SHSAT Math Quiz: Probability Of An Event
8 questions · exam conditions
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Probability Of An EventQuestion 1 of 8

A random integer from 10 through 30 inclusive is chosen. What is the probability the number is a multiple of 4?

320\dfrac{3}{20}
421\dfrac{4}{21}
521\dfrac{5}{21}
621\dfrac{6}{21}
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SHSAT Math Quiz

SHSAT Math Quiz: Probability Of An Event

Practice Probability Of An Event 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 Probability Of An Event, 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

A random integer from 10 through 30 inclusive is chosen. What is the probability the number is a multiple of 4?

  1. 320\dfrac{3}{20}
  2. 421\dfrac{4}{21}
  3. 521\dfrac{5}{21} (correct answer)
  4. 621\dfrac{6}{21}
Explanation: When you encounter probability questions involving "choosing from a range," you need to identify the total number of possible outcomes and the number of favorable outcomes. First, let's find the total number of integers from 10 through 30 inclusive. This is 30−10+1=2130 - 10 + 1 = 21 integers. (Remember to add 1 when counting inclusive ranges!) Next, identify which numbers in this range are multiples of 4. Start with the smallest multiple of 4 that's at least 10: that's 12. Then continue: 12, 16, 20, 24, 28. The next multiple would be 32, which exceeds our range. So we have exactly 5 multiples of 4. Therefore, the probability is favorable outcomestotal outcomes=521\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{5}{21}. Looking at the wrong answers: Choice A gives 320\frac{3}{20}, which likely comes from miscounting either the multiples (getting 3 instead of 5) or the total range (getting 20 instead of 21). Choice B shows 421\frac{4}{21} — this gets the denominator right but undercounts the multiples of 4, perhaps by missing one of the five. Choice D gives 621\frac{6}{21}, which overcounts the multiples, possibly by including 8 (too small) or 32 (too large). Study tip: Always double-check your counting in probability problems. For ranges, use the formula "end - start + 1" for inclusive counting. For multiples, systematically list them out rather than trying to calculate — it prevents errors and takes just seconds.

Question 2

A committee of 4 people is randomly selected from a group of 6 men and 4 women. What is the probability that the committee contains exactly 2 men and 2 women?

  1. 37\frac{3}{7} (correct answer)
  2. 60210\frac{60}{210}
  3. 47\frac{4}{7}
  4. \frac{90/210}
Explanation: The number of ways to choose exactly 2 men from 6 and 2 women from 4 is C(6,2)×C(4,2) = 15×6 = 90. The total number of ways to choose 4 people from 10 is C(10,4) = 210. So probability = 90/210 = 3/7. Choice B shows the calculation 60/210, which results from an error (possibly C(6,2)×C(4,2) = 15×4 = 60). Choice C gives 4/7, which might result from a computational error. Choice D shows 90/210 without simplification in an unusual format.

Question 3

A fair six-sided die is rolled once. What is the probability that the number rolled is greater than 4?

  1. 13\dfrac{1}{3} (correct answer)
  2. 12\dfrac{1}{2}
  3. 23\dfrac{2}{3}
  4. 56\dfrac{5}{6}
Explanation: This is a basic probability question involving favorable outcomes versus total possible outcomes. When working with dice problems, always start by identifying what outcomes satisfy your condition. A fair six-sided die has outcomes: 1, 2, 3, 4, 5, and 6. You need to find which numbers are greater than 4. Only 5 and 6 satisfy this condition, giving you 2 favorable outcomes out of 6 total possible outcomes. The probability is favorable outcomestotal outcomes=26=13\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{2}{6} = \frac{1}{3}, which is choice A. Let's examine why the other answers are incorrect. Choice B (12\frac{1}{2}) would be correct if you were looking for numbers greater than 3, since that would give you outcomes 4, 5, and 6 (3 out of 6 = 12\frac{1}{2}). Choice C (23\frac{2}{3}) represents finding numbers less than or equal to 4 (outcomes 1, 2, 3, 4), which is actually the complement of what you want. Choice D (56\frac{5}{6}) would mean 5 favorable outcomes, but there's no reasonable interpretation of "greater than 4" that yields 5 outcomes on a standard die. Strategy tip: In probability questions, always clearly identify your sample space (all possible outcomes) and count your favorable outcomes carefully. Double-check by asking yourself if your answer makes intuitive sense—since most numbers on a die (4 out of 6) are NOT greater than 4, the probability should be relatively small, making 13\frac{1}{3} reasonable.

Question 4

Two fair coins are tossed. What is the probability that exactly one head appears?

  1. 14\dfrac14
  2. 12\dfrac12 (correct answer)
  3. 34\dfrac34
  4. 11
Explanation: When you encounter probability questions involving multiple events, always start by listing all possible outcomes systematically. This ensures you don't miss any possibilities or count anything twice. For two coin tosses, each coin can land either heads (H) or tails (T). The complete sample space contains four equally likely outcomes: HH, HT, TH, and TT. Since the coins are fair, each outcome has probability 14\frac{1}{4}. To find the probability of exactly one head, identify which outcomes satisfy this condition. Looking at our list: HH has two heads (doesn't qualify), HT has exactly one head (qualifies), TH has exactly one head (qualifies), and TT has zero heads (doesn't qualify). So two outcomes out of four meet our criteria, giving us 24=12\frac{2}{4} = \frac{1}{2}. Choice A (14\frac{1}{4}) represents the probability of any single specific outcome, like getting HT in that exact order. This misses that there are two ways to get exactly one head. Choice C (34\frac{3}{4}) would be the probability of getting at least one head, which includes HH, HT, and TH - three outcomes instead of two. Choice D (1) suggests certainty, which would only apply if every possible outcome resulted in exactly one head. Since we can get zero heads (TT) or two heads (HH), this is clearly wrong. Remember: in probability problems with multiple events, always enumerate all possibilities first. This prevents calculation errors and helps you spot what each answer choice might represent.

Question 5

A bag contains 4 red marbles, 3 blue marbles, and 5 green marbles. If one marble is selected at random, what is the probability it is not blue?

  1. 312\dfrac{3}{12}
  2. 512\dfrac{5}{12}
  3. 912\dfrac{9}{12} (correct answer)
  4. 1012\dfrac{10}{12}
Explanation: When you encounter probability questions involving "not" events, remember that it's often easier to find what you want by subtracting from 1, or by directly counting the favorable outcomes. Let's find the probability that a randomly selected marble is not blue. First, count the total marbles: 4 red + 3 blue + 5 green = 12 marbles total. For "not blue" marbles, we want either red or green marbles. That's 4 red + 5 green = 9 marbles that are not blue. So the probability is 912\frac{9}{12}, which is answer choice C. You could also solve this using the complement rule: P(not blue) = 1 - P(blue) = 1−312=1212−312=9121 - \frac{3}{12} = \frac{12}{12} - \frac{3}{12} = \frac{9}{12}. Looking at the wrong answers: Choice A gives 312\frac{3}{12}, which is actually the probability that the marble IS blue - this represents the complement confusion. Choice B gives 512\frac{5}{12}, which only counts green marbles, missing the red ones that are also "not blue." Choice D gives 1012\frac{10}{12}, which incorrectly includes an extra marble, possibly from miscounting the total or the favorable outcomes. Strategy tip: For "not" probability questions, you can either count directly (as we did with 9 non-blue marbles) or use complements (1 minus the unwanted probability). Both methods should give the same answer, so use whichever feels more natural to you as a quick check.

Question 6

A bag contains 15 marbles: 6 red, 5 blue, and 4 green. If two marbles are drawn without replacement, what is the probability that both marbles are the same color?

  1. 31105\frac{31}{105} (correct answer)
  2. 37105\frac{37}{105}
  3. 1135\frac{11}{35}
  4. 27\frac{2}{7}
Explanation: To find the probability both marbles are the same color, calculate: P(both red) + P(both blue) + P(both green). P(both red) = (6/15)(5/14) = 30/210 = 1/7. P(both blue) = (5/15)(4/14) = 20/210 = 2/21. P(both green) = (4/15)(3/14) = 12/210 = 2/35. Converting to common denominator 105: 15/105 + 10/105 + 6/105 = 31/105. Choice B incorrectly adds the individual probabilities as fractions without considering proper combination. Choice C results from calculation errors in the common denominator. Choice D represents only the probability of drawing two red marbles.

Question 7

A box contains 8 identical balls except for color: 3 are white, 3 are black, and 2 are gray. Three balls are drawn simultaneously. What is the probability that all three balls are different colors?

  1. 928\frac{9}{28} (correct answer)
  2. 1856\frac{18}{56}
  3. 36112\frac{36}{112}
  4. 314\frac{3}{14}
Explanation: To get all different colors, we need exactly 1 white, 1 black, and 1 gray. The number of ways to choose this is C(3,1)×C(3,1)×C(2,1) = 3×3×2 = 18. The total number of ways to choose 3 balls from 8 is C(8,3) = 56. So probability = 18/56 = 9/28. Choice B shows 18/56 without simplification. Choice C represents a computational error (doubling the numerator and denominator). Choice D results from a different calculation error, possibly confusing the selection process.

Question 8

In a standard deck of 52 cards, what is the probability of drawing either a face card (Jack, Queen, or King) or a card from the suit of hearts, but not both?

  1. 1952\frac{19}{52}
  2. 926\frac{9}{26} (correct answer)
  3. 2352\frac{23}{52}
  4. 2152\frac{21}{52}
Explanation: This asks for P(face card OR heart) - P(face card AND heart). There are 12 face cards total and 13 hearts. The overlap consists of 3 cards (Jack, Queen, King of hearts). We want cards that are face cards but not hearts (9 cards) plus hearts that are not face cards (10 cards). So 9 + 10 = 19 cards total, giving probability 19/52 = 9/26. Choice A gives 19/52 without simplification. Choice C incorrectly adds all face cards and hearts without subtracting overlap. Choice D represents a common computational error in handling the overlap.