ISEE Lower Level Quantitative Reasoning Quiz: Calculating Probability

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 Quantitative Reasoning.

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.

Question 1

1. (\frac{4}{15})
2. (\frac{11}{30}) (correct answer)
3. (\frac{11}{19})
4. (\frac{19}{30})

Explanation: Let C be the number of cats and D be the number of dogs. We are given that (D = C + 8) and (D + C = 30). We can substitute the first equation into the second: ((C + 8) + C = 30). This simplifies to (2C + 8 = 30). Subtract 8 from both sides: (2C = 22). Divide by 2: (C = 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: (\frac{11}{30}).

Question 2

1. (\frac{1}{3})
2. (\frac{1}{2}) (correct answer)
3. (\frac{2}{3})
4. (\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 (\frac{3}{6}), which simplifies to (\frac{1}{2}).

Question 3

1. (\frac{1}{5})
2. (\frac{13}{50})
3. (\frac{4}{5}) (correct answer)
4. (\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 = 40) crayons are not blue. The probability is (\frac{40}{50}), which simplifies to (\frac{4}{5}). Method 2: Find the probability of picking a blue crayon, which is (\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 - \frac{1}{5} = \frac{4}{5}).

Question 4

1. (\frac{1}{2})
2. (\frac{7}{15}) (correct answer)
3. (\frac{8}{15})
4. (\frac{7}{8})

Explanation: First, find the total number of slips for weekend days (Saturday and Sunday). This is (10 + 6 = 16). Next, find the number of slips for weekdays by subtracting the weekend slips from the total: (30 - 16 = 14). The probability of choosing a weekday is the number of weekday slips divided by the total number of slips, which is (\frac{14}{30}). This fraction can be simplified by dividing both the numerator and the denominator by 2: (\frac{14 \div 2}{30 \div 2} = \frac{7}{15}).

Question 5

1. (\frac{1}{3})
2. (\frac{1}{4})
3. (\frac{5}{12}) (correct answer)
4. (\frac{7}{12})

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

Question 6

1. (\frac{1}{3})
2. (\frac{5}{12}) (correct answer)
3. (\frac{7}{12})
4. (\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 = 14). The number of green marbles is the total minus the red and blue marbles: (24 - 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 (\frac{10}{24}). This fraction must be simplified. Both 10 and 24 are divisible by 2, so (\frac{10}{24} = \frac{5}{12}).

Question 7

1. (\frac{5}{12})
2. (\frac{12}{25})
3. (\frac{13}{25}) (correct answer)
4. (\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 = 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: (\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 8

1. (\frac{1}{3}) (correct answer)
2. (\frac{1}{2})
3. (\frac{2}{3})
4. (\frac{12}{24})

Explanation: Let Y be the number of yellow gumballs and R be the number of red gumballs. We know that (R = 2Y) and (R + Y = 36). Substitute the first equation into the second: (2Y + Y = 36), which simplifies to (3Y = 36). Dividing by 3, we find (Y = 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: (\frac{12}{36}). This fraction simplifies to (\frac{1}{3}).

Question 9

1. (\frac{1}{4})
2. (\frac{1}{10})
3. (\frac{7}{20}) (correct answer)
4. (\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 = 7). The probability is the number of favorable outcomes divided by the total number of outcomes, which is (\frac{7}{20}).

Question 10

1. (\frac{1}{4})
2. (\frac{2}{20})
3. (\frac{7}{20}) (correct answer)
4. (\frac{7}{5})

Explanation: First, find the total number of sections on the spinner. We are told the probability of landing on red is (\frac{1}{4}) and that there are 5 red sections. This means that 5 is (\frac{1}{4}) of the total number of sections. So, the total number of sections is (5 \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 = 7). The probability of landing on blue is the number of blue sections divided by the total number of sections, which is (\frac{7}{20}).

Question 11

1. (\frac{2}{3}) (correct answer)
2. (\frac{7}{12})
3. (\frac{1}{3})
4. (\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 $\frac{8}{12} = \frac{2}{3}$. Choice A ($\frac{2}{3}$) is correct. Choice B ($\frac{7}{12}$) likely comes from miscounting and getting 7 months instead of 8. Choice C ($\frac{1}{3}$) represents the probability of months without 'R'. Choice D ($\frac{3}{4}$) 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 12

1. (\frac{2}{25})
2. (\frac{1}{10}) (correct answer)
3. (\frac{3}{25})
4. (\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: (\frac{5}{50}). This fraction simplifies to (\frac{1}{10}).

Question 13

1. (\frac{1}{7})
2. (\frac{1}{5})
3. (\frac{1}{3})
4. (\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 $\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{4}{21}$, which is answer choice D. Now for the wrong answers: Choice A ($\frac{1}{7}$) equals $\frac{3}{21}$, suggesting someone found only 3 numbers, possibly missing one of the multiples of 6. Choice B ($\frac{1}{5}$) equals $\frac{4.2}{21}$, which doesn't correspond to any whole number of tiles. Choice C ($\frac{1}{3}$) equals $\frac{7}{21}$, 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 14

1. (\frac{1}{8})
2. (\frac{1}{4})
3. (\frac{3}{4})
4. (\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 $\frac{3}{8}$. Looking at the wrong answers: Choice A ($\frac{1}{8}$) would be correct if you were looking for the probability of landing on exactly one specific number, like just 6. Choice B ($\frac{1}{4}$) 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 ($\frac{3}{4}$) 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 15

1. (\frac{1}{2})
2. (\frac{3}{5})
3. (\frac{2}{5})
4. (\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 $45 - 15 = 30$ tickets. Team C sold twice as many as Team B, so Team C sold $2 \times 30 = 60$ tickets. The total tickets sold by all three teams is $45 + 30 + 60 = 135$ 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: $\frac{45}{135} = \frac{1}{3}$. This confirms answer choice D. Looking at the wrong answers: Choice A ($\frac{1}{2}$) would mean Team A sold half of all tickets, but 45 is not half of 135. Choice B ($\frac{3}{5}$) represents $\frac{81}{135}$, which is much larger than Team A's actual contribution. Choice C ($\frac{2}{5}$) equals $\frac{54}{135}$, 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 16

1. (\frac{5}{12})
2. (\frac{1}{2}) (correct answer)
3. (\frac{7}{12})
4. (\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: (\frac{6}{12}). This fraction simplifies to (\frac{1}{2}).

Question 17

1. (\frac{2}{5})
2. (\frac{22}{45}) (correct answer)
3. (\frac{3}{5})
4. (\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 = 22) boys in total. Next, find the total number of students in the combined group. Ms. Anya's class has (10 + 15 = 25) students. Mr. Ben's class has (12 + 8 = 20) students. The total number of students is (25 + 20 = 45). The probability of choosing a boy is the total number of boys divided by the total number of students: (\frac{22}{45}).

Question 18

1. (\frac{1}{5}) (correct answer)
2. (\frac{8}{25})
3. (\frac{5}{13})
4. (\frac{1}{2})

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

Question 19

1. (\frac{7}{18})
2. (\frac{8}{18})
3. (\frac{7}{17}) (correct answer)
4. (\frac{8}{17})

Explanation: Initially, there are (6 + 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 = 7). The total number of socks remaining is (18 - 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 (\frac{7}{17}).

Question 20

1. (\frac{1}{3})
2. (\frac{2}{3})
3. (\frac{4}{9})
4. (\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 = 14). Next, find the total number of coins in the pocket: (6 + 8 + 4 = 18). The probability is the number of favorable outcomes divided by the total number of outcomes: (\frac{14}{18}). This fraction simplifies by dividing both numerator and denominator by 2: (\frac{14 \div 2}{18 \div 2} = \frac{7}{9}).