Praxis Math Quiz: Compute Simple Probability
20 questions · exam conditions
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Compute Simple ProbabilityQuestion 1 of 20

A spinner is divided into three sections: Red, which covers 1/2 of the area; Blue, which covers 1/3 of the area; and Green, which covers 1/6. If the spinner is spun twice, what is the probability that it lands on Red on the first spin and on Blue on the second spin?

1/9
1/6
1/4
5/6
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Praxis Math Quiz

Praxis Math Quiz: Compute Simple Probability

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

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 spinner is divided into three sections: Red, which covers 1/2 of the area; Blue, which covers 1/3 of the area; and Green, which covers 1/6. If the spinner is spun twice, what is the probability that it lands on Red on the first spin and on Blue on the second spin?

  1. 1/9
  2. 1/6 (correct answer)
  3. 1/4
  4. 5/6
Explanation: The two spins are independent events. The probability of the first event (landing on Red) is 1/2. The probability of the second event (landing on Blue) is 1/3. To find the probability of both events occurring in sequence, we multiply their individual probabilities: P(Red then Blue) = P(Red) × P(Blue) = (1/2) × (1/3) = 1/6.

Question 2

A bag contains two standard fair coins and one two-headed coin. A single coin is chosen at random from the bag and then flipped once. What is the probability that the result is heads?

  1. 1/2
  2. 3/4
  3. 2/3 (correct answer)
  4. 5/6
Explanation: This problem involves two stages. First, picking a coin, then flipping it. There are two paths to getting heads. Path 1: Pick a fair coin (P=2/3) and flip heads (P=1/2). The probability of this path is (2/3) * (1/2) = 1/3. Path 2: Pick the two-headed coin (P=1/3) and flip heads (P=1). The probability of this path is (1/3) * 1 = 1/3. The total probability of getting heads is the sum of the probabilities of these two mutually exclusive paths: 1/3 + 1/3 = 2/3.

Question 3

At a conference, 40% of attendees are from the East Coast and 60% are from the West Coast. Of the East Coast attendees, 70% use Brand A laptops. Of the West Coast attendees, 50% use Brand A laptops. What is the probability that a randomly selected conference attendee uses a Brand A laptop?

  1. 0.58 (correct answer)
  2. 0.60
  3. 0.62
  4. 0.70
Explanation: This is a weighted probability problem. We calculate the proportion of each group that uses Brand A and add them. Proportion from East Coast using Brand A: 0.40 (attendees) * 0.70 (laptop users) = 0.28. Proportion from West Coast using Brand A: 0.60 (attendees) * 0.50 (laptop users) = 0.30. The total probability is the sum of these two proportions: 0.28 + 0.30 = 0.58.

Question 4

A restaurant offers a lunch special with a choice of one sandwich (3 options), one side (4 options), and one drink (2 options). If a person randomly selects one item from each category, what is the probability they select the chicken sandwich, the fries, and the soda?

  1. 1/9
  2. 1/12
  3. 1/24 (correct answer)
  4. 1/3
Explanation: The choices are independent. The probability of choosing the chicken sandwich is 1/3. The probability of choosing the fries is 1/4. The probability of choosing the soda is 1/2. To find the probability of this specific combination being selected, multiply the individual probabilities: (1/3) * (1/4) * (1/2) = 1/24. This can also be seen as 1 favorable outcome out of 342=24 total possible meal combinations.

Question 5

A box contains 20 tickets. Some tickets are winners and some are losers. The probability of drawing a winning ticket is 1/5. If 5 losing tickets are removed from the box, what is the new probability of drawing a winning ticket?

  1. 1/4
  2. 1/3
  3. 4/15 (correct answer)
  4. 1/5
Explanation: Initially, P(winning) = 1/5. With 20 total tickets, the number of winning tickets is (1/5) * 20 = 4. The number of losing tickets is 20 - 4 = 16. After removing 5 losing tickets, the number of winning tickets remains 4, but the total number of tickets is now 20 - 5 = 15. The new probability of drawing a winning ticket is the number of winning tickets divided by the new total: 4/15.

Question 6

A bookshelf holds 8 mystery novels, 7 science fiction novels, and 5 historical fiction novels. If a reader randomly picks two books to read, without replacement, what is the probability that they are both science fiction novels?

  1. 21/190 (correct answer)
  2. 49/400
  3. 7/20
  4. 7/10
Explanation: There are a total of 8 + 7 + 5 = 20 books. The probability that the first book selected is science fiction is 7/20. After one science fiction novel is removed, there are 19 books left, and 6 of them are science fiction. The probability that the second book is also science fiction is 6/19. The probability of both events happening is the product: (7/20) * (6/19) = 42/380, which simplifies to 21/190.

Question 7

A password is two characters long. The first character must be a letter from the set {A, B, C, D}, and the second character must be a digit from the set {1, 2, 3, 4, 5}. If a password is generated randomly, what is the probability that the first character is a vowel and the second character is an even number?

  1. 1/20
  2. 1/10 (correct answer)
  3. 3/10
  4. 1/4
Explanation: The two events are independent. The first event is selecting a vowel from {A, B, C, D}. The only vowel is A, so there is 1 favorable outcome out of 4. P(vowel) = 1/4. The second event is selecting an even number from {1, 2, 3, 4, 5}. The even numbers are 2 and 4, so there are 2 favorable outcomes out of 5. P(even) = 2/5. The probability of both events occurring is the product: (1/4) * (2/5) = 2/20 = 1/10.

Question 8

A spinner has four equal sections labeled A, B, C, and D. The spinner is spun, and a standard 6-sided die is rolled. What is the probability that the spinner lands on a consonant and the die shows a number less than 3?

  1. 1/4 (correct answer)
  2. 1/2
  3. 1/8
  4. 1/12
Explanation: The consonants on the spinner are B, C, and D. So, the probability of landing on a consonant is 3/4. The numbers on the die less than 3 are 1 and 2. So, the probability of rolling a number less than 3 is 2/6, which simplifies to 1/3. Since these are independent events, we multiply their probabilities: (3/4) * (1/3) = 3/12 = 1/4.

Question 9

A high school has 300 students, of whom 60 are juniors. A total of 75 students are in the school band. If there are exactly 15 juniors in the band, what is the probability that a student selected at random from the entire school is a junior in the band?

  1. 1/20 (correct answer)
  2. 1/5
  3. 1/4
  4. 4/5
Explanation: The question asks for the probability of selecting a student who meets two criteria: being a junior AND being in the band. The problem states that there are 15 such students. The total number of students from which to choose is 300. Therefore, the probability is the number of favorable outcomes divided by the total number of outcomes: 15/300, which simplifies to 1/20.

Question 10

A target consists of two concentric circles. The inner circle has a radius of 2 inches and the outer circle has a radius of 5 inches. A dart is thrown and lands randomly on the target. What is the probability that the dart lands in the ring between the two circles, and not in the inner circle?

  1. 4/25
  2. 2/5
  3. 3/5
  4. 21/25 (correct answer)
Explanation: This is a geometric probability problem. The probability is the ratio of the favorable area to the total area. The total area is the area of the large circle: A_total = π(525^2) = 25π. The area of the inner circle is A_inner = π(222^2) = 4π. The area of the ring between the circles is the total area minus the inner area: A_ring = 25π - 4π = 21π. The probability of landing in the ring is (Area of Ring) / (Total Area) = (21π) / (25π) = 21/25.

Question 11

There are 20 sealed envelopes. Of these, 8 contain a $5 bill and 12 contain a $10 bill. The first person picks an envelope and it contains a $10 bill. Without replacement, what is the probability the second person also picks an envelope with a $10 bill?

  1. 3/5
  2. 12/19
  3. 11/19 (correct answer)
  4. 11/20
Explanation: This is a conditional probability problem. The first event has already occurred and has changed the conditions for the second event. After the first person picks a $10 bill, there are 19 envelopes remaining in total. Since one $10 bill was removed, there are now 12 - 1 = 11 envelopes with $10 bills left. Therefore, the probability that the second person picks a $10 bill is 11/19.

Question 12

A jar contains only red and blue marbles. The ratio of red marbles to blue marbles is 3:5. If a marble is drawn at random, what is the probability that it is NOT red?

  1. 3/8
  2. 3/5
  3. 5/8 (correct answer)
  4. 2/5
Explanation: A ratio of 3:5 means that for every 3 red marbles, there are 5 blue marbles. The total number of 'parts' in the ratio is 3 + 5 = 8. The probability of drawing a red marble is 3/8. The probability of drawing a marble that is NOT red is the same as the probability of drawing a blue marble. There are 5 parts blue out of a total of 8 parts. Thus, the probability is 5/8.

Question 13

A batch of 50 light bulbs contains 4 defective bulbs. If two light bulbs are selected at random without replacement, what is the probability that at least one of the selected bulbs is NOT defective?

  1. 6/1225
  2. 207/245
  3. 23/25
  4. 1219/1225 (correct answer)
Explanation: The easiest way to solve 'at least one' problems is to calculate the probability of the complementary event and subtract it from 1. The complement of 'at least one is not defective' is 'both are defective'. The probability that the first bulb is defective is 4/50. The probability that the second is also defective is 3/49. P(both defective) = (4/50) * (3/49) = 12/2450 = 6/1225. Therefore, the probability of at least one not being defective is 1 - (6/1225) = 1219/1225.

Question 14

Two fair coins are tossed. What is the probability that at least one head appears?

  1. 1/4
  2. 1/2
  3. 2/3
  4. 3/4 (correct answer)
Explanation: The possible outcomes when tossing two coins are HH, HT, TH, TT. There are 4 equally likely outcomes. The outcomes with at least one head are HH, HT, and TH. There are 3 favorable outcomes. Therefore, the probability is 3/4. Alternatively, one could calculate the probability of the complement event (no heads, which is TT) which is 1/4, and subtract this from 1: 1 - 1/4 = 3/4.

Question 15

A fair 6-sided die is rolled three times. What is the probability that a '1' is rolled on the first roll, a '2' on the second roll, and a '3' on the third roll?

  1. 1/2
  2. 1/6
  3. 1/18
  4. 1/216 (correct answer)
Explanation: Each die roll is an independent event with a probability of 1/6 for any specific outcome. The probability of rolling a '1' is 1/6. The probability of rolling a '2' is 1/6. The probability of rolling a '3' is 1/6. To find the probability of all three events happening in this specific sequence, we multiply their individual probabilities: (1/6) * (1/6) * (1/6) = 1/216.

Question 16

Two events, A and B, are independent. The probability of event A occurring is 0.6 and the probability of event B occurring is 0.3. What is the probability that neither event A nor event B occurs?

  1. 0.10
  2. 0.18
  3. 0.28 (correct answer)
  4. 0.42
Explanation: If P(A) = 0.6, then the probability of A not occurring, P(not A), is 1 - 0.6 = 0.4. If P(B) = 0.3, then the probability of B not occurring, P(not B), is 1 - 0.3 = 0.7. Since events A and B are independent, their complements (not A and not B) are also independent. The probability that neither occurs is the product of their individual probabilities: P(not A and not B) = P(not A) * P(not B) = 0.4 * 0.7 = 0.28.

Question 17

In a class of 28 students, 18 have a dog, 12 have a cat, and 5 have both a dog and a cat. What is the probability that a randomly selected student has a dog but does not have a cat?

  1. 13/28 (correct answer)
  2. 9/14
  3. 1/4
  4. 3/7
Explanation: We are given that 18 students have a dog. Within this group, 5 students also have a cat. To find the number of students who have a dog but not a cat, we subtract the overlap from the total number of dog owners: 18 - 5 = 13. These 13 students have only a dog. The total number of students is 28. Therefore, the probability of selecting a student with a dog but no cat is 13/28.

Question 18

Two dice are rolled. What is the probability that at least one of the dice shows a 6?

  1. 1/36
  2. 1/6
  3. 11/36 (correct answer)
  4. 25/36
Explanation: The easiest way to solve this is to find the probability of the complementary event: that neither die shows a 6. For a single die, the probability of not rolling a 6 is 5/6. For two dice, the probability that neither shows a 6 is (5/6) * (5/6) = 25/36. The probability that at least one die shows a 6 is 1 minus the probability that neither shows a 6: 1 - 25/36 = 11/36.

Question 19

From a standard 52-card deck, two cards are drawn without replacement. What is the probability that the first card is a heart and the second card is a spade?

  1. 1/16
  2. 13/204 (correct answer)
  3. 13/25
  4. 1/4
Explanation: The probability that the first card is a heart is 13/52, or 1/4. After drawing one heart, there are 51 cards remaining in the deck. The number of spades is still 13. So, the probability that the second card is a spade is 13/51. The probability of both events occurring in sequence is the product of their probabilities: (13/52) * (13/51) = (1/4) * (13/51) = 13/204.

Question 20

A number is randomly selected from the set of integers from 1 to 50, inclusive. What is the probability that the number selected is a perfect square?

  1. 1/10
  2. 7/50 (correct answer)
  3. 4/25
  4. 1/5
Explanation: The total number of outcomes is 50. We need to count the number of perfect squares between 1 and 50. These are 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, and 7^2=49. The next perfect square, 8^2=64, is too large. There are 7 perfect squares in the set. Therefore, the probability is the number of favorable outcomes divided by the total number of outcomes, which is 7/50.