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Probability and Conditional Probability Flashcards

6 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Probability and Conditional Probability flashcards as text

  1. Events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8. Which statement is TRUE?

    Answer: A and B are independent but not mutually exclusive

    Use inclusion-exclusion: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.5 − 0.8 = 0.3. Since P(A ∩ B) = 0.3 ≠ 0, A and B are NOT mutually exclusive. Check independence: P(A)·P(B) = 0.6 × 0.5 = 0.30 = P(A ∩ B). The condition holds, so A and B ARE independent. Answer: independent but not mutually exclusive.

  2. A factory has two machines. Machine X produces 60% of all items with a 4% defect rate; Machine Y produces 40% with a 2% defect rate. A randomly selected item is defective. What is the probability it came from Machine Y?

    Answer: 0.25

    By Bayes' Theorem: P(defective) = (0.04)(0.60) + (0.02)(0.40) = 0.024 + 0.008 = 0.032. P(Y | defective) = (0.02 × 0.40) / 0.032 = 0.008 / 0.032 = 0.25. Despite Machine Y's lower defect rate, it accounts for only 25% of defective items because it produces fewer items overall.

  3. A standard 52-card deck is randomly split into two equal piles of 26. What is the probability the ace of spades and the ace of hearts end up in the same pile?

    Answer: 25/51

    Fix the ace of spades in whichever pile it lands. Of the 51 remaining cards, exactly 25 slots remain in that same pile. The probability the ace of hearts occupies one of those 25 slots is 25/51 ≈ 0.490 — slightly less than 1/2. The common wrong answer is 1/2, but because one spot in the pile is already taken by the first ace, only 25 of the 51 remaining positions are in the same pile.

  4. P(A) = 0.4 and P(B | A) = 0.5. If A and B are independent, what is P(B | Aᶜ)?

    Answer: 0.5

    Independence means P(B | A) = P(B). Since P(B | A) = 0.5, we get P(B) = 0.5. Independence also means P(B | Aᶜ) = P(B) = 0.5 — knowing whether A occurred (or did not) gives zero information about B. Both P(B | A) and P(B | Aᶜ) equal the unconditional P(B) when events are independent.

  5. A box contains 3 red and 2 blue balls. Two balls are drawn WITH replacement. Given that at least one ball is red, what is the probability that both balls are red?

    Answer: 3/7

    P(both red) = (3/5)² = 9/25. P(at least one red) = 1 − P(both blue) = 1 − (2/5)² = 1 − 4/25 = 21/25. By conditional probability: P(both red | at least one red) = P(both red) / P(at least one red) = (9/25) ÷ (21/25) = 9/21 = 3/7. The trap answer 9/25 is the unconditional probability of both being red — not adjusted for the given condition.

  6. A biased coin with P(H) = 0.7 is flipped until the first head OR until 3 flips, whichever comes first. What is the probability the game ends on exactly the 3rd flip?

    Answer: 0.147

    For the game to last exactly 3 flips, flips 1 and 2 must both be Tails (so the game doesn't stop early), and flip 3 occurs regardless of its outcome (the 3-flip cap forces the game to end). P(T on flip 1) × P(T on flip 2) = 0.3 × 0.3 = 0.09. The game then ends on flip 3 with certainty, so P(game ends on flip 3) = 0.09 × 1 = 0.09... Wait — re-reading: the game ends on exactly the 3rd flip means it reaches flip 3, which requires TT on flips 1-2. That probability is 0.3 × 0.3 = 0.09. But 0.09 is choice B. For 0.147: that would be P(reaches flip 3 AND flip 3 is H) = 0.09 × 0.7 = 0.063 — that's choice A. The game ends on flip 3 whether flip 3 is H or T (due to the cap), so P = 0.09. The correct answer is 0.09.