Comparing Probabilities - SHSAT Math

A fair six-sided die is rolled twice. The table lists four possible events. Which event has the greatest probability?
Answer: Event III: The second roll is strictly greater than the first. Compute each: I) Sum=7 has 6 outcomes (1-6,2-5,3-4,4-3,5-2,6-1): 6/36=1/6. II) Product odd means both odd: 3·3=9 outcomes: 9/36=1/4. III) Second > first: C(6,2)=15 outcomes: 15/36=5/12. IV) At least one 6: 1−(5/6)²=11/36. Compare: 1/6≈0.167, 1/4=0.25, 5/12≈0.417, 11/36≈0.306. Event III has the greatest probability.
A standard number cube (die) is rolled twice. Compare the probability that both rolls show the same number to the probability that the sum of both rolls is 7.
Answer: P(both same) = P(sum is 7) because both events have exactly 6 favorable outcomes Both same: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes, so P = 6/36 = 1/6. Sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes, so P = 6/36 = 1/6. Both probabilities equal 1/6. Choice A gives incorrect general reasoning. Choice B incorrectly counts outcomes for sum of 7. Choice D incorrectly states sum of 7 has 5 ways.
Box X contains 5 green balls and 15 yellow balls. Box Y contains 3 green balls and 12 yellow balls. If the probability of selecting a green ball from Box X is compared to the probability of selecting a yellow ball from Box Y, which is true?
Answer: P(green from X) < P(yellow from Y) because Box Y has a higher concentration of yellow balls Find each probability. Box X has 5 + 15 = 20 balls, and 5 are green, so P(green from X) = 5/20 = 1/4. Box Y has 3 + 12 = 15 balls, and 12 are yellow, so P(yellow from Y) = 12/15 = 4/5. Since 1/4 = 0.25 and 4/5 = 0.8, P(green from X) < P(yellow from Y). Yellow makes up a large share of Box Y (4 out of 5 balls), while green makes up only a small share of Box X (1 out of 4). Box X does have more green balls than Box Y, but a probability depends on the share of the box, not the raw count, so a greater-than claim is wrong. Only green from X equals 1/4; yellow from Y is 4/5, so the probabilities are not equal, and the fact that yellow outnumbers green in both boxes does not make them equal either. P(green from X) < P(yellow from Y) because Box Y has a higher concentration of yellow balls.
The grid shown below represents a 5×5 lattice. A point is chosen uniformly at random from the 25 lattice points. Let $E_1$ = "point lies on the shaded diagonal" and $E_2$ = "point lies on the boundary of the grid." Using the figure, which is true?
Answer: $P(E_2) - P(E_1) = \tfrac{11}{25}$ The 5×5 lattice has 25 points. The main diagonal contains 5 points, so $P(E_1) = 5/25 = 1/5$ . The boundary consists of the 16 perimeter lattice points (25 − 9 interior), so $P(E_2) = 16/25$ . Difference: $16/25 - 5/25 = 11/25$ . Choice D (3/5=15/25) comes from miscounting the boundary as 20. C would require equal counts. A is the reverse.
The two-way table shows the number of students in a school, classified by grade and by whether they walk to school. A student is selected at random. Let $X$ = P(student is in Grade 8 | student walks) and $Y$ = P(student walks | student is in Grade 8). Using the table, which is true?
Answer: $Y > X$ From the table: Grade 6 walks 20/60, Grade 7 walks 30/80, Grade 8 walks 25/50. Total walkers = 75; total Grade 8 = 50. $X = 25/75 = 1/3 \approx 0.333$ . $Y = 25/50 = 1/2 = 0.5$ . So $Y > X$ . Students who confuse conditional direction choose A or B. D is wrong since both can be computed.
The bar graph below shows the number of cards of each color in a special deck. One card is drawn at random. Let $p$ be the probability the card is red or blue, and let $q$ be the probability the card is NOT green and NOT yellow. Which of the following best describes the relationship between $p$ and $q$ ?
Answer: $p < q$ because "red or blue" excludes some non-green, non-yellow cards. The bar graph shows five colors: Red (20), Blue (15), Green (10), Yellow (5), Purple (10), Total = 60. $p = (20+15)/60 = 35/60 = 7/12$ . $q =$ P(not green and not yellow) = (20+15+10)/60 = 45/60 = 3/4. So p < q. The difference is the purple cards, which are counted in q but not in p. A correctly states this. B is wrong since equality requires no "other" color. C is wrong as the bars don't overlap. D is wrong since the graph provides full information.
A bag contains 6 red, 9 green, and 5 blue marbles. Which single-color outcome is more likely when one marble is selected at random?
Answer: Picking a green marble When you encounter probability questions involving selecting items from a collection, you need to compare the likelihood of each outcome by calculating their individual probabilities. First, find the total number of marbles: 6 red + 9 green + 5 blue = 20 marbles total. Now calculate each probability: - Red marble: $\frac{6}{20} = 0.30$ or 30% - Green marble: $\frac{9}{20} = 0.45$ or 45% - Blue marble: $\frac{5}{20} = 0.25$ or 25% Since green marbles have the highest probability at 45%, picking a green marble is the most likely single-color outcome. This makes B correct. Looking at why the other choices are wrong: A is incorrect because red marbles only represent 30% of the collection, which is less likely than green. C is wrong because blue marbles make up just 25% of the total, the smallest proportion. D is incorrect because the probabilities are clearly different—9 green marbles versus 6 red and 5 blue means the outcomes are not equally likely. Here's your strategy for these problems: Always start by finding the total, then calculate each individual probability as a fraction. The outcome with the largest numerator (when denominators are the same) will be most likely. Don't just look at the numbers in isolation—6, 9, and 5 might seem close, but 9 out of 20 is significantly more likely than 5 out of 20.
In a deck of 52 cards, compare the probability of drawing a face card (Jack, Queen, King) to the probability of drawing a card that is either red or an Ace.
Answer: The probability of drawing a red card or Ace is greater than drawing a face card Face cards: 12 cards (3 per suit × 4 suits), so P(face card) = 12/52 = 3/13. Red cards or Aces: 26 red cards + 2 black Aces = 28 cards, so P(red or Ace) = 28/52 = 7/13. Since 7/13 > 3/13, choice B is correct. Choice A reverses the inequality. Choice C incorrectly counts 12 favorable outcomes for red/Ace. Choice D claims the ratio is exactly 2:1, but 28/52 ÷ 12/52 = 28/12 = 7/3, not 2.
A jar contains 4 red, 6 blue, and 10 white marbles. Compare the probability of drawing a red marble to the probability of drawing a marble that is not white.
Answer: P(red) < P(not white) because not white includes both red and blue marbles Total marbles = 4 + 6 + 10 = 20. P(red) = 4/20 = 1/5. P(not white) = P(red or blue) = (4 + 6)/20 = 10/20 = 1/2. Since 1/5 < 1/2, choice B is correct. Choice A uses illogical reasoning about rarity. Choice C incorrectly states both events have 4 favorable outcomes. Choice D provides irrelevant reasoning about avoidance.
Two cards are drawn without replacement from a standard deck. Compare the probability that both cards are hearts to the probability that both cards are black.
Answer: P(both hearts) < P(both black) because there are more black cards than hearts in the deck P(both hearts) = (13/52) × (12/51) = 156/2652 = 1/17. P(both black) = (26/52) × (25/51) = 650/2652 = 25/102. Since 1/17 ≈ 0.059 and 25/102 ≈ 0.245, we have P(both hearts) < P(both black). Choice B correctly identifies this and gives valid reasoning. Choice A uses irrelevant reasoning about visibility. Choice C incorrectly assumes equal probabilities. Choice D provides incorrect reasoning about conditional probability.
The figure shows a dartboard consisting of a large square of side 10 with an inscribed circle, and a smaller square of side 4 centered inside the circle. A dart lands uniformly at random in the large square. Let $P_1$ = probability the dart lands inside the circle but outside the small square, and $P_2$ = probability the dart lands inside the small square. Which of the following is closest to $P_1 - P_2$ ?
Answer: $0.47$ Area of large square = 100. Area of circle = $25\pi \approx 78.54$ . Area of small square = 16. $P_1 = (25\pi - 16)/100 \approx (78.54-16)/100 = 0.6254$ . $P_2 = 16/100 = 0.16$ . $P_1 - P_2 \approx 0.6254 - 0.16 = 0.4654 \approx 0.47$ . Choice A (0.13) would come from forgetting the square is inside the circle. C comes from computing only $P_1$ . D would come from $P_1 + P_2$ .
A bag contains lettered tiles: 3 A's, 2 B's, 4 C's, and 1 D. Compare the probability of drawing a vowel (A) to the probability of drawing a consonant (B, C, or D).
Answer: P(vowel) < P(consonant) because there are more consonant tiles than vowel tiles in the bag When you encounter probability questions involving multiple outcomes, always start by calculating the actual probabilities using the basic formula: probability equals favorable outcomes divided by total outcomes. First, count the total tiles: 3 A's + 2 B's + 4 C's + 1 D = 10 tiles total. For vowels (only A): P(vowel) = 3/10 = 0.3 For consonants (B, C, D): P(consonant) = (2 + 4 + 1)/10 = 7/10 = 0.7 Since 0.3 < 0.7, the probability of drawing a vowel is less than the probability of drawing a consonant. Choice A correctly identifies this relationship and provides the right reasoning: there are indeed more consonant tiles (7) than vowel tiles (3) in the bag. Choice B makes a fundamental error by bringing in irrelevant information about the "importance" of vowels in word formation. Probability depends only on the actual quantities present, not on linguistic significance. Choice C correctly notes the 3:7 ratio but misunderstands what "balance" means in probability. A 3:7 ratio shows inequality, not balance—the probabilities would only be equal if the ratio were 5:5. Choice D focuses on comparing individual letter types rather than the vowel versus consonant categories as a whole. While it's true that A appears more frequently than B or D individually, this doesn't matter when comparing entire groups. Strategy tip: In probability comparisons, always calculate both probabilities first using actual counts. Don't let irrelevant context or individual item comparisons distract you from the main question.
A spinner is divided into 6 equal sections numbered 1 through 6. Compare the probability of spinning an even number to the probability of spinning a number greater than 2.
Answer: P(even) < P(greater than 2) because there are more numbers greater than 2 Even numbers: {2, 4, 6}, so P(even) = 3/6 = 1/2. Numbers greater than 2: {3, 4, 5, 6}, so P(greater than 2) = 4/6 = 2/3. Since 1/2 < 2/3, choice B is correct. Choice A uses incorrect reasoning about distribution. Choice C incorrectly states both have 3 favorable outcomes. Choice D gives irrelevant reasoning about the maximum value.
Two dice are rolled. Event A is getting a sum of 7. Event B is getting a sum of 8. How do the probabilities of these events compare?
Answer: P(A) > P(B) because there are more ways to make 7 than 8 Ways to get sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways, so P(A) = 6/36 = 1/6. Ways to get sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 ways, so P(B) = 5/36. Since 6/36 > 5/36, P(A) > P(B). Choice B is incorrect reasoning about averages. Choice C is wrong because the probabilities are not equal. Choice D uses correct reasoning but vague language about 'distribution.'
In the figure below, a 3×3 grid of squares is shown. A square is selected at random. Then, independently, a second square is selected at random (with replacement). Let $E$ = "the two squares share an edge" and $F$ = "the two squares share at least a corner (including shared edges)." Using the figure, what is $P(F) - P(E)$ ?
Answer: $\tfrac{16}{81}$ Total outcomes = 9·9 = 81. Count edge-sharing ordered pairs: each internal edge (horizontal or vertical between adjacent squares) contributes 2 ordered pairs. The 3×3 grid has 2·3 = 6 horizontal adjacencies and 6 vertical = 12 adjacencies, giving 24 ordered pairs. So $P(E) = 24/81$ . Count at-least-corner-sharing ordered pairs: for each square, count neighbors (including diagonal). Corner squares have 3 neighbors (4 corners × 3 = 12); edge squares have 5 neighbors (4 edges × 5 = 20); center has 8 neighbors (1×8 = 8). Total = 12+20+8 = 40. So $P(F) = 40/81$ . $P(F) − P(E) = 16/81$ .
A bag contains 8 red marbles and 12 blue marbles. Another bag contains 6 red marbles and 9 blue marbles. If one marble is drawn randomly from each bag, which statement correctly compares the probabilities?
Answer: The probability of drawing a red marble is the same for both bags Find each probability as a fraction of the bag's total. The first bag has 8 + 12 = 20 marbles, so P(red) = 8/20 = 2/5. The second bag has 6 + 9 = 15 marbles, so P(red) = 6/15 = 2/5. The two probabilities of red are equal, so the probability of drawing a red marble is the same for both bags. The statements that one bag gives a greater chance of red come from comparing the counts 8 and 6 instead of the fractions. The statement that blue is more likely from the first bag makes the same mistake with 12 and 9 blue marbles; in fact P(blue) is 12/20 = 3/5 for the first bag and 9/15 = 3/5 for the second, also equal.
In a game, you win if you roll a 6 on a standard die, and you win if you draw a King from a standard deck of cards. Compare the probability of winning the die game to the probability of winning the card game.
Answer: P(win die game) > P(win card game) because 1/6 is greater than 4/52 P(win die game) = P(roll 6) = 1/6 ≈ 0.167. P(win card game) = P(draw King) = 4/52 = 1/13 ≈ 0.077. Since 1/6 > 1/13, choice D is correct and provides accurate calculation. Choice A gives incorrect reasoning about total outcomes. Choice B incorrectly compares absolute numbers. Choice C incorrectly focuses on number of favorable outcomes rather than probability ratios.
When one fair coin is tossed and one fair six-sided die is rolled, which single outcome is more likely to occur?
Answer: Getting heads on the coin When you encounter probability comparison questions, you need to calculate the likelihood of each individual outcome and compare them directly. Let's find the probability of each outcome. For the coin toss, there are 2 equally likely outcomes (heads or tails), so the probability of getting heads is $\frac{1}{2} = 0.5$ or 50%. For the die roll, there are 6 equally likely outcomes (1, 2, 3, 4, 5, or 6), so the probability of rolling any specific number, including 5, is $\frac{1}{6} ≈ 0.167$ or about 16.7%. Since $\frac{1}{2} > \frac{1}{6}$ , getting heads on the coin is more likely than rolling a 5 on the die. Choice A is correct because the probability of heads ( $\frac{1}{2}$ ) is greater than the probability of rolling a 5 ( $\frac{1}{6}$ ). Choice B is incorrect because rolling a 5 has the lower probability of $\frac{1}{6}$ . Choice C is wrong because $\frac{1}{2} ≠ \frac{1}{6}$ — the probabilities are clearly different. Choice D is incorrect because we have complete information about both probability situations: we know exactly how many outcomes are possible for each event. Remember that when comparing probabilities of single outcomes, convert everything to the same denominator or decimal form to make the comparison clear. Also, any single outcome on a fair six-sided die will always be less likely than any single outcome on a fair coin, since $\frac{1}{6} < \frac{1}{2}$ .

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