Math 3 Quiz: Computing Probabilities

20 questions · exam conditions

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Computing ProbabilitiesQuestion 1 of 20

A pharmaceutical company tests a new drug on patients. The drug is effective for 85% of patients who have the target condition and ineffective for 95% of patients who do not have the target condition. If 2% of the population has the target condition, what is the probability that a patient who tests positive for drug effectiveness actually has the target condition?

0.85

0.258

0.147

0.02

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Math 3 Quiz

Math 3 Quiz: Computing Probabilities

Practice Computing Probabilities in Math 3 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 Computing Probabilities, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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

0.85
0.258 (correct answer)
0.147
0.02

Explanation: Using Bayes' theorem: P(condition | positive test) = P(positive | condition) × P(condition) / P(positive test). P(positive test) = 0.85×0.02 + 0.05×0.98 = 0.017 + 0.049 = 0.066. Therefore, P(condition | positive test) = (0.85×0.02) / 0.066 = 0.017 / 0.066 ≈ 0.258. Choice A incorrectly uses the sensitivity rate. Choice C miscalculates the denominator. Choice D incorrectly uses the base rate of the condition.

Question 2

A bag contains 8 red marbles, 6 blue marbles, and 4 green marbles. Two marbles are drawn without replacement. What is the probability that the first marble is red and the second marble is blue?

$\frac{8}{51}$
$\frac{48}{306}$ (correct answer)
$\frac{14}{18}$
$\frac{48}{324}$

Explanation: For the first marble to be red: P(red) = 8/18. After removing one red marble, there are 17 marbles left, with 6 still blue. P(blue|red first) = 6/17. The probability of both events is (8/18) × (6/17) = 48/306. Choice A uses the wrong denominator. Choice C treats this as drawing with replacement. Choice D multiplies by 18² instead of accounting for the changing total.

Question 3

A fair six-sided die is rolled three times. What is the probability of getting exactly two even numbers?

$\frac{9}{216}$
$\frac{27}{216}$
$\frac{54}{216}$ (correct answer)
$\frac{108}{216}$

Explanation: This is a binomial probability with n=3, k=2, and p=1/2 (probability of even). The number of ways to choose 2 positions out of 3 is C(3,2) = 3. Each arrangement has probability (1/2)²(1/2)¹ = 1/8. Total probability = 3 × (1/8) = 3/8 = 54/216. Choice A uses only one arrangement. Choice B miscalculates the combinations. Choice D doubles the correct answer.

Question 4

A committee of 4 people is selected from a group of 6 men and 5 women. What is the probability that the committee contains exactly 2 men and 2 women?

$\frac{150}{330}$
$\frac{60}{165}$
$\frac{150}{495}$ (correct answer)
$\frac{30}{165}$

Explanation: Total ways to select 4 from 11 people: C(11,4) = 495. Ways to select 2 men from 6: C(6,2) = 15. Ways to select 2 women from 5: C(5,2) = 10. Favorable outcomes: 15 × 10 = 150. Probability = 150/495. Choice A uses wrong total. Choice B simplifies incorrectly. Choice D uses wrong numerator calculation.

Question 5

In a game, a player wins if they roll a sum of 7 or 11 with two fair dice, and loses if they roll a sum of 2, 3, or 12. For any other sum, the game continues. What is the probability that the game ends on the first roll?

$\frac{8}{36}$
$\frac{10}{36}$
$\frac{12}{36}$ (correct answer)
$\frac{14}{36}$

Explanation: Winning combinations (sum = 7 or 11): 6 ways for sum of 7, 2 ways for sum of 11 = 8 ways. Losing combinations (sum = 2, 3, or 12): 1 way for sum of 2, 2 ways for sum of 3, 1 way for sum of 12 = 4 ways. Game ends on first roll: 8 + 4 = 12 ways. Probability = 12/36. Choice A counts only winning outcomes. Choice B miscounts some combinations. Choice D includes outcomes that continue the game.

Question 6

A fair coin is flipped 5 times. What is the probability of getting at least 3 heads?

$\frac{10}{32}$
$\frac{16}{32}$ (correct answer)
$\frac{20}{32}$
$\frac{26}{32}$

Explanation: At least 3 heads means exactly 3, 4, or 5 heads. P(exactly 3) = C(5,3)/2^5 = 10/32. P(exactly 4) = C(5,4)/2^5 = 5/32. P(exactly 5) = C(5,5)/2^5 = 1/32. Total: (10 + 5 + 1)/32 = 16/32. Choice A gives only P(exactly 3). Choice C miscalculates the combinations. Choice D includes outcomes with 2 heads.

Question 7

A restaurant serves 3 types of appetizers, 4 types of main courses, and 2 types of desserts. A customer orders one item from each category. If the customer randomly selects items and appetizer A is chosen, what is the probability that main course M₂ is also chosen?

$\frac{1}{24}$
$\frac{1}{12}$
$\frac{1}{6}$
$\frac{1}{4}$ (correct answer)

Explanation: Since the selection of appetizer A is already given, and the choices are independent, the probability of selecting main course M₂ is simply 1/4 (one choice out of 4 main courses). The appetizer choice doesn't affect the main course probability. Choice A uses the total number of meal combinations. Choice B incorrectly combines probabilities. Choice C uses the wrong denominator.

Question 8

A medical test correctly identifies a disease 95% of the time when the disease is present and correctly identifies no disease 90% of the time when the disease is absent. If 2% of the population has the disease, what is the probability that a person who tests positive actually has the disease?

0.162 (correct answer)
0.190
0.238
0.950

Explanation: Using Bayes' theorem: P(disease|positive) = P(positive|disease)×P(disease)/P(positive). P(positive) = 0.95×0.02 + 0.10×0.98 = 0.019 + 0.098 = 0.117. Therefore P(disease|positive) = (0.95×0.02)/0.117 = 0.019/0.117 ≈ 0.162. Choice B uses incorrect denominators. Choice C forgets to account for false positives. Choice D confuses this with the test sensitivity.

Question 9

Events X and Y are independent with P(X) = 0.4 and P(Y) = 0.3. What is P(X' ∩ Y'), where X' and Y' represent the complements of X and Y?

0.42 (correct answer)
0.58
0.70
0.88

Explanation: Since X and Y are independent, X' and Y' are also independent. P(X') = 1 - 0.4 = 0.6 and P(Y') = 1 - 0.3 = 0.7. Therefore, P(X' ∩ Y') = P(X') × P(Y') = 0.6 × 0.7 = 0.42. Choice B gives P(X ∪ Y)'. Choice C gives only P(Y'). Choice D gives P(X') + P(Y') - P(X')P(Y'), which is incorrect for intersection.

Question 10

A company's quality control department tests products in batches. Historical data shows that 15% of batches contain defective items. When a batch contains defective items, the test correctly identifies it 90% of the time. When a batch contains no defective items, the test incorrectly flags it as defective 5% of the time.

If the test flags a batch as defective, what is the probability that the batch actually contains defective items?

0.727
0.760 (correct answer)
0.850
0.900

Explanation: Using Bayes' theorem: P(defective|flagged) = P(flagged|defective)×P(defective)/P(flagged). P(flagged) = 0.90×0.15 + 0.05×0.85 = 0.135 + 0.0425 = 0.1775. Therefore P(defective|flagged) = (0.90×0.15)/0.1775 = 0.135/0.1775 ≈ 0.760. Choice A uses incorrect calculation of P(flagged). Choice C confuses this with the base rate. Choice D confuses this with test sensitivity.

Question 11

In a certain high school, 65% of students play sports, 40% participate in clubs, and 25% do both. If a student is randomly selected, what is the probability that the student plays sports given that they participate in clubs?

$\frac{1}{4}$
$\frac{5}{8}$ (correct answer)
$\frac{25}{40}$
$\frac{25}{65}$

Explanation: This is a conditional probability problem. P(sports|clubs) = P(sports and clubs)/P(clubs) = 0.25/0.40 = 5/8. Choice A incorrectly uses P(both)/P(total). Choice C appears correct but doesn't simplify the fraction properly. Choice D uses P(both)/P(sports) instead of the correct conditional probability formula.

Question 12

A survey of 200 college students found that 120 students drink coffee, 80 students drink tea, and 30 students drink both coffee and tea.

Based on the survey data, if a student is selected at random, what is the probability that the student drinks coffee or tea but not both?

$\frac{170}{200}$
$\frac{140}{200}$ (correct answer)
$\frac{110}{200}$
$\frac{200}{200}$

Explanation: Students who drink only coffee: 120 - 30 = 90. Students who drink only tea: 80 - 30 = 50. Students who drink coffee or tea but not both: 90 + 50 = 140. Probability = 140/200. Choice A includes those who drink both. Choice C miscalculates the exclusive categories. Choice D assumes all students drink at least one beverage.

Question 13

A quality control inspector tests electronic components from two production lines. Line A has a 12% defect rate, and Line B has an 8% defect rate. If 60% of all components come from Line A and 40% come from Line B, what is the probability that a randomly selected defective component came from Line A?

0.60
0.69 (correct answer)
0.72
0.12

Explanation: This requires Bayes' theorem. First, find P(defective) = P(defective|A)×P(A) + P(defective|B)×P(B) = 0.12×0.60 + 0.08×0.40 = 0.072 + 0.032 = 0.104. Then P(A|defective) = P(defective|A)×P(A) / P(defective) = (0.12×0.60) / 0.104 = 0.072 / 0.104 ≈ 0.69. Choice A incorrectly uses the overall proportion from Line A. Choice C uses an incorrect calculation mixing the defect rates. Choice D incorrectly uses just the defect rate of Line A.

Question 14

In a card game, players draw cards without replacement from a standard 52-card deck. If the first player draws 2 cards and both are hearts, what is the probability that the next card drawn by the second player is also a heart?

$\frac{11}{50}$ (correct answer)
$\frac{13}{52}$
$\frac{11}{52}$
$\frac{12}{50}$

Explanation: After the first player draws 2 hearts, there are 11 hearts remaining out of 50 total cards left in the deck. The probability is 11/50. Choice B incorrectly uses the original probability before any cards were drawn. Choice C uses the correct numerator but wrong denominator, forgetting that 2 cards were already drawn. Choice D incorrectly assumes only 1 heart was drawn instead of 2.

Question 15

Two fair six-sided dice are rolled. Given that the sum is greater than 8, what is the probability that both dice show the same number?

$\frac{2}{36}$
$\frac{1}{6}$
$\frac{6}{36}$
$\frac{2}{10}$ (correct answer)

Explanation: This is a conditional probability problem where you need to find the probability of an event (both dice same) given that another event (sum > 8) has already occurred. When you see "given that" in a probability question, use the formula: P(A|B) = P(A and B) / P(B). First, identify all outcomes where the sum is greater than 8. These are: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6). That's 10 total outcomes where the sum exceeds 8. Next, find outcomes where both conditions are met: sum > 8 AND both dice show the same number. Looking at our list, only (5,5) and (6,6) satisfy both conditions - that's 2 outcomes. Therefore, the conditional probability is

\frac{2}{10} = \frac{1}{5}

. Answer A (

\frac{2}{36}

) incorrectly uses the total sample space of 36 instead of the reduced sample space of 10 outcomes where sum > 8. Answer B (

\frac{1}{6}

) likely comes from incorrectly thinking there are 6 ways to get matching dice and dividing by 36, then simplifying wrong. Answer C (

\frac{6}{36}

) represents the total probability of getting matching dice without the given condition, ignoring the "sum > 8" constraint entirely. Remember: conditional probability problems require you to work within the restricted sample space defined by the given condition. Always identify this smaller set of favorable outcomes first, then find your specific event within that set.

Question 16

A spinner has three sections: red (probability 0.5), blue (probability 0.3), and yellow (probability 0.2). The spinner is spun 4 times independently. What is the probability that red appears exactly twice, blue appears exactly once, and yellow appears exactly once?

$\frac{4!}{2! \cdot 1! \cdot 1!} \times (0.5)^2 \times (0.3)^1 \times (0.2)^1$ (correct answer)
$(0.5)^2 \times (0.3)^1 \times (0.2)^1$
$4 \times (0.5)^2 \times (0.3)^1 \times (0.2)^1$
$\binom{4}{2} \times (0.5)^2 \times (0.3)^1 \times (0.2)^1$

Explanation: This is a multinomial probability. The number of ways to arrange 2 reds, 1 blue, and 1 yellow in 4 spins is 4!/(2!×1!×1!) = 12. Each specific arrangement has probability (0.5)²×(0.3)×(0.2). Total probability = 12×(0.5)²×(0.3)×(0.2) = 0.018. Choice B omits the multinomial coefficient. Choice C uses 4 instead of the correct multinomial coefficient 12. Choice D uses the binomial coefficient for only one color, ignoring the constraint on the other colors.

Question 17

At a school carnival, the probability of winning the ring toss is 0.3, and the probability of winning the basketball shot is 0.4. These events are independent. Maria plays both games exactly once. What is the probability that she wins exactly one of the two games?

0.42
0.58 (correct answer)
0.70
0.12

Explanation: To win exactly one game, Maria either wins ring toss and loses basketball, or loses ring toss and wins basketball. P(exactly one win) = P(ring toss win AND basketball loss) + P(ring toss loss AND basketball win) = (0.3×0.6) + (0.7×0.4) = 0.18 + 0.28 = 0.58. Choice A incorrectly calculates P(win ring toss OR win basketball) = 0.3 + 0.4 - 0.12. Choice C adds the individual probabilities without considering independence. Choice D gives the probability of winning both games.

Question 18

In a survey of 200 students, 120 play soccer, 80 play tennis, and 30 play both sports. If a student is selected at random and is known to play tennis, what is the probability that this student also plays soccer?

$\frac{30}{200}$
$\frac{30}{120}$
$\frac{30}{80}$ (correct answer)
$\frac{50}{80}$

Explanation: This is a conditional probability problem. P(plays soccer | plays tennis) = P(plays both) / P(plays tennis) = 30/80. We're given that the student plays tennis, so our sample space is reduced to the 80 tennis players, of whom 30 also play soccer. Choice A gives the unconditional probability of playing both sports. Choice B incorrectly uses soccer players as the denominator. Choice D incorrectly calculates tennis-only players in the numerator.

Question 19

In a lottery game, a player wins if they match at least 3 out of 5 numbers drawn from the set {1, 2, 3, ..., 20} without replacement. If a player's ticket has the numbers {2, 7, 13, 16, 19}, what is the probability that they win (match at least 3 numbers)?

$1 - \frac{\binom{15}{5} + \binom{15}{4} \times \binom{5}{1} + \binom{15}{3} \times \binom{5}{2}}{\binom{20}{5}}$
$\frac{\binom{5}{3} + \binom{5}{4} + \binom{5}{5}}{\binom{20}{5}}$
$\frac{\binom{5}{3}}{\binom{20}{5}}$
$\frac{\binom{5}{3} \times \binom{15}{2} + \binom{5}{4} \times \binom{15}{1} + \binom{5}{5}}{\binom{20}{5}}$ (correct answer)

Explanation: When you encounter a lottery probability problem involving "at least" scenarios, think about using hypergeometric distribution. You're drawing without replacement from a finite population, where some items are "successes" (your numbers) and others are "failures" (remaining numbers). The correct approach requires calculating the probability of exactly 3, 4, or 5 matches separately, then adding them. For each case, you multiply the ways to choose matching numbers from your 5 ticket numbers by the ways to choose non-matching numbers from the remaining 15 numbers, all divided by the total ways to choose 5 numbers from 20. For exactly 3 matches:

\binom{5}{3} \times \binom{15}{2}

(choose 3 from your 5 numbers, 2 from the other 15) For exactly 4 matches:

\binom{5}{4} \times \binom{15}{1}

(choose 4 from your 5 numbers, 1 from the other 15) For exactly 5 matches:

\binom{5}{5} \times \binom{15}{0} = \binom{5}{5}

(all your numbers match) Answer D correctly sums these three scenarios in the numerator, with

\binom{20}{5}

as the denominator representing total possible draws. Answer A uses the complement approach (1 minus probability of 0, 1, or 2 matches) but has calculation errors. Answer B ignores the fact that you must also select from the non-matching numbers—it only counts ways to choose from your ticket. Answer C only calculates the probability of exactly 3 matches, missing 4 and 5 matches entirely. Remember: for hypergeometric problems, always account for both the "success" group and "failure" group when calculating each scenario, then sum all favorable outcomes.

Question 20

A box contains 5 red balls and 3 blue balls. Three balls are drawn without replacement. What is the probability that exactly 2 are red?

$\frac{C(5,2) \times C(3,1)}{C(8,3)}$ (correct answer)
$\frac{5^2 \times 3}{8^3}$
$\frac{C(5,2) \times C(8,1)}{C(8,3)}$
$\frac{C(3,2) \times C(5,1)}{C(8,3)}$

Explanation: This is a hypergeometric distribution. We need exactly 2 red balls from 5 red balls: C(5,2). We need exactly 1 blue ball from 3 blue balls: C(3,1). Total ways to choose 3 balls from 8: C(8,3). Probability = [C(5,2) × C(3,1)]/C(8,3). Choice B treats draws as independent. Choice C incorrectly selects from all 8 balls for the blue requirement. Choice D reverses the red and blue selections.