Math 1 Quiz: Simple Event Probabilities

11 questions · exam conditions

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Simple Event ProbabilitiesQuestion 1 of 11

A standard deck of 52 cards is shuffled. Three cards are dealt face up in sequence without replacement. What is the probability that the third card is a heart, given that exactly one of the first two cards is a heart?

\frac{25}{102}

\frac{13}{50}

\frac{11}{50}

\frac{12}{50}

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Math 1 Quiz

Math 1 Quiz: Simple Event Probabilities

Practice Simple Event Probabilities in Math 1 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 Simple Event Probabilities, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

$\frac{25}{102}$
$\frac{13}{50}$
$\frac{11}{50}$
$\frac{12}{50}$ (correct answer)

Explanation: This is a conditional probability problem where you need to find the probability of an event given specific information about previous events. When you see "given that" in a probability question, you're working within a restricted sample space. Let's break down what "exactly one of the first two cards is a heart" means. This can happen in two ways: the first card is a heart and the second isn't, or the first card isn't a heart and the second is. For the first scenario (heart, then non-heart): After drawing a heart first, there are 12 hearts left among 51 remaining cards. The probability the second card isn't a heart is

\frac{39}{51}

. Now there are 50 cards left, with 12 still being hearts. So the probability the third card is a heart is

\frac{12}{50}

. For the second scenario (non-heart, then heart): After drawing a non-heart first, there are still 13 hearts among 51 remaining cards. The probability the second card is a heart is

\frac{13}{51}

. Now there are 50 cards left, with 12 being hearts. Again, the probability the third card is a heart is

\frac{12}{50}

. In both valid scenarios, the probability that the third card is a heart is

\frac{12}{50}

, making D correct. Choice A (

\frac{25}{102}

) incorrectly uses 102 as a denominator, suggesting confusion about the remaining deck size. Choice B (

\frac{13}{50}

) assumes all 13 hearts are still available. Choice C (

\frac{11}{50}

) incorrectly assumes two hearts have been removed. Remember: in conditional probability problems, carefully track what information restricts your sample space and how previous events affect the remaining possibilities.

Question 2

In a simulation of 1000 trials, event A occurred 340 times, event B occurred 280 times, and both events A and B occurred together 95 times. Based on this simulation, what is the probability that event A occurs given that event B has already occurred?

$\frac{95}{1000}$
$\frac{95}{340}$
$\frac{95}{280}$ (correct answer)
$\frac{245}{1000}$

Explanation: This is conditional probability P(A|B) = P(A and B)/P(B). From the simulation: P(A and B) = 95/1000 and P(B) = 280/1000. Therefore P(A|B) = (95/1000)/(280/1000) = 95/280. Choice A gives P(A and B) instead of conditional probability. Choice B incorrectly uses P(A) in denominator instead of P(B). Choice D appears to subtract 95 from 340 but uses wrong total.

Question 3

A simulation is run where a fair coin is flipped until either 3 heads in a row occur or 3 tails in a row occur, whichever happens first. Based on the theoretical analysis of this process, what is the probability that the sequence ends with 3 heads in a row?

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

Explanation: By symmetry, since the coin is fair and we're looking for either 3 consecutive heads OR 3 consecutive tails (whichever comes first), the probability of ending with 3 heads must equal the probability of ending with 3 tails. Since these are the only two possible outcomes and they're mutually exclusive and exhaustive, we have P(3 heads) + P(3 tails) = 1. By symmetry, P(3 heads) = P(3 tails), so 2×P(3 heads) = 1, which gives P(3 heads) = 1/2. Choice A (1/4) might be chosen by students who incorrectly think about the probability of getting exactly HHH in the first 3 flips. Choice B (3/8) might arise from flawed calculations involving the probability of various short sequences. Choice D (5/8) might be chosen by students who make computational errors in more complex approaches to this problem.

Question 4

A quality control inspector at a factory tests electronic components. Historical data shows that 2% of components are defective. The inspector uses a testing device that correctly identifies defective components 95% of the time and correctly identifies non-defective components 98% of the time.

If the testing device indicates that a randomly selected component is defective, what is the probability that the component is actually defective?

$\frac{19}{117}$
$\frac{95}{117}$
$\frac{19}{98}$
$\frac{95}{193}$ (correct answer)

Explanation: This is a Bayes' theorem problem. Let D = component is defective, T = test indicates defective. We want P(D|T). Given: P(D) = 0.02, P(T|D) = 0.95, P(T|D') = 0.02 (since P(correct identification of non-defective) = 0.98). Using Bayes' theorem: P(D|T) = P(T|D)×P(D) / [P(T|D)×P(D) + P(T|D')×P(D')]. P(T|D)×P(D) = 0.95 × 0.02 = 0.019; P(T|D')×P(D') = 0.02 × 0.98 = 0.0196. So P(D|T) = 0.019 / (0.019 + 0.0196) = 0.019 / 0.0386 = 19/38.6 = 190/386 = 95/193. Choice A might result from calculation errors in the Bayes formula. Choice B incorrectly uses 95/117, possibly confusing the sensitivity of the test with the posterior probability. Choice C might arise from using incorrect denominators in the Bayes calculation.

Question 5

Consider the sample space of all possible outcomes when rolling two distinguishable six-sided dice. Event A is 'the sum of the dice is even' and Event B is 'at least one die shows a 1'. What is P(A|B)?

$\frac{5}{11}$ (correct answer)
$\frac{6}{11}$
$\frac{1}{2}$
$\frac{7}{11}$

Explanation: First, find the outcomes in event B (at least one die shows 1). These are: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1). That's 11 outcomes. Now find outcomes in both A and B (sum is even AND at least one die shows 1): (1,1) sum=2 ✓, (1,3) sum=4 ✓, (1,5) sum=6 ✓, (3,1) sum=4 ✓, (5,1) sum=6 ✓. That's 5 outcomes. Therefore P(A|B) = P(A∩B)/P(B) = 5/11. Choice B (6/11) might result from miscounting the favorable outcomes. Choice C (1/2) might be chosen by students who incorrectly think the conditional probability equals the unconditional probability P(A). Choice D (7/11) could arise from including outcomes like (1,2), (1,4), (1,6) which have odd sums.

Question 6

A standard deck of 52 cards is shuffled, and cards are drawn one by one without replacement until a face card (Jack, Queen, or King) is drawn. What is the probability that exactly 4 cards are drawn in this process?

$\frac{40 \times 39 \times 38 \times 12}{52 \times 51 \times 50 \times 49}$ (correct answer)
$\frac{40 \times 39 \times 38 \times 37 \times 12}{52 \times 51 \times 50 \times 49 \times 48}$
$\frac{40^3 \times 12}{52 \times 51 \times 50 \times 49}$
$\frac{\binom{40}{3} \times 12}{\binom{52}{4}}$

Explanation: For exactly 4 cards to be drawn, the first 3 cards must be non-face cards, and the 4th card must be a face card. There are 40 non-face cards and 12 face cards in a standard deck. P(1st card is non-face) = 40/52; P(2nd card is non-face | 1st was non-face) = 39/51; P(3rd card is non-face | first 2 were non-face) = 38/50; P(4th card is face | first 3 were non-face) = 12/49. The probability is the product: (40/52) × (39/51) × (38/50) × (12/49) = (40×39×38×12)/(52×51×50×49). Choice B is incorrect because it includes a 5th draw with 37/48, but we stop after 4 cards. Choice C is incorrect because it uses 40³ instead of the sequential probability 40×39×38. Choice D uses combinations which would be appropriate for a different type of problem where order doesn't matter, but here the sequence of draws matters.

Question 7

A fair six-sided die is rolled three times. What is the probability of getting at least one 6, given that the sum of the three rolls is exactly 15?

$\frac{9}{10}$ (correct answer)
$\frac{7}{10}$
$\frac{8}{10}$
$\frac{6}{10}$

Explanation: First find all ways to get sum = 15 with three dice: (3,6,6), (4,5,6), (5,5,5), (6,3,6), (6,4,5), (6,5,4), (6,6,3), (5,4,6), (5,6,4), (4,6,5). That's 10 total outcomes. Of these, only (5,5,5) has no 6, so 9 outcomes have at least one 6. P(at least one 6 | sum = 15) = 9/10. Choice B miscounts outcomes with at least one 6. Choice C miscounts total favorable outcomes. Choice D significantly undercounts.

Question 8

A bag contains 5 red, 3 blue, and 7 green marbles. Two marbles are drawn without replacement. If the first marble drawn is red, what is the probability that the second marble is also red?

$\frac{5}{14}$
$\frac{5}{15}$
$\frac{4}{15}$
$\frac{4}{14}$ (correct answer)

Explanation: When you encounter probability problems involving "without replacement," the key insight is that each draw changes the total number of items available for subsequent draws. Since the first marble drawn is red, you now have a reduced bag containing 4 red marbles (one was already drawn), 3 blue marbles, and 7 green marbles. This gives you a total of 14 marbles remaining in the bag. The probability that the second marble is red equals the number of red marbles left divided by the total marbles left:

\frac{4}{14}

. This matches answer choice D. Let's examine why the other options are incorrect. Choice A (

\frac{5}{14}

) uses the original number of red marbles (5) instead of accounting for the fact that one red marble was already removed. Choice B (

\frac{5}{15}

) makes the same mistake with the numerator and also uses the original total (15) instead of the reduced total. Choice C (

\frac{4}{15}

) correctly reduces the red marbles from 5 to 4 but incorrectly uses the original total of 15 marbles instead of recognizing that one marble has been removed. The most common trap in "without replacement" problems is forgetting to update both the favorable outcomes and the total possible outcomes after each draw. Always ask yourself: "What's left in the bag?" Remember that conditional probability problems require you to work with the new, reduced sample space created by the given condition.

Question 9

A lottery system uses balls numbered 1 through 30. Five balls are drawn without replacement. What is the probability that the smallest number drawn is 6?

$\frac{C(25,4)}{C(30,5)}$
$\frac{C(24,4)}{C(30,5)}$ (correct answer)
$\frac{C(24,4)}{C(29,4)}$
$\frac{C(25,5)}{C(30,5)}$

Explanation: For the smallest number to be 6, we must: (1) include ball 6, and (2) choose the remaining 4 balls from numbers 7-30 (which is 24 balls), and (3) not choose any balls from 1-5. This gives us C(24,4) favorable outcomes out of C(30,5) total outcomes. Choice A incorrectly allows choosing from numbers 6-30 for the remaining 4 balls, which would make 6 not necessarily the smallest. Choice C uses wrong denominator. Choice D chooses 5 balls from 7-31, which doesn't include the required ball 6.

Question 10

A spinner has 8 equal sections numbered 1 through 8. Sarah spins twice and records the sum of the two numbers. What is the probability that the sum is greater than 10 but not equal to 16?

$\frac{15}{64}$
$\frac{21}{64}$
$\frac{20}{64}$ (correct answer)
$\frac{16}{64}$

Explanation: Total outcomes: 8² = 64. Sums greater than 10: (3,8), (4,7), (4,8), (5,6), (5,7), (5,8), (6,5), (6,6), (6,7), (6,8), (7,4), (7,5), (7,6), (7,7), (7,8), (8,3), (8,4), (8,5), (8,6), (8,7), (8,8) = 21 outcomes. But we exclude sum = 16, which only occurs with (8,8) = 1 outcome. So 21 - 1 = 20 favorable outcomes. Probability = 20/64. Choice A miscounts favorable outcomes. Choice B includes the sum of 16. Choice D undercounts significantly.

Question 11

Two fair coins are flipped simultaneously 4 times. What is the probability of getting exactly 3 outcomes where both coins show the same face (either both heads or both tails)?

$\frac{1}{4}$ (correct answer)
$\frac{3}{16}$
$\frac{1}{8}$
$\frac{5}{16}$

Explanation: Each simultaneous flip has 4 equally likely outcomes: HH, HT, TH, TT. The probability of both coins showing the same face is P(HH or TT) = 2/4 = 1/2. We want exactly 3 successes in 4 trials with success probability 1/2. Using binomial probability: C(4,3)×(1/2)³×(1/2)¹ = 4×(1/2)⁴ = 4/16 = 1/4. Choice B uses wrong binomial coefficient. Choice C miscalculates the success probability. Choice D incorrectly includes the case of 4 successes.