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Simple Probability & Complements - Pre Algebra

  1. Two dice are rolled simultaneously. Given that the sum is greater than 8, what is the probability that both dice show the same number?
    Answer: 210\frac{2}{10} 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.
  2. A fair coin is flipped 4 times. What is the probability of getting at most 1 head?
    Answer: 516\frac{5}{16} 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.
  3. 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?
    Answer: 12\frac{1}{2} 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.
  4. In a game, the probability of winning on any single attempt is 27\frac{2}{7}. If Alex plays the game three times, what is the probability that he wins exactly once?
    Answer: 150343\frac{150}{343} 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.
  5. 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?
    Answer: 511\frac{5}{11} 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.

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