TACHS Quiz: Simple Probability
13 questions · exam conditions
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Simple ProbabilityQuestion 1 of 13

A day of the week is selected at random. What is the probability that the day chosen is a weekend day (Saturday or Sunday)?

17\frac{1}{7}
27\frac{2}{7}
37\frac{3}{7}
57\frac{5}{7}
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TACHS Quiz

TACHS Quiz: Simple Probability

Practice Simple Probability in TACHS 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 Simple Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for TACHS.

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 day of the week is selected at random. What is the probability that the day chosen is a weekend day (Saturday or Sunday)?

  1. 17\frac{1}{7}
  2. 27\frac{2}{7} (correct answer)
  3. 37\frac{3}{7}
  4. 57\frac{5}{7}
Explanation: When you encounter probability questions, remember that probability equals the number of favorable outcomes divided by the total number of possible outcomes. Here, you're selecting one day from the seven days of the week, so there are 7 total possible outcomes. The favorable outcomes are weekend days: Saturday and Sunday. That gives you 2 favorable outcomes. Using the probability formula: P(weekend day)=favorable outcomestotal outcomes=27P(\text{weekend day}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{2}{7} This confirms that answer B is correct. Let's examine why the other choices are wrong. Answer A gives 17\frac{1}{7}, which would be the probability of selecting one specific day (like only Saturday or only Sunday), but the question asks for either weekend day. Answer C shows 37\frac{3}{7}, which might result from incorrectly counting three days as weekend days—perhaps mistakenly including Friday. Answer D presents 57\frac{5}{7}, which represents the probability of selecting a weekday (Monday through Friday), which is the opposite of what the question asks for. The key insight is carefully identifying what constitutes a "favorable outcome." Weekend days specifically means Saturday and Sunday—exactly two days out of seven. For probability questions on the TACHS, always start by clearly counting your favorable outcomes and total possible outcomes. Write them down if needed. Many students rush and miscount, especially when dealing with groups like "weekend days" or "weekdays."

Question 2

An integer is chosen at random from 11 through 1010, inclusive. What is the probability that the integer is prime?

  1. 14\frac{1}{4}
  2. 25\frac{2}{5} (correct answer)
  3. 35\frac{3}{5}
  4. 45\frac{4}{5}
Explanation: When you encounter probability questions, you need to identify the favorable outcomes and divide by the total possible outcomes. Here, you're looking for the probability of selecting a prime number from the integers 1 through 10. First, let's identify all prime numbers in this range. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Going through each number: 1 is not prime (by definition), 2 is prime, 3 is prime, 4 is not prime (divisible by 2), 5 is prime, 6 is not prime (divisible by 2 and 3), 7 is prime, 8 is not prime (divisible by 2), 9 is not prime (divisible by 3), and 10 is not prime (divisible by 2 and 5). So the prime numbers from 1 to 10 are: 2, 3, 5, and 7. That's 4 prime numbers out of 10 total numbers, giving us a probability of 410=25\frac{4}{10} = \frac{2}{5}. Looking at the wrong answers: A) 14\frac{1}{4} might result from incorrectly thinking there are only 2 or 3 primes in this range. C) 35\frac{3}{5} could come from miscounting and finding 6 primes instead of 4. D) 45\frac{4}{5} represents finding 8 primes, which suggests a fundamental misunderstanding of what makes a number prime. Remember to systematically check each number for primality by testing if it has any divisors other than 1 and itself. Don't rush through the counting—prime identification is where most errors occur in probability questions involving primes.

Question 3

A spinner is divided into 5 equal sections numbered 11 through 55. What is the probability that a single spin lands on a number divisible by 33?

  1. 14\frac{1}{4}
  2. 15\frac{1}{5} (correct answer)
  3. 25\frac{2}{5}
  4. 35\frac{3}{5}
Explanation: When you encounter probability questions involving spinners or similar scenarios, remember that probability equals the number of favorable outcomes divided by the total number of possible outcomes. First, identify what numbers from 1 through 5 are divisible by 3. To be divisible by 3, a number must be evenly divided by 3 with no remainder. Checking each number: 1 ÷ 3 = 0.33... (not divisible), 2 ÷ 3 = 0.67... (not divisible), 3 ÷ 3 = 1 (divisible), 4 ÷ 3 = 1.33... (not divisible), and 5 ÷ 3 = 1.67... (not divisible). Only the number 3 is divisible by 3. Since there's 1 favorable outcome (landing on 3) out of 5 total possible outcomes, the probability is 15\frac{1}{5}. Choice A (14\frac{1}{4}) incorrectly uses 4 as the denominator, perhaps from mistakenly excluding one section or confusing this with a different scenario. Choice C (25\frac{2}{5}) suggests there are 2 numbers divisible by 3, which might come from incorrectly thinking 6 is on the spinner or miscounting. Choice D (35\frac{3}{5}) uses 3 as the numerator, possibly confusing the number we're looking for (3) with the count of favorable outcomes. Study tip: For divisibility problems, always check each number systematically. Don't assume patterns—actually verify which numbers meet the criteria. Also, remember that in basic probability, your denominator should always equal the total number of equally likely outcomes.

Question 4

A student is chosen at random from a class of 15 boys and 10 girls. What is the probability that the student selected is a girl?

  1. 13\frac{1}{3}
  2. 25\frac{2}{5} (correct answer)
  3. 35\frac{3}{5}
  4. 23\frac{2}{3}
Explanation: This is a basic probability question where you need to find the likelihood of selecting a specific type of student from the total group. Probability always equals the number of favorable outcomes divided by the total number of possible outcomes. First, let's find the total number of students: 15 boys + 10 girls = 25 students total. Since we want the probability of selecting a girl, the number of favorable outcomes is 10 (the number of girls). Therefore, the probability is 1025=25\frac{10}{25} = \frac{2}{5}, which is answer choice B. Let's examine why the other answers are incorrect. Choice A (13\frac{1}{3}) might result from incorrectly using only part of the class size in your calculation. Choice C (35\frac{3}{5}) equals 1525\frac{15}{25}, which would be the probability of selecting a boy, not a girl—this is a common trap where students calculate the opposite of what's asked. Choice D (23\frac{2}{3}) doesn't correspond to any logical calculation with the given numbers and likely represents a random guess or computational error. When solving probability problems on the TACHS, always identify three key elements: the total number of possible outcomes, the number of favorable outcomes, and exactly what event you're calculating the probability for. Double-check that you're answering the right question—test makers often include the probability of the complementary event as a distractor.

Question 5

A fair six-sided die is rolled once. What is the probability that the number shown is greater than 44?

  1. 16\frac{1}{6}
  2. 13\frac{1}{3} (correct answer)
  3. 12\frac{1}{2}
  4. 23\frac{2}{3}
Explanation: When you encounter probability questions involving dice, remember that probability equals the number of favorable outcomes divided by the total number of possible outcomes. A standard six-sided die has faces numbered 1, 2, 3, 4, 5, and 6. Since we want numbers greater than 4, we need to identify which outcomes satisfy this condition. The numbers greater than 4 are 5 and 6 — that's 2 favorable outcomes. The total number of possible outcomes when rolling once is 6. Therefore, the probability is 26=13\frac{2}{6} = \frac{1}{3}, which is answer choice B. Let's examine why the other options are incorrect. Choice A gives 16\frac{1}{6}, which would be correct if only one number were greater than 4, but both 5 and 6 qualify. Choice C suggests 12\frac{1}{2}, which would mean 3 numbers are greater than 4 — perhaps someone incorrectly included 4 itself or miscounted the outcomes. Choice D gives 23\frac{2}{3}, which would indicate 4 favorable outcomes out of 6, suggesting someone might have counted numbers greater than or equal to 3 (which would be 3, 4, 5, 6). Strategy tip: For dice probability problems, always list out the specific outcomes that satisfy the condition before calculating. This prevents counting errors and helps you avoid the common trap of confusing "greater than" with "greater than or equal to." Double-check by ensuring your favorable outcomes plus unfavorable outcomes equal the total possible outcomes.

Question 6

A fair coin is flipped once. What is the probability that it lands heads up?

  1. 14\frac{1}{4}
  2. 13\frac{1}{3}
  3. 12\frac{1}{2} (correct answer)
  4. 23\frac{2}{3}
Explanation: When you encounter probability questions involving coins, dice, or other "fair" objects, you're working with basic probability where all outcomes are equally likely. The fundamental formula is: probability = (number of favorable outcomes) ÷ (total number of possible outcomes). A fair coin has exactly two possible outcomes when flipped: heads or tails. Since we want the probability of getting heads, there's only one favorable outcome (heads) out of two total possible outcomes. Therefore, the probability is 12\frac{1}{2}, making choice C correct. Let's examine why the other answers don't work. Choice A (14\frac{1}{4}) would suggest there are four equally likely outcomes, but a coin only has two sides. This fraction might appear if you were calculating the probability of getting heads twice in a row, but that's not what this question asks. Choice B (13\frac{1}{3}) implies three equally likely outcomes, which also doesn't match a two-sided coin. Choice D (23\frac{2}{3}) suggests that favorable outcomes outnumber unfavorable ones, but getting heads is no more likely than getting tails on a fair coin. Remember this key strategy: for basic probability questions, always start by identifying all possible outcomes, then count how many of those outcomes satisfy what you're looking for. Fair coins, standard dice, and similar objects have predictable numbers of equally likely outcomes, so don't overthink these problems—stick to the straightforward counting approach.

Question 7

An integer is selected at random from −3-3 to 33 inclusive. What is the probability that the integer selected is a non-negative even number?

  1. 17\frac{1}{7}
  2. 27\frac{2}{7} (correct answer)
  3. 37\frac{3}{7}
  4. 47\frac{4}{7}
Explanation: When you encounter probability questions, start by identifying your sample space (all possible outcomes) and your favorable outcomes (what you're looking for). The integers from −3-3 to 33 inclusive are: −3,−2,−1,0,1,2,3-3, -2, -1, 0, 1, 2, 3. That's 7 total integers in your sample space. Now identify the non-negative even numbers in this range. Non-negative means greater than or equal to zero, so you're looking at 0,1,2,30, 1, 2, 3. Among these, the even numbers are 00 and 22. So you have 2 favorable outcomes. The probability is favorable outcomestotal outcomes=27\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{2}{7}, which is choice B. Let's see where the other answers come from. Choice A (17\frac{1}{7}) would result if you only counted one favorable outcome—perhaps forgetting that zero is even, or missing that zero is non-negative. Choice C (37\frac{3}{7}) might occur if you incorrectly included all non-negative numbers (0,1,20, 1, 2) instead of just the even ones. Choice D (47\frac{4}{7}) could happen if you counted all even numbers in the range (−2,0,2-2, 0, 2) plus one extra, forgetting the "non-negative" requirement. Remember that zero is both even and non-negative—these special properties of zero frequently appear in TACHS probability questions. Always carefully parse compound conditions like "non-negative even" by checking each requirement separately.

Question 8

One card is drawn at random from a standard deck. What is the probability that the card is either a heart or a spade?

  1. 14\frac{1}{4}
  2. 13\frac{1}{3}
  3. 12\frac{1}{2} (correct answer)
  4. 34\frac{3}{4}
Explanation: When you encounter probability questions involving "or," you're dealing with the addition principle. The key is identifying whether the events can happen simultaneously (which affects your calculation). In a standard deck, there are 52 cards total, with 13 cards in each of the four suits: hearts, diamonds, clubs, and spades. Since you want the probability of drawing either a heart OR a spade, you need to count all favorable outcomes. There are 13 hearts and 13 spades, giving you 26 favorable cards out of 52 total cards. This gives you 2652=12\frac{26}{52} = \frac{1}{2}. Notice that hearts and spades are mutually exclusive events—a single card cannot be both a heart and a spade simultaneously. This means you can simply add the probabilities: 1352+1352=2652=12\frac{13}{52} + \frac{13}{52} = \frac{26}{52} = \frac{1}{2}. Choice A (14\frac{1}{4}) represents the probability of drawing just one specific suit, like hearts only. Choice B (13\frac{1}{3}) doesn't correspond to any logical grouping in a standard deck. Choice D (34\frac{3}{4}) would be correct if you wanted three of the four suits, but the question asks for only two specific suits. The correct answer is C. Strategy tip: For "or" probability questions, always check if the events are mutually exclusive (can't happen together). If they are, add the individual probabilities. If not, you'll need to subtract the overlap to avoid double-counting.

Question 9

A two-digit whole number is chosen at random from 1010 through 9999, inclusive. What is the probability that the number selected is divisible by 55?

  1. 16\frac{1}{6}
  2. 15\frac{1}{5} (correct answer)
  3. 14\frac{1}{4}
  4. 13\frac{1}{3}
Explanation: When you see a probability question asking about numbers with specific properties, you need to count favorable outcomes and divide by total possible outcomes. First, let's find how many two-digit numbers exist from 10 through 99. These are: 10, 11, 12, ..., 99. That's 99−10+1=9099 - 10 + 1 = 90 total numbers. Next, identify which numbers are divisible by 5. A number is divisible by 5 if and only if it ends in 0 or 5. The two-digit numbers ending in 0 are: 10, 20, 30, 40, 50, 60, 70, 80, 90 (that's 9 numbers). The two-digit numbers ending in 5 are: 15, 25, 35, 45, 55, 65, 75, 85, 95 (that's 9 numbers). So we have 9+9=189 + 9 = 18 favorable outcomes. Therefore, the probability is 1890=15\frac{18}{90} = \frac{1}{5}, which is choice B. Looking at the wrong answers: Choice A (16\frac{1}{6}) would suggest only 15 favorable outcomes, perhaps from miscounting the multiples of 5. Choice C (14\frac{1}{4}) implies 22.5 favorable outcomes, which might come from incorrectly thinking every fourth number is divisible by 5. Choice D (13\frac{1}{3}) suggests 30 favorable outcomes, possibly from confusing divisibility rules or miscounting the total range. Remember: For divisibility by 5, focus only on the last digit. If it's 0 or 5, the entire number is divisible by 5. This makes counting much easier than checking each number individually.

Question 10

A box holds 6 red pens, 4 green pens, and 2 blue pens. If one pen is selected at random, what is the probability that the pen is not red?

  1. 13\frac{1}{3}
  2. 12\frac{1}{2} (correct answer)
  3. 23\frac{2}{3}
  4. 56\frac{5}{6}
Explanation: When you encounter probability questions asking for the likelihood that something does not happen, you're dealing with complement probability - one of the most reliable shortcuts in probability problems. First, let's find the total number of pens: 6 red + 4 green + 2 blue = 12 pens total. The probability that a pen is NOT red equals 1 minus the probability that it IS red. Since there are 6 red pens out of 12 total, the probability of selecting red is 612=12\frac{6}{12} = \frac{1}{2}. Therefore, the probability of NOT selecting red is 1−12=121 - \frac{1}{2} = \frac{1}{2}. You can verify this by counting directly: non-red pens include 4 green + 2 blue = 6 pens, so 612=12\frac{6}{12} = \frac{1}{2}. Looking at the wrong answers: Choice A (13\frac{1}{3}) might come from incorrectly using only one color of non-red pens, like 412=13\frac{4}{12} = \frac{1}{3} for just green pens. Choice C (23\frac{2}{3}) could result from miscounting the total as 9 instead of 12, giving 69=23\frac{6}{9} = \frac{2}{3}. Choice D (56\frac{5}{6}) might arise from forgetting to include one type of pen in your count. For complement probability questions, remember this formula: P(not A) = 1 - P(A). This approach often prevents counting errors and is usually faster than adding up all the "not" cases, especially when there are multiple categories to avoid.

Question 11

One card is drawn at random from a standard 52-card deck. What is the probability that the card is a face card (Jack, Queen, or King)?

  1. 14\frac{1}{4}
  2. 113\frac{1}{13}
  3. 313\frac{3}{13} (correct answer)
  4. 326\frac{3}{26}
Explanation: When you encounter probability questions involving card decks, always start by identifying what you're looking for and the total number of possible outcomes. A standard deck has 52 cards, and you need to find how many cards meet your criteria. Face cards are Jacks, Queens, and Kings. Since there are 4 suits (hearts, diamonds, clubs, spades), and each suit contains one Jack, one Queen, and one King, you have:
  • 4 Jacks + 4 Queens + 4 Kings = 12 face cards total
The probability formula is: P=favorable outcomestotal outcomes=1252=313P = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{12}{52} = \frac{3}{13} Choice A (14\frac{1}{4}) represents a common error where students might think face cards make up one-fourth of the deck. While 14=1352\frac{1}{4} = \frac{13}{52}, this would mean 13 face cards, but there are only 12. Choice B (113\frac{1}{13}) suggests only 4 face cards total, which would be true if you only counted one type (like just the Kings) instead of all three face card types. Choice D (326\frac{3}{26}) might result from incorrectly dividing by 26 instead of 52, perhaps thinking about half the deck or confusing red/black cards with the total. The correct answer is C: 313\frac{3}{13}. Study tip: For deck probability problems, always count systematically by suits. Remember the key numbers: 52 total cards, 4 suits, 13 cards per suit, and 12 face cards total. Write these down if needed during the test.

Question 12

A bag contains 4 blue marbles, 6 green marbles, and 5 yellow marbles. If one marble is selected at random, what is the probability that it is green?

  1. 13\frac{1}{3}
  2. 25\frac{2}{5} (correct answer)
  3. 12\frac{1}{2}
  4. 35\frac{3}{5}
Explanation: When you encounter probability questions, you're looking for the ratio of favorable outcomes to total possible outcomes. The key is to carefully count what you have and what you want. First, let's find the total number of marbles: 4 blue + 6 green + 5 yellow = 15 marbles total. Since you want the probability of selecting a green marble, and there are 6 green marbles, the probability is 615\frac{6}{15}. Simplifying this fraction by dividing both numerator and denominator by 3 gives us 25\frac{2}{5}, which is answer choice B. Let's examine why the other answers are incorrect. Choice A (13\frac{1}{3}) would be the result if you mistakenly thought there were only 3 colors and assumed equal probability for each color, ignoring the actual quantities. Choice C (12\frac{1}{2}) doesn't correspond to any logical calculation with these numbers. Choice D (35\frac{3}{5}) is a trap that might result from incorrectly using 10 as your denominator instead of 15, perhaps by miscounting the total marbles. Remember that probability problems on the TACHS often test your ability to organize information and perform accurate calculations under pressure. Always start by clearly identifying the total number of possible outcomes, then count the favorable outcomes. Double-check your arithmetic, especially when simplifying fractions, as the test makers frequently include unsimplified fractions as distractors.

Question 13

A letter is selected at random from the word STATISTICS. What is the probability that the chosen letter is T?

  1. 14\frac{1}{4}
  2. 13\frac{1}{3}
  3. 310\frac{3}{10} (correct answer)
  4. 25\frac{2}{5}
Explanation: When you encounter probability questions involving selecting letters from words, you need to identify how many favorable outcomes exist compared to the total number of possible outcomes. To find the probability that a randomly selected letter from STATISTICS is T, first count the total letters in the word: S-T-A-T-I-S-T-I-C-S gives us 10 letters total. Next, count how many times the letter T appears: looking through the word, T appears in positions 2, 4, and 7, so there are 3 T's. The probability equals the number of favorable outcomes (T's) divided by the total number of possible outcomes (all letters): 310\frac{3}{10}. This matches answer choice C. Now let's see why the other answers are incorrect. Answer A (14\frac{1}{4}) would mean T appears once out of every 4 letters, but we have 3 T's in 10 letters, not 1 in 4. Answer B (13\frac{1}{3}) suggests T makes up one-third of the letters, which would require about 3.3 T's out of 10 letters—close but not exact. Answer D (25\frac{2}{5}) equals 410\frac{4}{10}, which would mean 4 T's exist, but we only counted 3. The key strategy for letter probability problems is to carefully count each occurrence of your target letter—don't assume each letter appears only once. Many words contain repeated letters, and missing duplicates is a common error that leads to wrong answers on the TACHS.