Math 3 Quiz: Multi Step Probability Problems

13 questions · exam conditions

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Multi Step Probability ProblemsQuestion 1 of 13

A factory produces widgets with a 5% defect rate. If a quality inspector randomly selects 2 widgets independently, what is the probability that at least one widget is defective?

0.9025

0.10

0.05

0.0975

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Math 3 Quiz

Math 3 Quiz: Multi Step Probability Problems

Practice Multi Step Probability Problems 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 Multi Step Probability Problems, 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

A factory produces widgets with a 5% defect rate. If a quality inspector randomly selects 2 widgets independently, what is the probability that at least one widget is defective?

$0.9025$
$0.10$
$0.05$
$0.0975$ (correct answer)

Explanation: When you encounter probability questions involving "at least one" outcome, the most efficient approach is usually to use the complement rule: find the probability that none of the events occur, then subtract from 1. Here, we want the probability that at least one widget is defective. The complement is that both widgets are non-defective. Since widgets are selected independently and the defect rate is 5%, each widget has a 95% chance of being non-defective. The probability that both widgets are non-defective is:

0.95 \times 0.95 = 0.9025

Therefore, the probability that at least one is defective is:

1 - 0.9025 = 0.0975

Looking at the wrong answers: Choice A (

0.9025

) represents the probability that both widgets are non-defective - this is the complement of what we want, not the answer itself. Choice B (

0.10

) might tempt you if you mistakenly think "at least one defective" means doubling the single defect rate (

2 \times 0.05 = 0.10

), but this ignores the possibility of both being defective and doesn't account for proper probability calculations. Choice C (

0.05

) is simply the defect rate for a single widget, showing a failure to consider that we're selecting two widgets. Study tip: For "at least one" probability problems, always consider using the complement rule (1 minus the probability of none). It's usually much faster than calculating multiple scenarios directly, especially as the number of trials increases.

Question 2

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

$\frac{1}{5}$ (correct answer)
$\frac{2}{10}$
$\frac{1}{6}$
$\frac{3}{10}$

Explanation: Outcomes with sum > 8: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) = 10 outcomes. Among these, only (5,5) and (6,6) have matching dice = 2 outcomes. P(same|sum>8) = 2/10 = 1/5. Choice B is the unreduced fraction, C ignores the condition, D miscounts favorable outcomes.

Question 3

A medical test for a rare disease has a 95% accuracy rate for positive cases and a 98% accuracy rate for negative cases. If 0.2% of the population actually has the disease, what is the probability that a person who tests positive actually has the disease?

0.087 (correct answer)
0.950
0.792
0.196

Explanation: Using Bayes' theorem: P(Disease|Positive) = P(Positive|Disease) × P(Disease) / P(Positive). P(Positive) = 0.95 × 0.002 + 0.02 × 0.998 = 0.0019 + 0.01996 = 0.02186. Therefore P(Disease|Positive) = (0.95 × 0.002) / 0.02186 ≈ 0.087. Choice B ignores base rates, C uses incorrect conditional probability formula, D represents P(Positive|Disease).

Question 4

A game show has three doors. Behind one door is a car, behind the other two are goats. A contestant chooses a door. The host, who knows what's behind each door, opens a different door revealing a goat. The host then offers the contestant a chance to switch to the remaining unopened door.

If the contestant always switches, what is the probability of winning the car, given that the host opened a door with a goat?

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

Explanation: The host always opens a door with a goat (given), so this condition doesn't change the analysis. Initially, the contestant's door has probability 1/3 of having the car. The other two doors collectively have probability 2/3. When the host opens one of those doors (revealing a goat), the remaining unopened door gets the full 2/3 probability. Therefore, switching gives probability 2/3 of winning. Choice A assumes equal probability after host's action, C is the probability of the original choice, D incorrectly incorporates the given condition.

Question 5

A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn sequentially without replacement. Given that the first marble drawn is red, what is the probability that both marbles drawn are the same color?

$\frac{4}{7}$ (correct answer)
$\frac{5}{14}$
$\frac{25}{64}$
$\frac{3}{7}$

Explanation: Given that the first marble is red, there are 4 red marbles and 3 blue marbles remaining in the bag (7 total). For both marbles to be the same color, the second marble must also be red. The probability is

\frac{4}{7}

. Choice B incorrectly uses the original total of 8 marbles. Choice C incorrectly calculates the probability without the given condition. Choice D gives the probability that the second marble is blue.

Question 6

In a class of 30 students, 18 students play basketball and 20 students play soccer. If 12 students play both sports, what is the probability that a randomly selected student who plays basketball also plays soccer?

$\frac{3}{5}$
$\frac{2}{5}$
$\frac{2}{3}$ (correct answer)
$\frac{12}{30}$

Explanation: When you encounter problems about students playing multiple sports or belonging to overlapping groups, you're dealing with conditional probability. The key phrase here is "a randomly selected student who plays basketball also plays soccer" — this means you're finding the probability within the basketball-playing group only. This is a conditional probability problem: P(plays soccer | plays basketball). You need to find what fraction of basketball players also play soccer. Since 18 students play basketball and 12 of those also play soccer, the probability is

\frac{12}{18} = \frac{2}{3}

. You're essentially asking: "Of the 18 basketball players, how many also play soccer?" Now let's examine why the other answers are wrong. Choice A (

\frac{3}{5}

) would result from incorrectly using

\frac{18}{30}

, which gives the probability that any random student plays basketball, not the conditional probability we need. Choice B (

\frac{2}{5}

) comes from the error of using

\frac{12}{30}

and then simplifying, treating this as a simple probability rather than conditional. Choice D (

\frac{12}{30}

) represents the probability that any randomly selected student plays both sports, ignoring the condition that we're only considering basketball players. Remember this pattern: when you see "given that" or "who already" in probability questions, you're narrowing your sample space to just that subset. Always identify your denominator carefully — it's not always the total population, but rather the specific group mentioned in the condition.

Question 7

Events A and B are such that $P(A) = 0.6$ , $P(B) = 0.4$ , and $P(A \cap B) = 0.3$ . If event A occurs, what is the probability that event B also occurs?

$0.12$
$0.3$
$0.75$
$0.5$ (correct answer)

Explanation: When you see a question asking for the probability of one event given that another has occurred, you're dealing with conditional probability. The key phrase "If event A occurs, what is the probability that event B also occurs?" signals that you need

P(B|A)

. The conditional probability formula is

P(B|A) = \frac{P(A \cap B)}{P(A)}

. This makes intuitive sense: you're finding what fraction of event A's outcomes also include event B. Given the values

P(A) = 0.6

P(B) = 0.4

, and

P(A \cap B) = 0.3

, you can substitute directly:

P(B|A) = \frac{0.3}{0.6} = 0.5

. Looking at the wrong answers: Choice A (

0.12

) represents

P(A) \times P(B) = 0.6 \times 0.4 = 0.24

... wait, that's not even

0.12

. This might come from incorrectly multiplying

P(A \cap B) \times P(B) = 0.3 \times 0.4 = 0.12

, which has no meaningful interpretation. Choice B (

0.3

) is simply

P(A \cap B)

, showing confusion between joint probability and conditional probability. Choice C (

0.75

) could result from incorrectly calculating

\frac{P(B)}{P(A \cap B)} = \frac{0.4}{0.3} \approx 1.33

... actually, this doesn't yield

0.75

either, but it might come from

\frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.4} = 0.75

, which would be

P(A|B)

, not

P(B|A)

. Remember: conditional probability always uses the "given" event in the denominator. Watch for the direction of conditioning—

P(B|A)

and

P(A|B)

are generally different values.

Question 8

A medical test for a disease has a 95% accuracy rate for positive cases and a 90% accuracy rate for negative cases. If 2% of the population has the disease, what is the probability that a person who tests positive actually has the disease?

$\frac{95}{100}$
$\frac{19}{117}$ (correct answer)
$\frac{19}{98}$
$\frac{2}{100}$

Explanation: This is a classic conditional probability problem that requires Bayes' theorem. When you see medical testing scenarios, remember that you're looking for the probability of having the disease given a positive test result, not just the test's accuracy rate. Let's work through this systematically. Out of 10,000 people: 2% (200 people) have the disease, and 98% (9,800 people) don't. The test correctly identifies 95% of those with the disease:

0.95 \times 200 = 190

true positives. The test incorrectly identifies 10% of healthy people as positive:

0.10 \times 9,800 = 980

false positives. Total positive tests:

190 + 980 = 1,170

. Of these positive tests, only 190 are true positives. Therefore, the probability that someone who tests positive actually has the disease is

\frac{190}{1,170} = \frac{19}{117}

. Answer A (

\frac{95}{100}

) represents the test's sensitivity—the probability of testing positive if you have the disease. This is a common trap. Answer C (

\frac{19}{98}

) incorrectly uses only the diseased population as the denominator, ignoring false positives from the healthy population. Answer D (

\frac{2}{100}

) is simply the disease prevalence rate, not the conditional probability we need. Study tip: In Bayes' problems, always calculate the total number of positive results (true positives + false positives) for your denominator. The counterintuitive result here—that most positive tests are false positives when disease prevalence is low—frequently appears on standardized exams.

Question 9

A student takes two independent quizzes. The probability of passing the first quiz is $\frac{3}{4}$ , and the probability of passing the second quiz is $\frac{2}{3}$ . What is the probability that the student passes exactly one quiz?

$\frac{5}{12}$ (correct answer)
$\frac{7}{12}$
$\frac{1}{2}$
$\frac{17}{24}$

Explanation: The probability of passing exactly one quiz is P(pass first, fail second) + P(fail first, pass second) =

\frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{2}{3} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}

. Choice B incorrectly adds the individual passing probabilities. Choice C represents the average of the two passing probabilities. Choice D incorrectly calculates the probability of passing at least one quiz.

Question 10

A bag contains 8 red balls and 12 blue balls. Three balls are drawn without replacement. What is the probability that the second ball is red, given that exactly two of the three balls drawn are red?

$\frac{2}{3}$ (correct answer)
$\frac{8}{20}$
$\frac{1}{2}$
$\frac{7}{19}$

Explanation: Given exactly 2 reds among 3 balls, the possible patterns are RRB, RBR, BRR. By symmetry, each red position is equally likely among the patterns where exactly 2 are red. Since 2 out of 3 positions contain red balls, P(2nd is red | exactly 2 reds) = 2/3. Choice B is P(red) initially, C assumes equal likelihood without considering the constraint, D uses incorrect conditional calculation.

Question 11

In a deck of 52 cards, three cards are drawn sequentially without replacement. What is the probability that the first card is a heart, the second card is a spade, and the third card is also a spade?

$\frac{13 \cdot 13 \cdot 12}{52 \cdot 51 \cdot 50}$ (correct answer)
$\frac{13 \cdot 12 \cdot 11}{52 \cdot 51 \cdot 50}$
$\frac{13^3}{52^3}$
$\frac{13 \cdot 13 \cdot 11}{52 \cdot 51 \cdot 50}$

Explanation: The probability is

\frac{13}{52} \cdot \frac{13}{51} \cdot \frac{12}{50}

\frac{13 \cdot 13 \cdot 12}{52 \cdot 51 \cdot 50}

. There are 13 hearts initially, 13 spades after drawing a heart, and 12 spades remaining after drawing one spade. Choice B incorrectly assumes the first spade affects the second spade count. Choice C ignores the without replacement condition. Choice D incorrectly uses 11 instead of 12 for the final spade count.

Question 12

A coin is flipped three times. Given that at least two of the flips resulted in heads, what is the probability that all three flips resulted in heads?

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

Explanation: When you encounter a conditional probability problem like this, you need to identify what information you're given and what you're asked to find. This is a classic "given that" scenario where additional information changes the sample space. Let's list all possible outcomes when flipping a coin three times: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. That's 8 equally likely outcomes. The key phrase is "given that at least two flips resulted in heads." This condition eliminates some possibilities from our sample space. The outcomes with at least two heads are: HHH, HHT, HTH, and THH. So we now have 4 favorable outcomes instead of the original 8. Among these 4 remaining outcomes, only 1 shows all three heads (HHH). Therefore, the conditional probability is

\frac{1}{4}

, making C correct. Let's examine the wrong answers: A)

\frac{1}{2}

likely comes from incorrectly thinking that once you know two heads occurred, the third flip is independent with probability

\frac{1}{2}

. This ignores that we're looking at completed outcomes. B)

\frac{1}{3}

might result from incorrectly identifying only 3 outcomes with at least two heads, or confusing this with a different conditional probability setup. D)

\frac{3}{8}

probably comes from dividing the 3 outcomes with exactly two heads by the total original sample space of 8, which misses the point of conditional probability. Remember: conditional probability problems require you to recalculate based on the new, restricted sample space created by the given condition.

Question 13

A box contains 4 red balls, 3 green balls, and 3 blue balls. Two balls are drawn without replacement. What is the probability that the second ball is red, given that the first ball drawn was not red?

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

Explanation: This is a conditional probability problem involving drawing without replacement. When you see "given that" in a probability question, you're working with updated information that changes the sample space. Given that the first ball drawn was not red, it must have been either green or blue (6 balls total: 3 green + 3 blue). After removing one non-red ball, you're left with 9 balls total: 4 red balls and 5 non-red balls (either 2 green + 3 blue, or 3 green + 2 blue, depending on what was drawn first). The probability that the second ball is red equals the number of red balls remaining divided by the total balls remaining:

\frac{4}{9}

. Looking at the wrong answers: Choice A (

\frac{2}{5}

) incorrectly assumes only 5 balls remain total, perhaps by miscounting after the first draw. Choice C (

\frac{1}{3}

) might come from incorrectly thinking there are 3 possible outcomes (red, green, blue) and assuming they're equally likely, which ignores the different quantities of each color. Choice D (

\frac{3}{9}

) uses 9 as the correct denominator but 3 as the numerator, possibly confusing the number of each non-red color with the number of red balls. Remember that "without replacement" problems require you to update both the favorable outcomes and total possible outcomes after each draw. The key insight is recognizing that conditional probability creates a new, smaller sample space based on the given information.