TACHS Quiz: Computing Probability
4 questions · exam conditions
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Computing ProbabilityQuestion 1 of 4

If a month of the year is selected at random, what is the probability that the month has exactly 31 days?

712\dfrac{7}{12}
13\dfrac{1}{3}
512\dfrac{5}{12}
14\dfrac{1}{4}
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TACHS Quiz

TACHS Quiz: Computing Probability

Practice Computing 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 Computing 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

If a month of the year is selected at random, what is the probability that the month has exactly 31 days?

  1. 712\dfrac{7}{12} (correct answer)
  2. 13\dfrac{1}{3}
  3. 512\dfrac{5}{12}
  4. 14\dfrac{1}{4}
Explanation: When you encounter a probability question, you need to identify the total number of possible outcomes and the number of favorable outcomes. Here, you're selecting from 12 months, so the total possible outcomes is 12. To find the favorable outcomes, you need to count which months have exactly 31 days. Let's go through the calendar: January (31), February (28/29), March (31), April (30), May (31), June (30), July (31), August (31), September (30), October (31), November (30), and December (31). The months with exactly 31 days are January, March, May, July, August, October, and December—that's 7 months. The probability is therefore favorable outcomestotal outcomes=712\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{7}{12}, which is choice A. Looking at the wrong answers: Choice B (13\frac{1}{3}) equals 412\frac{4}{12}, which might result from miscounting and thinking only 4 months have 31 days. Choice C (512\frac{5}{12}) suggests someone counted only 5 months with 31 days, perhaps forgetting a couple of the summer months or mixing up which months have 30 versus 31 days. Choice D (14\frac{1}{4}) equals 312\frac{3}{12}, indicating a significant undercount of 31-day months. For probability problems involving calendar dates, always double-check your counting by going through each month systematically. A helpful memory trick: "30 days has September, April, June, and November"—all other months except February have 31 days.

Question 2

A fair number cube numbered 1 through 6 is rolled once. What is the probability that the number shown is not a multiple of 2 or 3?

  1. 13\dfrac{1}{3} (correct answer)
  2. 12\dfrac{1}{2}
  3. 23\dfrac{2}{3}
  4. 56\dfrac{5}{6}
Explanation: When you encounter probability questions asking for events that are "not" something, you're dealing with complement problems. The key is to first identify exactly which outcomes satisfy the condition, then calculate the probability. Let's identify which numbers from 1 to 6 are not multiples of 2 or 3. First, find the multiples of 2: {2, 4, 6}. Then the multiples of 3: {3, 6}. Combined, the multiples of 2 or 3 are: {2, 3, 4, 6}. This leaves only {1, 5} as numbers that are not multiples of 2 or 3. Since there are 2 favorable outcomes out of 6 possible outcomes, the probability is 26=13\frac{2}{6} = \frac{1}{3}, making A correct. Looking at the wrong answers: B) 12\frac{1}{2} represents 3 out of 6 outcomes, which might come from miscounting or confusing this with a simpler "not multiple of 2" or "not multiple of 3" question. C) 23\frac{2}{3} represents 4 out of 6 outcomes—this is actually the probability of the opposite event (rolling a multiple of 2 or 3), a classic complement confusion. D) 56\frac{5}{6} represents 5 out of 6 outcomes, suggesting the student may have incorrectly excluded only one number instead of properly identifying all multiples of 2 or 3. Remember: when a probability question uses "not," always identify what you're excluding first, then count what remains. Double-check by verifying that your favorable outcomes plus excluded outcomes equal the total possible outcomes.

Question 3

A spinner is divided into 5 equal sections labeled 1, 1, 2, 3, and 4. What is the probability that a single spin lands on an odd number?

  1. 35\dfrac{3}{5} (correct answer)
  2. 45\dfrac{4}{5}
  3. 12\dfrac{1}{2}
  4. 25\dfrac{2}{5}
Explanation: When you encounter probability questions involving spinners or similar scenarios, focus on identifying the favorable outcomes versus the total possible outcomes. Let's examine this spinner systematically. You have 5 equal sections labeled 1, 1, 2, 3, and 4. To find the probability of landing on an odd number, first identify which numbers are odd: 1 and 3 are odd, while 2 and 4 are even. Now count the favorable outcomes. Since there are two sections labeled "1" and one section labeled "3," you have 3 sections total that show odd numbers. The total number of sections is 5. Therefore, the probability is 35\frac{3}{5}, which is choice A. Let's examine why the other options are incorrect. Choice B (45\frac{4}{5}) might tempt you if you mistakenly counted 4 different odd outcomes, perhaps by thinking there are 4 distinct numbers (1, 2, 3, 4) and most are odd. Choice C (12\frac{1}{2}) could result from incorrectly reasoning that there are 2 types of numbers (odd and even) so the probability should be equal. Choice D (25\frac{2}{5}) might occur if you only counted the number of different odd values (1 and 3) rather than the actual sections containing odd numbers. Remember: in probability problems, always count the actual outcomes, not just the distinct values. If a spinner has repeated labels, each section counts as a separate outcome, even if the numbers are identical.

Question 4

A letter is chosen at random from the word COMPUTER\text{COMPUTER}. What is the probability that the letter selected is a vowel?

  1. 38\dfrac{3}{8} (correct answer)
  2. 27\dfrac{2}{7}
  3. 13\dfrac{1}{3}
  4. 58\dfrac{5}{8}
Explanation: When you encounter probability questions involving selecting items from a collection, you need to identify the favorable outcomes and divide by the total possible outcomes. Let's examine the word COMPUTER letter by letter: C-O-M-P-U-T-E-R. This gives us 8 total letters. Now identify the vowels (A, E, I, O, U): we have O, U, and E. That's 3 vowels out of 8 total letters. The probability of selecting a vowel is: number of vowelstotal number of letters=38\frac{\text{number of vowels}}{\text{total number of letters}} = \frac{3}{8} This confirms that choice A is correct. Choice B (27\frac{2}{7}) represents a double error: miscounting both the vowels (as 2 instead of 3) and the total letters (as 7 instead of 8). Perhaps someone missed one of the vowels and also miscounted the word length. Choice C (13\frac{1}{3}) likely comes from incorrectly thinking there's only 1 vowel, then somehow arriving at 3 total letters, or confusing this with the probability of NOT selecting a vowel from a different scenario. Choice D (58\frac{5}{8}) represents counting consonants instead of vowels. COMPUTER has 5 consonants (C, M, P, T, R), so this would be the probability of selecting a consonant, not a vowel. For probability questions, always clearly identify what you're looking for first, then systematically count both the favorable outcomes and total possible outcomes. Double-check by ensuring your numerator and denominator add up correctly when considering complementary events.