Middle School Math Quiz: Simple Probability And Complements

5 questions · exam conditions

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Simple Probability And ComplementsQuestion 1 of 5

In a certain school, 60% of students play sports, 40% are in the honor society, and 25% do both. If a student is selected at random, what is the probability that the student plays sports or is in the honor society, but not both?

\frac{1}{2}

\frac{3}{4}

\frac{1}{4}

\frac{7}{20}

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Middle School Math Quiz

Middle School Math Quiz: Simple Probability And Complements

Practice Simple Probability And Complements in Middle School 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 Simple Probability And Complements, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School 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

$\frac{1}{2}$ (correct answer)
$\frac{3}{4}$
$\frac{1}{4}$
$\frac{7}{20}$

Explanation: We want students who satisfy exactly one condition. Let S = plays sports, H = honor society. P(S) = 0.6, P(H) = 0.4, P(S ∩ H) = 0.25. We want P(S ∩ H') + P(S' ∩ H) = P(S only) + P(H only). P(S only) = P(S) - P(S ∩ H) = 0.6 - 0.25 = 0.35. P(H only) = P(H) - P(S ∩ H) = 0.4 - 0.25 = 0.15. Total: 0.35 + 0.15 = 0.50 = 1/2. Choice B (3/4) represents P(S ∪ H). Choice C (1/4) represents P(S ∩ H). Choice D (7/20) represents an incorrect calculation.

Question 2

A bag contains 12 marbles: 5 red, 4 blue, and 3 green. If two marbles are drawn without replacement, what is the probability that the second marble is red, given that the first marble drawn was NOT red?

$\frac{5}{11}$ (correct answer)
$\frac{5}{12}$
$\frac{4}{11}$
$\frac{7}{11}$

Explanation: If the first marble is NOT red, then it must be blue or green (7 marbles total). After removing one non-red marble, there are 11 marbles left, and all 5 red marbles remain. Therefore, P(second is red | first is not red) = 5/11. Choice B (5/12) incorrectly uses the original total. Choice C (4/11) confuses this with the probability involving blue marbles. Choice D (7/11) incorrectly represents the probability that the second marble is NOT red.

Question 3

In a game, the probability of winning on any single attempt is $\frac{2}{7}$ . If Alex plays the game three times, what is the probability that he wins exactly once?

$\frac{6}{343}$
$\frac{75}{343}$
$\frac{25}{343}$
$\frac{150}{343}$ (correct answer)

Explanation: This requires binomial probability. P(win) = 2/7, so P(lose) = 5/7. For exactly one win in three games: P = C(3,1) × (2/7)¹ × (5/7)² = 3 × (2/7) × (25/49) = 3 × 50/343 = 150/343. Choice A (6/343) omits the combination factor and uses wrong powers. Choice B (75/343) uses incorrect calculation of (5/7)². Choice C (25/343) represents the probability of exactly one specific sequence without accounting for the three different positions the win could occur.

Question 4

A fair coin is flipped 4 times. What is the probability of getting at most 1 head?

$\frac{1}{16}$
$\frac{4}{16}$
$\frac{5}{16}$ (correct answer)
$\frac{6}{16}$

Explanation: At most 1 head means 0 heads or exactly 1 head. P(0 heads) = C(4,0) × (1/2)⁴ = 1 × 1/16 = 1/16. P(exactly 1 head) = C(4,1) × (1/2)¹ × (1/2)³ = 4 × 1/16 = 4/16. Total: 1/16 + 4/16 = 5/16. Choice A (1/16) represents only 0 heads. Choice B (4/16) represents only exactly 1 head. Choice D (6/16) incorrectly adds an extra outcome.

Question 5

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

$\frac{2}{36}$
$\frac{2}{10}$ (correct answer)
$\frac{3}{10}$
$\frac{6}{36}$

Explanation: First, find all outcomes where 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, the outcomes where both dice show the same number are: (5,5) and (6,6) = 2 outcomes. Therefore, P(same number | sum > 8) = 2/10 = 1/5. Choice A (2/36) uses the total sample space instead of the conditional space. Choice C (3/10) incorrectly counts an extra double. Choice D (6/36) represents all possible doubles in the entire sample space.