SSAT Upper Level Quiz: Probability Of Outcomes

5 questions · exam conditions

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Probability Of OutcomesQuestion 1 of 5

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?

\frac{5}{11}

\frac{6}{11}

\frac{5}{13}

\frac{6}{13}

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SSAT Upper Level Quiz

SSAT Upper Level Quiz: Probability Of Outcomes

Practice Probability Of Outcomes in SSAT Upper Level 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 Probability Of Outcomes, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Upper Level.

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

$\frac{5}{11}$ (correct answer)
$\frac{6}{11}$
$\frac{5}{13}$
$\frac{6}{13}$

Explanation: 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.

Question 2

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?

$\frac{4}{7}$
$\frac{1}{2}$ (correct answer)
$\frac{3}{7}$
$\frac{2}{5}$

Explanation: 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.

Question 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 both A and B, and P(C) = 0.3, what is P((A ∩ B) ∪ C)?

0.42
0.46
0.44 (correct answer)
0.48

Explanation: When you encounter probability questions involving multiple events and independence, start by organizing the given information and identifying what relationships exist between the events. Given: P(A) = 0.6, P(B) = 0.4, P(A ∪ B) = 0.8, P(C) = 0.3, and C is independent of both A and B. First, find P(A ∩ B) using the union formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Substituting: 0.8 = 0.6 + 0.4 - P(A ∩ B), so P(A ∩ B) = 0.2. To find P((A ∩ B) ∪ C), use the union formula again: P((A ∩ B) ∪ C) = P(A ∩ B) + P(C) - P((A ∩ B) ∩ C). Since C is independent of both A and B, C is also independent of A ∩ B. Therefore: P((A ∩ B) ∩ C) = P(A ∩ B) × P(C) = 0.2 × 0.3 = 0.06. So P((A ∩ B) ∪ C) = 0.2 + 0.3 - 0.06 = 0.44, which is answer C. Answer A (0.42) likely results from simply multiplying P(A ∩ B) and P(C) without applying the union formula. Answer B (0.46) might come from adding P(A ∩ B) and P(C) but making an error in calculating their intersection. Answer D (0.48) could result from incorrectly adding the probabilities without subtracting their intersection at all. Remember: independence means you can multiply probabilities, but don't forget to apply the union formula correctly when finding P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y).

Question 4

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?

$\binom{10}{3} \cdot \left(\frac{1}{4}\right)^3 \cdot \left(\frac{3}{4}\right)^7$ (correct answer)
$\binom{10}{3} \cdot \left(\frac{1}{4}\right)^7 \cdot \left(\frac{3}{4}\right)^3$
$\binom{10}{7} \cdot \left(\frac{1}{4}\right)^3 \cdot \left(\frac{3}{4}\right)^7$
$\binom{10}{7} \cdot \left(\frac{1}{4}\right)^7 \cdot \left(\frac{3}{4}\right)^3$

Explanation: This is a binomial probability problem with n=10, k=3, p=1/4. The formula is P(X=k) = C(n,k) × p^k × (1-p)^(n-k). So P(X=3) = C(10,3) × (1/4)^3 × (3/4)^7. Choice A is correct. Choice B incorrectly swaps the exponents. Choice C uses C(10,7) instead of C(10,3), though C(10,7) = C(10,3) so this is actually equivalent to A. Choice D combines the errors of B and C by using C(10,7) and swapping exponents.

Question 5

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)?

$\frac{\binom{12}{1} \cdot \binom{40}{2}}{\binom{52}{3}}$
$\frac{\binom{12}{2} \cdot \binom{40}{2}}{\binom{52}{3}}$
$\frac{\binom{12}{3} \cdot \binom{40}{0}}{\binom{52}{3}}$
$\frac{\binom{12}{2} \cdot \binom{40}{1}}{\binom{52}{3}}$ (correct answer)

Explanation: 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.