ISEE Lower Level Quiz: Calculating Probability
20 questions · exam conditions
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Calculating ProbabilityQuestion 1 of 20

A box contains 30 slips of paper, each with a day of the week written on it. There are 10 slips for Saturday and 6 for Sunday. The remaining slips are for weekdays. If a slip is chosen at random, what is the probability it shows a weekday?

12\frac{1}{2}
715\frac{7}{15}
815\frac{8}{15}
78\frac{7}{8}
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ISEE Lower Level Quiz

ISEE Lower Level Quiz: Calculating Probability

Practice Calculating Probability in ISEE Lower Level 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 Calculating Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Lower Level.

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 box contains 30 slips of paper, each with a day of the week written on it. There are 10 slips for Saturday and 6 for Sunday. The remaining slips are for weekdays. If a slip is chosen at random, what is the probability it shows a weekday?

  1. 12\frac{1}{2}
  2. 715\frac{7}{15} (correct answer)
  3. 815\frac{8}{15}
  4. 78\frac{7}{8}
Explanation: First, find the total number of slips for weekend days (Saturday and Sunday). This is 10+6=1610 + 6 = 16. Next, find the number of slips for weekdays by subtracting the weekend slips from the total: 30−16=1430 - 16 = 14. The probability of choosing a weekday is the number of weekday slips divided by the total number of slips, which is 1430\frac{14}{30}. This fraction can be simplified by dividing both the numerator and the denominator by 2: 14÷230÷2=715\frac{14 \div 2}{30 \div 2} = \frac{7}{15}.

Question 2

A cookie jar contains 36 cookies. One-third of the cookies are chocolate chip, one-fourth are oatmeal, and the rest are peanut butter. If a cookie is chosen at random, what is the probability it is a peanut butter cookie?

  1. 13\frac{1}{3}
  2. 14\frac{1}{4}
  3. 512\frac{5}{12} (correct answer)
  4. 712\frac{7}{12}
Explanation: First, calculate the number of chocolate chip cookies: 13×36=12\frac{1}{3} \times 36 = 12. Next, calculate the number of oatmeal cookies: 14×36=9\frac{1}{4} \times 36 = 9. The total number of chocolate chip and oatmeal cookies is 12+9=2112 + 9 = 21. To find the number of peanut butter cookies, subtract this from the total: 36−21=1536 - 21 = 15. The probability of choosing a peanut butter cookie is the number of peanut butter cookies divided by the total number of cookies: 1536\frac{15}{36}. To simplify, divide the numerator and denominator by their greatest common factor, which is 3: 15÷336÷3=512\frac{15 \div 3}{36 \div 3} = \frac{5}{12}.

Question 3

A standard six-sided die is rolled once. What is the probability that the number rolled is an even number that is also a factor of 12?

  1. 13\frac{1}{3}
  2. 12\frac{1}{2} (correct answer)
  3. 23\frac{2}{3}
  4. 56\frac{5}{6}
Explanation: The possible outcomes when rolling a six-sided die are {1, 2, 3, 4, 5, 6}. We need to find the outcomes that satisfy two conditions: being an even number and being a factor of 12. The even numbers are {2, 4, 6}. Now we check which of these are also factors of 12. 12 is divisible by 2, 4, and 6. So, all three numbers {2, 4, 6} satisfy both conditions. There are 3 favorable outcomes. The total number of outcomes is 6. The probability is 36\frac{3}{6}, which simplifies to 12\frac{1}{2}.

Question 4

A number is chosen at random from the whole numbers between 1 and 50, inclusive. What is the probability that the number is a two-digit number and the sum of its digits is 5?

  1. 225\frac{2}{25}
  2. 110\frac{1}{10} (correct answer)
  3. 325\frac{3}{25}
  4. 150\frac{1}{50}
Explanation: The total number of possible outcomes is 50, as the numbers are from 1 to 50 inclusive. We need to find the numbers that meet two conditions: they must be two-digit numbers, and the sum of their digits must be 5. Let's list them: 14 (1+4=5), 23 (2+3=5), 32 (3+2=5), 41 (4+1=5), and 50 (5+0=5). There are 5 such numbers. The probability is the number of favorable outcomes divided by the total number of outcomes: 550\frac{5}{50}. This fraction simplifies to 110\frac{1}{10}.

Question 5

A bag contains 24 marbles in total. There are 8 red marbles and 6 blue marbles. The rest of the marbles are green. If a marble is chosen at random from the bag, what is the probability that it is green?

  1. 13\frac{1}{3}
  2. 512\frac{5}{12} (correct answer)
  3. 712\frac{7}{12}
  4. 1024\frac{10}{24}
Explanation: First, find the number of green marbles. There are 24 marbles in total, with 8 red and 6 blue. The number of red and blue marbles together is 8+6=148 + 6 = 14. The number of green marbles is the total minus the red and blue marbles: 24−14=1024 - 14 = 10 green marbles. The probability of choosing a green marble is the number of green marbles divided by the total number of marbles, which is 1024\frac{10}{24}. This fraction must be simplified. Both 10 and 24 are divisible by 2, so 1024=512\frac{10}{24} = \frac{5}{12}.

Question 6

In a class of 25 students, 12 are boys and the rest are girls. Of the boys, 5 have brown hair. Of the girls, 8 have brown hair. If the teacher calls on one student at random, what is the probability that the student has brown hair?

  1. 512\frac{5}{12}
  2. 1225\frac{12}{25}
  3. 1325\frac{13}{25} (correct answer)
  4. 813\frac{8}{13}
Explanation: The question asks for the probability that a randomly selected student has brown hair. We need to find the total number of students with brown hair and divide it by the total number of students. There are 5 boys with brown hair and 8 girls with brown hair. So, the total number of students with brown hair is 5+8=135 + 8 = 13. The total number of students in the class is 25. The probability is the number of students with brown hair divided by the total number of students: 1325\frac{13}{25}. The information that there are 12 boys is extra information not needed to find the total number of students with brown hair.

Question 7

A gumball machine contains only red and yellow gumballs. There are twice as many red gumballs as yellow gumballs. If the machine has 36 gumballs in total, what is the probability of getting a yellow gumball?

  1. 13\frac{1}{3} (correct answer)
  2. 12\frac{1}{2}
  3. 23\frac{2}{3}
  4. 1224\frac{12}{24}
Explanation: Let Y be the number of yellow gumballs and R be the number of red gumballs. We know that R=2YR = 2Y and R+Y=36R + Y = 36. Substitute the first equation into the second: 2Y+Y=362Y + Y = 36, which simplifies to 3Y=363Y = 36. Dividing by 3, we find Y=12Y = 12. So there are 12 yellow gumballs. The probability of getting a yellow gumball is the number of yellow gumballs divided by the total number of gumballs: 1236\frac{12}{36}. This fraction simplifies to 13\frac{1}{3}.

Question 8

A set of cards is numbered from 1 to 20, with one number per card. If you draw one card at random, what is the probability that the number on the card is a multiple of 4 or a multiple of 7?

  1. 14\frac{1}{4}
  2. 110\frac{1}{10}
  3. 720\frac{7}{20} (correct answer)
  4. 120\frac{1}{20}
Explanation: First, identify the total number of possible outcomes, which is 20 since there are 20 cards. Next, find the number of favorable outcomes. The multiples of 4 between 1 and 20 are 4, 8, 12, 16, and 20. There are 5 such multiples. The multiples of 7 between 1 and 20 are 7 and 14. There are 2 such multiples. Since there are no numbers that are multiples of both 4 and 7 in this range, we can add the number of favorable outcomes together: 5+2=75 + 2 = 7. The probability is the number of favorable outcomes divided by the total number of outcomes, which is 720\frac{7}{20}.

Question 9

A spinner is divided into equal sections of red, blue, and green. The probability of the spinner landing on red is 14\frac{1}{4}. There are 5 red sections. If there are 2 more blue sections than red sections, what is the probability of landing on blue?

  1. 14\frac{1}{4}
  2. 220\frac{2}{20}
  3. 720\frac{7}{20} (correct answer)
  4. 75\frac{7}{5}
Explanation: First, find the total number of sections on the spinner. We are told the probability of landing on red is 14\frac{1}{4} and that there are 5 red sections. This means that 5 is 14\frac{1}{4} of the total number of sections. So, the total number of sections is 5×4=205 \times 4 = 20. Next, find the number of blue sections. There are 2 more blue sections than red sections, so the number of blue sections is 5+2=75 + 2 = 7. The probability of landing on blue is the number of blue sections divided by the total number of sections, which is 720\frac{7}{20}.

Question 10

The name of one of the 12 months is picked at random. What is the probability that the name of the month contains the letter 'R'?

  1. 23\frac{2}{3} (correct answer)
  2. 712\frac{7}{12}
  3. 13\frac{1}{3}
  4. 34\frac{3}{4}
Explanation: When you encounter probability questions, remember that probability equals the number of favorable outcomes divided by the total number of possible outcomes. Let's identify which months contain the letter 'R'. Going through all 12 months systematically: January (no R), February (has R), March (has R), April (has R), May (no R), June (no R), July (no R), August (no R), September (has R), October (has R), November (has R), December (has R). Count them up: February, March, April, September, October, November, December, and August all contain 'R'. Wait—let me recheck August carefully: A-U-G-U-S-T. No 'R' there. So we have 8 months with 'R': February, March, April, September, October, November, December, and... actually, let me be more careful. The months with 'R' are: February, March, April, September, October, November, December—that's 7 months, plus January contains no 'R', May has no 'R', June has no 'R', July has no 'R', August has no 'R'. So 8 months contain 'R'. Actually, let me recount systematically: February, March, April, September, October, November, December, January (no), May (no), June (no), July (no), August (no). That's 8 months with 'R' out of 12 total. So the probability is 812=23\frac{8}{12} = \frac{2}{3}. Choice A (23\frac{2}{3}) is correct. Choice B (712\frac{7}{12}) likely comes from miscounting and getting 7 months instead of 8. Choice C (13\frac{1}{3}) represents the probability of months without 'R'. Choice D (34\frac{3}{4}) doesn't correspond to this count. Strategy tip: In probability problems, always list out all possibilities systematically to avoid counting errors, especially with familiar items like months where you might rush.

Question 11

For a school fundraiser, students sell raffle tickets. Team A sold 45 tickets. Team B sold 15 fewer tickets than Team A. Team C sold twice as many tickets as Team B. If one ticket is drawn from all the tickets sold by the three teams, what is the probability it was sold by Team A?

  1. 12\frac{1}{2}
  2. 35\frac{3}{5}
  3. 25\frac{2}{5}
  4. 13\frac{1}{3} (correct answer)
Explanation: When you encounter probability questions involving multiple groups, you need to find what fraction one group represents of the total. Start by calculating how many tickets each team sold, then find the total. Team A sold 45 tickets. Team B sold 15 fewer than Team A, so Team B sold 45−15=3045 - 15 = 30 tickets. Team C sold twice as many as Team B, so Team C sold 2×30=602 \times 30 = 60 tickets. The total tickets sold by all three teams is 45+30+60=13545 + 30 + 60 = 135 tickets. The probability that a randomly drawn ticket was sold by Team A is the number of tickets Team A sold divided by the total: 45135=13\frac{45}{135} = \frac{1}{3}. This confirms answer choice D. Looking at the wrong answers: Choice A (12\frac{1}{2}) would mean Team A sold half of all tickets, but 45 is not half of 135. Choice B (35\frac{3}{5}) represents 81135\frac{81}{135}, which is much larger than Team A's actual contribution. Choice C (25\frac{2}{5}) equals 54135\frac{54}{135}, which might tempt you if you miscalculated one of the teams' totals. Remember that probability questions often involve multi-step calculations where you must find individual quantities before determining the total. Always double-check your arithmetic for each team's contribution, and verify that your final fraction represents the correct group as a portion of the whole. Don't rush through the setup—careful organization of the given information prevents calculation errors.

Question 12

Ms. Anya's class has 10 boys and 15 girls. Mr. Ben's class has 12 boys and 8 girls. The two classes are combined for a school event. If one student is chosen at random from the combined group, what is the probability that the student is a boy?

  1. 25\frac{2}{5}
  2. 2245\frac{22}{45} (correct answer)
  3. 35\frac{3}{5}
  4. 2223\frac{22}{23}
Explanation: First, find the total number of boys in the combined group. Ms. Anya has 10 boys and Mr. Ben has 12 boys, so there are 10+12=2210 + 12 = 22 boys in total. Next, find the total number of students in the combined group. Ms. Anya's class has 10+15=2510 + 15 = 25 students. Mr. Ben's class has 12+8=2012 + 8 = 20 students. The total number of students is 25+20=4525 + 20 = 45. The probability of choosing a boy is the total number of boys divided by the total number of students: 2245\frac{22}{45}.

Question 13

A box contains 25 toy blocks. There are red cubes, blue cubes, red spheres, and blue spheres. There are 12 cubes in total, and 15 red blocks in total. If there are 7 red cubes, what is the probability that a randomly chosen block is a blue sphere?

  1. 15\frac{1}{5} (correct answer)
  2. 825\frac{8}{25}
  3. 513\frac{5}{13}
  4. 12\frac{1}{2}
Explanation: This problem requires several steps. The total number of blocks is 25. Total cubes = 12, so total spheres = 25−12=1325 - 12 = 13. Total red blocks = 15, so total blue blocks = 25−15=1025 - 15 = 10. We are given there are 7 red cubes. We can find the number of blue cubes: Total cubes - Red cubes = 12−7=512 - 7 = 5 blue cubes. Now we can find the number of blue spheres: Total blue blocks - Blue cubes = 10−5=510 - 5 = 5 blue spheres. The probability of choosing a blue sphere is the number of blue spheres divided by the total number of blocks: 525\frac{5}{25}. This simplifies to 15\frac{1}{5}.

Question 14

A drawer contains 6 blue socks, 8 black socks, and 4 white socks. Liam takes out one black sock and puts it on. He then reaches into the drawer again without looking. What is the probability he will now pull out another black sock?

  1. 718\frac{7}{18}
  2. 818\frac{8}{18}
  3. 717\frac{7}{17} (correct answer)
  4. 817\frac{8}{17}
Explanation: Initially, there are 6+8+4=186 + 8 + 4 = 18 socks in the drawer. After Liam takes out one black sock, both the number of black socks and the total number of socks decrease by one. The number of black socks remaining is 8−1=78 - 1 = 7. The total number of socks remaining is 18−1=1718 - 1 = 17. The probability of pulling out another black sock is the new number of black socks divided by the new total number of socks, which is 717\frac{7}{17}.

Question 15

A pet shelter has only cats and dogs. There are 8 more dogs than cats. If there are 30 animals in total, what is the probability that a randomly chosen animal is a cat?

  1. 415\frac{4}{15}
  2. 1130\frac{11}{30} (correct answer)
  3. 1119\frac{11}{19}
  4. 1930\frac{19}{30}
Explanation: Let C be the number of cats and D be the number of dogs. We are given that D=C+8D = C + 8 and D+C=30D + C = 30. We can substitute the first equation into the second: (C+8)+C=30(C + 8) + C = 30. This simplifies to 2C+8=302C + 8 = 30. Subtract 8 from both sides: 2C=222C = 22. Divide by 2: C=11C = 11. There are 11 cats. The total number of animals is 30. The probability of choosing a cat is the number of cats divided by the total number of animals: 1130\frac{11}{30}.

Question 16

A box contains 50 crayons. There are 15 red, 10 blue, 12 green, and the rest are yellow. What is the probability of randomly picking a crayon that is NOT blue?

  1. 15\frac{1}{5}
  2. 1350\frac{13}{50}
  3. 45\frac{4}{5} (correct answer)
  4. 110\frac{1}{10}
Explanation: There are two ways to solve this. Method 1: Find the number of crayons that are not blue. The total is 50, and 10 are blue, so 50−10=4050 - 10 = 40 crayons are not blue. The probability is 4050\frac{40}{50}, which simplifies to 45\frac{4}{5}. Method 2: Find the probability of picking a blue crayon, which is 1050=15\frac{10}{50} = \frac{1}{5}. The probability of the event NOT happening is 1 minus the probability of the event happening. So, the probability of not picking a blue crayon is 1−15=451 - \frac{1}{5} = \frac{4}{5}.

Question 17

A bag contains tiles numbered from 10 to 30, inclusive. If a tile is drawn at random, what is the probability that the number on the tile is divisible by both 2 and 3?

  1. 17\frac{1}{7}
  2. 15\frac{1}{5}
  3. 13\frac{1}{3}
  4. 421\frac{4}{21} (correct answer)
Explanation: When you see a probability question asking for numbers divisible by multiple conditions, you need to find numbers that satisfy all conditions simultaneously. A number divisible by both 2 and 3 must be divisible by 6 (since 2 and 3 are coprime, their least common multiple is 2 × 3 = 6). First, identify all tiles in the bag: numbers 10 through 30, inclusive. That's 30 - 10 + 1 = 21 total tiles. Next, find which numbers from 10 to 30 are divisible by 6. Divide each endpoint by 6: 10 ÷ 6 = 1.67 and 30 ÷ 6 = 5. So you need multiples of 6 from the 2nd multiple (12) through the 5th multiple (30). These are: 12, 18, 24, and 30. That's 4 favorable outcomes. The probability is favorable outcomestotal outcomes=421\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{4}{21}, which is answer choice D. Now for the wrong answers: Choice A (17\frac{1}{7}) equals 321\frac{3}{21}, suggesting someone found only 3 numbers, possibly missing one of the multiples of 6. Choice B (15\frac{1}{5}) equals 4.221\frac{4.2}{21}, which doesn't correspond to any whole number of tiles. Choice C (13\frac{1}{3}) equals 721\frac{7}{21}, which might result from incorrectly counting numbers divisible by 2 OR 3, rather than both. Strategy tip: When a problem asks for divisibility by multiple numbers, find their least common multiple first. This converts a complex condition into a simpler "divisible by one number" problem.

Question 18

A spinner is divided into 8 equal sections, numbered 1 through 8. What is the probability of the spinner landing on a number that is at least 6?

  1. 18\frac{1}{8}
  2. 14\frac{1}{4}
  3. 34\frac{3}{4}
  4. 38\frac{3}{8} (correct answer)
Explanation: When you encounter probability questions involving spinners or dice, you're looking for the ratio of favorable outcomes to total possible outcomes. The key is carefully identifying which outcomes satisfy the given condition. This spinner has 8 equal sections numbered 1 through 8, so there are 8 total possible outcomes. You need to find the probability of landing on "a number that is at least 6." The phrase "at least 6" means 6 or greater, so the favorable outcomes are 6, 7, and 8. That's 3 favorable outcomes out of 8 total possible outcomes, giving you a probability of 38\frac{3}{8}. Looking at the wrong answers: Choice A (18\frac{1}{8}) would be correct if you were looking for the probability of landing on exactly one specific number, like just 6. Choice B (14\frac{1}{4}) represents 2 out of 8 outcomes - perhaps you miscounted and only considered 6 and 7, forgetting that 8 also qualifies as "at least 6." Choice C (34\frac{3}{4}) represents 6 out of 8 outcomes, which would happen if you mistakenly found numbers that are NOT at least 6 (1, 2, 3, 4, 5) and confused favorable with unfavorable outcomes. Remember that "at least" means "greater than or equal to," so always include the boundary number itself. When solving probability problems, write out all favorable outcomes to avoid miscounting, and double-check that your fraction uses the correct total number of possible outcomes in the denominator.

Question 19

Each letter of the word QUANTITATIVE is written on a separate tile and placed in a bag. If one tile is drawn at random, what is the probability that it is a vowel?

  1. 512\frac{5}{12}
  2. 12\frac{1}{2} (correct answer)
  3. 712\frac{7}{12}
  4. 611\frac{6}{11}
Explanation: First, count the total number of letters in the word QUANTITATIVE. There are 12 letters. Next, count the number of vowels (A, E, I, O, U). The vowels in QUANTITATIVE are U, A, I, A, I, E. There are 6 vowels. The probability of drawing a vowel is the number of vowels divided by the total number of letters: 612\frac{6}{12}. This fraction simplifies to 12\frac{1}{2}.

Question 20

Leo's pocket contains 6 quarters, 8 dimes, and 4 nickels. He does not have any pennies. If he pulls one coin out without looking, what is the probability that the coin is worth more than 5 cents?

  1. 13\frac{1}{3}
  2. 23\frac{2}{3}
  3. 49\frac{4}{9}
  4. 79\frac{7}{9} (correct answer)
Explanation: First, identify which coins are worth more than 5 cents. Quarters (25 cents) and dimes (10 cents) are worth more than 5 cents. Nickels are worth exactly 5 cents, so they are not included. The number of favorable outcomes is the sum of quarters and dimes: 6+8=146 + 8 = 14. Next, find the total number of coins in the pocket: 6+8+4=186 + 8 + 4 = 18. The probability is the number of favorable outcomes divided by the total number of outcomes: 1418\frac{14}{18}. This fraction simplifies by dividing both numerator and denominator by 2: 14÷218÷2=79\frac{14 \div 2}{18 \div 2} = \frac{7}{9}.