PSAT Math Quiz: Probability

20 questions · exam conditions

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ProbabilityQuestion 1 of 20

A standard 52-card deck is shuffled. What is the probability that the first card dealt is a heart or a king?

\dfrac{1}{4}

\dfrac{4}{13}

\dfrac{1}{3}

\dfrac{5}{13}

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PSAT Math Quiz

PSAT Math Quiz: Probability

Practice Probability in PSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

A standard 52-card deck is shuffled. What is the probability that the first card dealt is a heart or a king?

$\dfrac{1}{4}$
$\dfrac{4}{13}$ (correct answer)
$\dfrac{1}{3}$
$\dfrac{5}{13}$

Explanation: 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.

Question 2

A fair coin is flipped twice. What is the probability of obtaining exactly one head?

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

Explanation: 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.

Question 3

A fair six-sided die is rolled three times. What is the probability that the sum of the three rolls equals $4$ ?

$\dfrac{1}{36}$
$\dfrac{1}{72}$ (correct answer)
$\dfrac{1}{54}$
$\dfrac{1}{216}$

Explanation: When you encounter probability questions involving multiple events, you need to find the total number of favorable outcomes and divide by the total possible outcomes. First, let's establish the sample space. Rolling a six-sided die three times gives us

6^3 = 216

total possible outcomes. Now we need to find how many ways three dice can sum to 4. Since each die shows at least 1, the minimum possible sum is 3. To get a sum of 4, we need exactly one additional pip distributed among the three dice. This means exactly one die shows 2, and the other two show 1. The possible outcomes are: (2,1,1), (1,2,1), and (1,1,2). That's exactly 3 favorable outcomes. Therefore, the probability is

\frac{3}{216} = \frac{1}{72}

, which is choice B. Let's examine why the other answers are wrong: A)

\frac{1}{36}

assumes only 36 total outcomes, which would be correct for rolling two dice, not three. C)

\frac{1}{54}

might result from incorrectly calculating the total outcomes as

6 \times 9 = 54

or making an error in reducing the fraction. D)

\frac{1}{216}

suggests only one way to achieve a sum of 4, ignoring that the die showing 2 can be in any of the three positions. Strategy tip: For dice probability problems, systematically list all possible combinations that satisfy the condition, then count carefully. Don't forget that order matters—(2,1,1) is different from (1,2,1).

Question 4

A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If one marble is selected at random, what is the probability that the marble selected is not blue?

$\dfrac{1}{5}$
$\dfrac{3}{10}$
$\dfrac{7}{10}$ (correct answer)
$\dfrac{4}{5}$

Explanation: When you encounter probability questions asking for something "not" happening, you're dealing with complementary probability. The key insight is that the probability of an event not occurring equals 1 minus the probability of it occurring. First, let's find the total number of marbles: 5 red + 3 blue + 2 green = 10 marbles total. To find the probability that a marble is not blue, you can approach this two ways. The direct method: count all non-blue marbles. There are 5 red + 2 green = 7 non-blue marbles. So P(not blue) =

\dfrac{7}{10}

. Alternatively, use the complement rule: P(not blue) = 1 - P(blue) = 1 -

\dfrac{3}{10}

\dfrac{7}{10}

. Both methods give the same answer: C. Looking at the wrong answers: A)

\dfrac{1}{5}

might result from incorrectly using only one color (like just green marbles) out of the total. B)

\dfrac{3}{10}

is actually the probability of selecting blue—this is the classic trap of forgetting the "not" in the question. D)

\dfrac{4}{5}

could come from miscounting marbles or using an incorrect total. Strategy tip: For "not" probability questions, always double-check whether you're finding what the question actually asks for. Choice B being the probability of the opposite event is an extremely common wrong answer on standardized tests. When in doubt, use both the direct counting method and the complement rule (1 - probability of the event) to verify your answer.

Question 5

A fair coin is flipped 3 times. What is the probability of getting at least one head?

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

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 6

A box contains 4 math books and 6 history books. Two books are chosen at random without replacement. What is the probability that the second book chosen is a math book given that the first book chosen is a history book?

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

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 7

A jar contains 6 red marbles, 5 blue marbles, and 4 green marbles (15 total). One marble is chosen at random. Given that the marble is not green, what is the probability it is blue?

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

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 8

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?

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

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 9

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)?

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

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 10

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?

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

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 11

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?

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

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 12

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?

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

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 13

A jar contains 6 red marbles, 4 blue marbles, and 5 green marbles (15 total). One marble is selected at random. Given that the marble is not blue, what is the probability it is green? Express your answer as a fraction in simplest form.

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

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 14

A box contains 4 light bulbs, exactly 2 of which are defective. Two bulbs are selected at random without replacement. What is the probability that at least one of the bulbs selected is defective?

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

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 15

In a survey, $60\%$ of respondents own a smartphone and $30\%$ own a tablet. If $20\%$ own both a smartphone and a tablet, what is the probability that a randomly selected respondent owns a smartphone or a tablet?

$0.5$
$0.6$
$0.7$ (correct answer)
$0.8$

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 16

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?

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

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) = $\frac{5}{10} \times \frac{5}{9} = \frac{25}{90}$

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.

Question 17

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?

$\dfrac{3}{20}$
$\dfrac{1}{8}$
$\dfrac{9}{56}$
$\dfrac{3}{28}$ (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 18

An online game randomly generates a two–character code. The first character is one of the 5 vowels $\{A,E,I,O,U\}$ and the second character is a digit from 0 to 9. Repeats are allowed. What is the probability that the code generated is either "E7" or begins with "O"?

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

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 19

A class of 15 students includes 9 boys and 6 girls. The teacher randomly arranges all 15 students in a single-file line. What is the probability that the first student in line is a girl and the last student in line is a boy?

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

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 20

A bag contains 5 chocolate candies, 3 strawberry candies, and 2 lemon candies. One candy is selected at random. What is the probability the candy is not lemon?

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

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.