SHSAT Math Quiz: Comparing Probabilities

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Comparing ProbabilitiesQuestion 1 of 18

When one fair coin is tossed and one fair six-sided die is rolled, which single outcome is more likely to occur?

Getting heads on the coin

Rolling a 5 on the die

The two outcomes are equally likely

There is not enough information to decide

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SHSAT Math Quiz

SHSAT Math Quiz: Comparing Probabilities

Practice Comparing Probabilities in SHSAT Math 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 Comparing Probabilities, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

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

When one fair coin is tossed and one fair six-sided die is rolled, which single outcome is more likely to occur?

Getting heads on the coin (correct answer)
Rolling a 5 on the die
The two outcomes are equally likely
There is not enough information to decide

Explanation: 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}

Question 2

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?

Picking a red marble
Picking a green marble (correct answer)
Picking a blue marble
All single colors are equally likely

Explanation: 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.

Question 3

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?

$X = Y$
$X > Y$
$Y > X$ (correct answer)
$X$ and $Y$ cannot both be determined from the table.

Explanation: 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.

Question 4

A fair six-sided die is rolled twice. The table lists four possible events. Which event has the greatest probability?

Event I: The sum of the two rolls equals 7.
Event II: The product of the two rolls is odd.
Event III: The second roll is strictly greater than the first. (correct answer)
Event IV: At least one roll shows a 6.

Explanation: 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.

Question 5

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?

The probability of drawing a red marble from the first bag is greater than from the second bag
The probability of drawing a red marble from the second bag is greater than from the first bag
The probability of drawing a red marble is the same for both bags (correct answer)
The probability of drawing a blue marble is the same for both bags

Explanation: First bag: P(red) = 8/20 = 2/5 = 0.4. Second bag: P(red) = 6/15 = 2/5 = 0.4. The probabilities are equal. Choice A is wrong because 8/20 is not greater than 6/15. Choice B is wrong because 6/15 is not greater than 8/20. Choice D is wrong because P(blue) in first bag = 12/20 = 3/5 while P(blue) in second bag = 9/15 = 3/5, which are equal, but this doesn't address the red marble comparison asked in the stem.

Question 6

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?

P(green from X) > P(yellow from Y) because Box X has more green balls total
P(green from X) < P(yellow from Y) because yellow balls outnumber green balls in both boxes
P(green from X) = P(yellow from Y) because both probabilities equal 1/4
P(green from X) < P(yellow from Y) because Box Y has a higher concentration of yellow balls (correct answer)

Explanation: P(green from X) = 5/20 = 1/4. P(yellow from Y) = 12/15 = 4/5. Since 1/4 = 0.25 and 4/5 = 0.8, we have 1/4 < 4/5. Choice D correctly identifies this relationship and gives valid reasoning. Choice A focuses on absolute numbers rather than probability. Choice B gives irrelevant reasoning. Choice C incorrectly calculates P(yellow from Y) as 1/4.

Question 7

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.

P(win die game) > P(win card game) because dice have fewer total outcomes than cards
P(win die game) < P(win card game) because there are more Kings than sides showing 6
P(win die game) = P(win card game) because both games have exactly one favorable outcome per attempt
P(win die game) > P(win card game) because 1/6 is greater than 4/52 (correct answer)

Explanation: 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.

Question 8

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).

P(vowel) < P(consonant) because there are more consonant tiles than vowel tiles in the bag (correct answer)
P(vowel) > P(consonant) because vowels are more important in forming words than consonants
P(vowel) = P(consonant) because the ratio of vowels to consonants balances at 3:7
P(vowel) > P(consonant) because there are 3 A tiles but only 1 or 2 of each consonant type

Explanation: 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.

Question 9

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.

P(both same) > P(sum is 7) because matching outcomes are more likely than specific sums
P(both same) < P(sum is 7) because there are more ways to get sum of 7
P(both same) = P(sum is 7) because both events have exactly 6 favorable outcomes (correct answer)
P(both same) > P(sum is 7) because identical outcomes occur in exactly 6 ways while sum of 7 occurs in 5 ways

Explanation: 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.

Question 10

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 $P(F) - P(E)$ ?

$\tfrac{16}{81}$ (correct answer)
$\tfrac{20}{81}$
$\tfrac{24}{81}$
$\tfrac{8}{81}$

Explanation: 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

Question 11

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 $q$ ?

$p < q$ because "red or blue" excludes some non-green, non-yellow cards. (correct answer)
$p = q$ only if the deck contains exactly four colors.
$p > q$ because "red or blue" includes cards that may also be green or yellow.
$p$ and $q$ cannot be compared without more information.

Explanation: 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.

Question 12

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?

$P(E_1) > P(E_2)$
$P(E_2) - P(E_1) = \tfrac{11}{25}$ (correct answer)
$P(E_1) = P(E_2)$
$P(E_2) - P(E_1) = \tfrac{3}{5}$

Explanation: 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.

Question 13

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 $P_1 - P_2$ ?

$0.13$
$0.47$ (correct answer)
$0.63$
$0.79$

Explanation: 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

Question 14

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.

The probability of drawing a face card is greater than drawing a red card or Ace
The probability of drawing a red card or Ace is greater than drawing a face card (correct answer)
Both probabilities are equal since each involves 12 favorable outcomes
The probability of drawing a red card or Ace is exactly twice the probability of drawing a face card

Explanation: 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.

Question 15

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.

P(even) > P(greater than 2) because even numbers are more spread out
P(even) < P(greater than 2) because there are more numbers greater than 2 (correct answer)
P(even) = P(greater than 2) because both events have exactly 3 favorable outcomes
P(even) < P(greater than 2) because the maximum even number is 6

Explanation: 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.

Question 16

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.

P(red) > P(not white) because red marbles are less common than white marbles
P(red) < P(not white) because not white includes both red and blue marbles (correct answer)
P(red) = P(not white) because both events involve exactly 4 favorable outcomes
P(red) > P(not white) because there are more white marbles to avoid than red marbles to select

Explanation: 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.

Question 17

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?

P(A) > P(B) because there are more ways to make 7 than 8 (correct answer)
P(A) < P(B) because 8 is closer to the average sum of 7.5
P(A) = P(B) because both sums are equally likely with two dice
P(A) > P(B) because 7 appears more frequently in the middle of the distribution

Explanation: 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.'

Question 18

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.

P(both hearts) > P(both black) because hearts are red cards which stand out more
P(both hearts) < P(both black) because there are more black cards than hearts in the deck (correct answer)
P(both hearts) = P(both black) because both involve drawing two cards of the same type
P(both hearts) > P(both black) because the second heart is easier to draw after the first

Explanation: 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.