ISEE Middle Level Quiz: Basic Probability

20 questions · exam conditions

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Basic ProbabilityQuestion 1 of 20

A bag contains only red, blue, and green marbles. The probability of selecting a red marble is $\frac{1}{4}$ , and the probability of selecting a blue marble is $\frac{1}{3}$ . If there are 10 green marbles in the bag, what is the total number of marbles in the bag?

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ISEE Middle Level Quiz

ISEE Middle Level Quiz: Basic Probability

Practice Basic Probability in ISEE Middle 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 Basic Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Middle 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

12
17
24 (correct answer)
30

Explanation: First, find the probability of selecting a green marble. The sum of the probabilities for all outcomes must be 1. The probability of selecting a red or blue marble is

\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}

. Therefore, the probability of selecting a green marble is

1 - \frac{7}{12} = \frac{5}{12}

. Let T be the total number of marbles. We know that the number of green marbles is 10, so

\frac{5}{12} \times T = 10

. To find T, we can multiply both sides by

\frac{12}{5}

T = 10 \times \frac{12}{5} = \frac{120}{5} = 24

. The total number of marbles is 24.

Question 2

A bag has 2 red, 6 blue, 2 green marbles; what is the probability of red?

1/2
1/5
2/12
2/10 (correct answer)

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of drawing a red marble from a bag with 2 red, 6 blue, and 2 green marbles based on the sample space of 10 marbles. Choice D is correct because it accurately calculates the probability as 2/10, using the 2 red marbles out of 10 total. Choice A is incorrect because it assumes half are red, leading to 1/2. This error often occurs when students overlook the actual counts. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 3

A spinner with 5 equal sections (labeled 1 to 5) is spun 50 times. The experimental results show it landed on section 2 a total of 12 times. What is the absolute difference between the theoretical probability and the experimental probability of landing on section 2?

$\frac{1}{25}$ (correct answer)
$\frac{1}{5}$
$\frac{6}{25}$
$\frac{11}{25}$

Explanation: The theoretical probability of landing on any one of the 5 equal sections is

\frac{1}{5}

. The experimental probability is based on the results of the experiment. It landed on section 2 a total of 12 times out of 50 spins, so the experimental probability is

\frac{12}{50}

. To find the difference, we calculate

|\frac{1}{5} - \frac{12}{50}|

. First, find a common denominator:

\frac{1}{5} = \frac{10}{50}

. The difference is

|\frac{10}{50} - \frac{12}{50}| = |-\frac{2}{50}| = \frac{2}{50}

, which simplifies to

\frac{1}{25}

Question 4

In a class of 30 students, 18 play soccer and 15 play basketball. Every student plays at least one of these two sports. If a student is chosen at random, what is the probability that the student plays only basketball?

$\frac{1}{10}$
$\frac{2}{5}$ (correct answer)
$\frac{1}{2}$
$\frac{4}{5}$

Explanation: Let S be the set of students who play soccer and B be the set of students who play basketball. We know |S| = 18, |B| = 15, and |S U B| = 30. The number of students who play both sports is |S ∩ B| = |S| + |B| - |S U B| = 18 + 15 - 30 = 3. The number of students who play only basketball is the number of students in B minus the number of students who play both: |B| - |S ∩ B| = 15 - 3 = 12. The total number of students is 30. The probability of choosing a student who plays only basketball is

\frac{12}{30} = \frac{2}{5}

Question 5

A fair six-sided die is rolled and a spinner with four equal sections (Red, Blue, Green, Yellow) is spun. What is the probability that the die shows a prime number and the spinner does not land on Green?

$\frac{1}{4}$
$\frac{3}{8}$ (correct answer)
$\frac{1}{2}$
$\frac{3}{4}$

Explanation: The prime numbers on a six-sided die are 2, 3, and 5. There are 3 favorable outcomes out of 6 total outcomes. So, the probability of rolling a prime number is

\frac{3}{6} = \frac{1}{2}

. The spinner has four equal sections. Not landing on Green means landing on Red, Blue, or Yellow. There are 3 favorable outcomes out of 4 total outcomes. The probability of not landing on Green is

\frac{3}{4}

. Since the two events are independent, we multiply their probabilities:

\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}

Question 6

Compare the quantities in Column A and Column B.

Column A: The probability of rolling a sum of 8 with two fair six-sided dice. Column B: The probability of drawing a face card (Jack, Queen, or King) from a standard 52-card deck.

The quantity in Column A is greater.
The quantity in Column B is greater. (correct answer)
The two quantities are equal.
The relationship cannot be determined from the information given.

Explanation: Column A: The total number of outcomes when rolling two dice is

6 \times 6 = 36

. The combinations that sum to 8 are (2,6), (3,5), (4,4), (5,3), and (6,2). There are 5 favorable outcomes. The probability is

\frac{5}{36}

. Column B: A standard deck has 52 cards. There are 3 face cards (Jack, Queen, King) in each of the 4 suits, so there are

3 \times 4 = 12

face cards in total. The probability is

\frac{12}{52} = \frac{3}{13}

. To compare

\frac{5}{36}

and

\frac{3}{13}

, we can find a common denominator or cross-multiply. Cross-multiplication gives

5 \times 13 = 65

for Column A and

3 \times 36 = 108

for Column B. Since

65 < 108

, the quantity in Column B is greater.

Question 7

A bag contains 4 red marbles and 6 blue marbles. A fair coin is tossed. If the coin lands on heads, a marble is drawn from the bag. If the coin lands on tails, no marble is drawn. What is the probability that a red marble is drawn?

$\frac{1}{5}$ (correct answer)
$\frac{2}{5}$
$\frac{1}{2}$
$\frac{9}{10}$

Explanation: This is a two-step probability problem. For a red marble to be drawn, two events must happen: the coin must land on heads, AND a red marble must be drawn. The probability of the coin landing on heads is

\frac{1}{2}

. If the coin is heads, a marble is drawn from the bag with 4 red and 6 blue marbles (10 total). The probability of drawing a red marble in this case is

\frac{4}{10} = \frac{2}{5}

. The overall probability of both events happening is the product of their individual probabilities:

P(\text{Heads and Red}) = P(\text{Heads}) \times P(\text{Red}|\text{Heads}) = \frac{1}{2} \times \frac{4}{10} = \frac{4}{20} = \frac{1}{5}

. If the coin is tails, no marble is drawn, so the probability of drawing a red marble is 0. The total probability is

\frac{1}{5} + 0 = \frac{1}{5}

Question 8

A spinner is divided into three sections: Red, Blue, and Green. The probability of landing on Red is $\frac{1}{2}$ , and the probability of landing on Blue is $\frac{1}{3}$ . The spinner is spun twice. What is the probability of landing on Red first and then Green second?

$\frac{1}{12}$ (correct answer)
$\frac{1}{9}$
$\frac{1}{6}$
$\frac{2}{3}$

Explanation: First, we need to find the probability of landing on Green. Since the probabilities of all outcomes must sum to 1, we have

P(\text{Green}) = 1 - P(\text{Red}) - P(\text{Blue}) = 1 - \frac{1}{2} - \frac{1}{3}

. Finding a common denominator, this is

\frac{6}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6}

. The two spins are independent events. To find the probability of landing on Red first AND Green second, we multiply their individual probabilities:

P(\text{Red then Green}) = P(\text{Red}) \times P(\text{Green}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

Question 9

A fair coin is tossed three times. What is the probability of getting at least one head?

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

Explanation: The total number of possible outcomes when tossing a coin three times is

2 \times 2 \times 2 = 8

. The outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The event 'at least one head' is the complement of the event 'no heads'. The only outcome with no heads is TTT. The probability of getting no heads is

\frac{1}{8}

. Therefore, the probability of getting at least one head is

1 - P(\text{no heads}) = 1 - \frac{1}{8} = \frac{7}{8}

Question 10

In a board game, you roll two dice; what is the probability of sum 7?

6/36
1/36
7/36
1/6 (correct answer)

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of rolling a sum of 7 with two dice based on the sample space of 36 possible outcomes. Choice D is correct because it accurately calculates the probability as 1/6, using the 6 favorable outcomes for sum 7 out of 36 total rolls. Choice B is incorrect because it counts only one specific pair, leading to 1/36. This error often occurs when students overlook multiple ways to achieve the sum. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 11

A 100-page book has a winning ticket placed on a random page. What is the probability that the winning page number is a multiple of 7 but not a multiple of 5?

$\frac{1}{10}$
$\frac{3}{25}$ (correct answer)
$\frac{7}{50}$
$\frac{1}{5}$

Explanation: First, find the number of multiples of 7 from 1 to 100. This is

\lfloor \frac{100}{7} \rfloor = 14

. Next, we need to exclude the numbers that are also multiples of 5. A number that is a multiple of both 7 and 5 is a multiple of their least common multiple, which is 35. The multiples of 35 from 1 to 100 are 35 and 70. There are 2 such numbers. The number of pages that are multiples of 7 but not 5 is

14 - 2 = 12

. The total number of pages is 100. So, the probability is

\frac{12}{100} = \frac{3}{25}

Question 12

A random number generator selects an integer from -5 to 5, inclusive. What is the probability that the selected number is positive and even?

$\frac{1}{5}$
$\frac{2}{11}$ (correct answer)
$\frac{3}{11}$
$\frac{5}{11}$

Explanation: The set of integers from -5 to 5, inclusive, is {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}. To find the total number of integers, we calculate

5 - (-5) + 1 = 11

. The numbers in this set that are positive and even are {2, 4}. There are 2 favorable outcomes. Therefore, the probability is

\frac{2}{11}

Question 13

A box contains 20 balls. The ratio of red balls to blue balls is 3:2. Four red balls and one blue ball are added to the box. What is the new probability of picking a blue ball at random?

$\frac{2}{5}$
$\frac{8}{25}$
$\frac{9}{25}$ (correct answer)
$\frac{9}{20}$

Explanation: Initially, there are 20 balls with a red to blue ratio of 3:2. This means there are

3+2=5

parts. The value of one part is

20 \div 5 = 4

. So, there are

3 \times 4 = 12

red balls and

2 \times 4 = 8

blue balls. Then, 4 red balls and 1 blue ball are added. The new number of red balls is

12 + 4 = 16

. The new number of blue balls is

8 + 1 = 9

. The new total number of balls is

20 + 4 + 1 = 25

. The new probability of picking a blue ball is

\frac{\text{new number of blue balls}}{\text{new total number of balls}} = \frac{9}{25}

Question 14

You draw 1 card from a 52-card deck; what is the probability of a heart?

1/13
1/2
1/4 (correct answer)
1/52

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of drawing a heart from a 52-card deck based on the sample space of 52 cards. Choice C is correct because it accurately calculates the probability as 1/4, using the 13 hearts out of 52 cards. Choice A is incorrect because it counts only one specific heart, leading to 1/13. This error often occurs when students overlook the total number of cards in a suit. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 15

You flip two coins for a warm-up; what is the probability of at least one head?

1/4
3/4 (correct answer)
1/2
4/4

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of getting at least one head when flipping two coins based on the sample space of 4 outcomes. Choice B is correct because it accurately calculates the probability as 3/4, using the 3 favorable outcomes out of 4 total flips. Choice A is incorrect because it calculates the probability of both tails, leading to 1/4. This error often occurs when students overlook complementary counting. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 16

You draw 1 card from a shuffled deck; what is the probability of a heart?

1/2
1/4 (correct answer)
1/52
1/13

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of drawing a heart from a shuffled deck based on the sample space of 52 cards. Choice B is correct because it accurately calculates the probability as 1/4, using the 13 hearts out of 52 cards. Choice A is incorrect because it assumes half the deck are hearts, leading to 1/2. This error often occurs when students overlook the four suits. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 17

A bag has 5 red, 3 blue, 2 green marbles; what is the probability of red?

1/2 (correct answer)
5/10
3/10
5/12

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of drawing a red marble from a bag with 5 red, 3 blue, and 2 green marbles based on the sample space of 10 marbles. Choice A is correct because it accurately calculates the probability as 1/2, using the 5 red marbles out of 10 total. Choice C is incorrect because it uses the blue marbles instead, leading to 3/10. This error often occurs when students overlook the color specified in the question. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 18

You flip two fair coins; what is the probability of at least one head?

1/4
1/2
3/4 (correct answer)
2/4

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of getting at least one head when flipping two fair coins based on the sample space of 4 outcomes. Choice C is correct because it accurately calculates the probability as 3/4, using the 3 favorable outcomes out of 4 total flips. Choice A is incorrect because it calculates the probability of both tails instead, leading to 1/4. This error often occurs when students overlook complementary events. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 19

In math club, you roll two fair dice; what is the probability of sum 7?

1/6 (correct answer)
5/36
6/36
1/12

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to determine the probability of rolling a sum of 7 with two fair dice based on the sample space of 36 possible outcomes. Choice A is correct because it accurately calculates the probability as 1/6, using the 6 favorable outcomes for sum 7 out of 36 total rolls. Choice D is incorrect because it halves the favorable outcomes, leading to 1/12. This error often occurs when students overlook dice distinguishability. To help students, encourage them to carefully list all possible outcomes and use clear diagrams or tables to visualize probabilities. Practice converting between fractions, decimals, and percentages, and emphasize checking calculations for accuracy.

Question 20

During recess, you flip two fair coins: {HH, HT, TH, TT} with 1/4 each. Is $P(\text{at least one head})$ greater than, less than, or equal to $P(\text{two heads})$ ?

Greater than (correct answer)
Less than
Equal to
Cannot be determined

Explanation: This question tests middle school quantitative reasoning skills, specifically solving basic probability problems. Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total number of possible outcomes. In this scenario, students are asked to compare P(at least one head) with P(two heads) when flipping two fair coins. Choice A is correct because P(at least one head) = 3/4 (outcomes HH, HT, TH) is greater than P(two heads) = 1/4 (outcome HH only). Choice C (equal to) is incorrect because students might confuse "at least one" with "exactly one," not realizing that "at least one" includes the case of two heads. To help students, emphasize the meaning of "at least" in probability, which includes the specified amount and anything more. Practice comparing probabilities and understanding subset relationships between events.