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.

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