Probability of Outcomes - SSAT Upper Level Quantitative

1. In a certain game, a player wins if they roll a sum of 7 or 11 with two dice, and loses if they roll a sum of 2, 3, or 12. For any other sum, they continue rolling until they either roll their initial sum again (win) or roll a 7 (lose). If a player's first roll is a sum of 8, what is the probability they will eventually win the game?
   Answer: $\frac{5}{11}$ Since the first roll is 8, the player must continue rolling until they get either 8 (win) or 7 (lose). Ways to get sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 ways. Ways to get sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways. All other outcomes are ignored. So the player wins if they roll 8 before rolling 7. The probability is 5/(5+6) = 5/11. Choice B reverses the numerator and denominator components. Choice C uses wrong denominator of 13. Choice D combines the errors of B and C.
2. A multiple choice test has 10 questions, each with 4 answer choices. If a student randomly guesses on each question, what is the probability that they get exactly 3 questions correct?
   Answer: $\binom{10}{3} \cdot \left(\frac{1}{4}\right)^3 \cdot \left(\frac{3}{4}\right)^7$ This is a binomial probability with n = 10 questions, k = 3 correct, and probability p = 1/4 of guessing any one question correctly. The probability of exactly k successes is C(n, k) × $p^k$ × $(1 − p)^{n − k}$, so P(exactly 3) = C(10, 3) × $(1/4)^3$ × $(3/4)^7$. The expression with $(1/4)^7$ × $(3/4)^3$ swaps the exponents and gives the probability of exactly 7 correct. The expression $(1/4)^3$ × $(3/4)^7$ with no coefficient is the probability of one particular order, such as the first 3 right and the last 7 wrong; it leaves out the C(10, 3) = 120 ways to choose which 3 questions are right. The expression C(10, 3) × $(1/4)^3$ leaves out the requirement that the other 7 answers are wrong.
3. Two events A and B are such that P(A) = 0.6, P(B) = 0.4, and P(A ∪ B) = 0.8. If event C is independent of the event A ∩ B, and P(C) = 0.3, what is P((A ∩ B) ∪ C)?
   Answer: 0.44 Start with the union formula to find the overlap of A and B: P(A ∪ B) = P(A) + P(B) − P(A ∩ B), so 0.8 = 0.6 + 0.4 − P(A ∩ B) and P(A ∩ B) = 0.2. Now apply the union formula to the events A ∩ B and C: P((A ∩ B) ∪ C) = P(A ∩ B) + P(C) − P((A ∩ B) ∩ C). Because C is independent of A ∩ B, the overlap is a product: P((A ∩ B) ∩ C) = 0.2 × 0.3 = 0.06. So P((A ∩ B) ∪ C) = 0.2 + 0.3 − 0.06 = 0.44. The tempting wrong answers subtract the wrong overlap. Adding 0.2 and 0.3 with no subtraction gives 0.5, which is too large; 0.48 and 0.46 subtract only 0.02 or 0.04, and 0.42 subtracts 0.08, none of which equals 0.2 × 0.3. Independence lets you multiply to get the overlap, but you still need the full union formula.
4. Three cards are drawn without replacement from a standard 52-card deck. What is the probability that exactly two of the cards are face cards (Jacks, Queens, or Kings)?
   Answer: $\frac{\binom{12}{2} \cdot \binom{40}{1}}{\binom{52}{3}}$ When you encounter probability questions involving drawing cards "without replacement," you're dealing with combinations where order doesn't matter. The key is identifying what you're selecting and what you're selecting from. In a standard deck, there are 12 face cards (4 Jacks + 4 Queens + 4 Kings) and 40 non-face cards. Since you want exactly 2 face cards out of 3 draws, you need to select 2 face cards from the 12 available AND 1 non-face card from the 40 available. The correct probability uses the multiplication principle: $\frac{\binom{12}{2} \cdot \binom{40}{1}}{\binom{52}{3}}$. The numerator counts favorable outcomes (2 face cards AND 1 non-face card), while the denominator counts all possible ways to draw 3 cards from 52. Choice A uses $\binom{12}{1}$ and $\binom{40}{2}$, which would give you exactly 1 face card and 2 non-face cards—the opposite of what you want. Choice B uses $\binom{40}{2}$, meaning you're selecting 2 non-face cards, but you only want 1 non-face card since you need exactly 2 face cards total. Choice C uses $\binom{12}{3}$ and $\binom{40}{0}$, which calculates the probability of drawing exactly 3 face cards (all face cards, no non-face cards). The trap here is misreading what "exactly two face cards" means. Remember: if you want exactly 2 face cards out of 3 total cards, the remaining card must be a non-face card. Always verify that your combination selections add up to your total number of draws.
5. A fair coin is flipped repeatedly until either 3 consecutive heads or 3 consecutive tails appear. What is the probability that the sequence ends with 3 consecutive heads?
   Answer: $\frac{1}{2}$ This problem tests your understanding of probability in sequential events with symmetry. When dealing with coin flip sequences that end based on symmetric conditions (3 consecutive heads OR 3 consecutive tails), look for underlying symmetry in the problem structure. Since we're flipping a fair coin, heads and tails are equally likely on each flip. The key insight is that the stopping condition treats heads and tails completely symmetrically - we stop when we get 3 of either outcome in a row. Because the coin is fair and the stopping rules are identical for both outcomes, the probability of ending with 3 heads must equal the probability of ending with 3 tails. Let P(H) = probability of ending with 3 consecutive heads, and P(T) = probability of ending with 3 consecutive tails. By symmetry, P(H) = P(T). Since these are the only two possible outcomes when the sequence ends, we have P(H) + P(T) = 1. Therefore: P(H) + P(H) = 1, which gives us P(H) = $\frac{1}{2}$. Answer choice A ($\frac{4}{7}$) and C ($\frac{3}{7}$) likely come from complex state-based calculations that contain errors in tracking transition probabilities. Answer choice D ($\frac{2}{5}$) might result from incorrectly assuming unequal weighting of outcomes or miscounting possible sequences. When you encounter probability problems involving fair coins or dice with symmetric stopping conditions, first check whether symmetry can solve the problem immediately. This approach is much faster than setting up complex probability trees or state diagrams.

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