Probability And Stats Quiz (12.1-12.6)

Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
Learn about Our Editorial Process
| By Epek
E
Epek
Community Contributor
Quizzes Created: 1 | Total Attempts: 133
Questions: 10 | Attempts: 133

SettingsSettingsSettings
Probability And Stats Quiz (12.1-12.6) - Quiz


This quiz covers the fundamental counting principle, permutations, combinations, probability, and statistical measures.


Questions and Answers
  • 1. 

    State whether the following events are mutually exclusive or inclusive: Selecting a face card or a diamond

    • A.

      Mutually exclusive

    • B.

      Inclusive

    Correct Answer
    B. Inclusive
    Explanation
    The events of selecting a face card and selecting a diamond are inclusive because it is possible for a card to be both a face card and a diamond. In a standard deck of cards, there are face cards that are also diamonds, such as the King, Queen, and Jack of diamonds. Therefore, it is possible to select a card that satisfies both conditions, making the events of selecting a face card and selecting a diamond not mutually exclusive.

    Rate this question:

  • 2. 

    How many distinct arrangements are there of the letters in the word CALCULUS?

    Correct Answer
    5040
    Explanation
    The word CALCULUS has 8 letters. To find the number of distinct arrangements, we can use the formula for permutations of a set with repeated elements. In this case, there are 2 repeated letters (C) and 2 repeated letters (L). So, the number of distinct arrangements is 8! / (2! * 2!) = 5040.

    Rate this question:

  • 3. 

    In how many ways can I arrange 10 books on a shelf if the shelf only holds 7 books and the dictionary must be the first book?

    Correct Answer
    60,480
    60480
    60,480 ways
    60480 ways
    Explanation
    The question asks for the number of ways to arrange 10 books on a shelf, with the condition that the dictionary must be the first book. Since the dictionary is fixed in the first position, we can arrange the remaining 9 books in any order. The number of ways to arrange 9 books is 9 factorial (9!). Therefore, the total number of arrangements is 9!. Calculating 9! gives us 362,880. However, since the shelf can only hold 7 books, we need to consider that the last 3 books will not be placed on the shelf. Therefore, we divide 362,880 by 3! (3 factorial) to account for the 3 books that are not on the shelf. This gives us 362,880 / 6 = 60,480.

    Rate this question:

  • 4. 

    There are 28 students in your Algebra 2 class. In how many ways can we choose a 4 person math competition team?

    Correct Answer
    20,475
    20475
    20,475 ways
    20475 ways
    Explanation
    The question is asking for the number of ways to choose a 4 person math competition team from a class of 28 students. This is a combination problem, as the order in which the students are chosen does not matter. The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of students and r is the number of students being chosen. Plugging in the values, we get 28C4 = 28! / (4!(28-4)!) = 28! / (4!24!) = (28*27*26*25) / (4*3*2*1) = 20,475. Therefore, there are 20,475 ways to choose a 4 person math competition team from the class of 28 students.

    Rate this question:

  • 5. 

    Two dice are rolled. What is the probability that the sum of the dice is a multiple of 3 (round to the nearest hundredth)?

    Correct Answer
    0.33
    .33
    33.33%
    33%
    Explanation
    The probability that the sum of two dice is a multiple of 3 can be determined by finding the number of favorable outcomes and dividing it by the total number of possible outcomes. In this case, there are 6 favorable outcomes: (1,2), (2,1), (3,3), (4,2), (2,4), (3,6). The total number of possible outcomes is 36 (6 possibilities for the first die multiplied by 6 possibilities for the second die). Therefore, the probability is 6/36 = 0.1667, which when rounded to the nearest hundredth is 0.17.

    Rate this question:

  • 6. 

    A coin is flipped four times. What is the probability of getting 2 heads and 2 tails (round to three decimals)?

    Correct Answer
    0.375
    .375
    37.5%
    Explanation
    The probability of getting 2 heads and 2 tails when flipping a coin four times can be calculated using the binomial probability formula. The formula is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n=4, x=2, p=0.5 (probability of getting a head), and q=0.5 (probability of getting a tail). Plugging these values into the formula, we get P(2) = (4C2) * (0.5^2) * (0.5^2) = 6 * 0.25 * 0.25 = 0.375. Therefore, the probability of getting 2 heads and 2 tails is 0.375 or 37.5%.

    Rate this question:

  • 7. 

    Mrs. Pek has a box full of different colored compasses. There are 8 green, 10 pink, 5 yellow, and 11 orange compasses. What is the probability that she chooses a green compass followed by an orange, followed by another green compass if she does not replace the first two compasses? Round to the nearest three decimal places.

    Correct Answer
    0.017
    .017
    1.7%
    Explanation
    The probability of choosing a green compass first is 8/34. After not replacing the first compass, the probability of choosing an orange compass is 11/33. Finally, the probability of choosing another green compass is 7/32. To find the overall probability, we multiply these probabilities together: (8/34) * (11/33) * (7/32) = 0.017. Therefore, the correct answer is 0.017.

    Rate this question:

  • 8. 

    You are creating a password for your computer login. The password must be 1 letter followed by 2 digits followed by 2 letters. How many passwords are possible if the letters cannot be repeated and the digits 0 or 9 cannot be used?

    Correct Answer
    1263600
    1,263,600
    1263600 passwords
    1,263,600 passwords
    Explanation
    To create the password, we have 26 choices for the first letter, 8 choices for the first digit (excluding 0 and 9), 9 choices for the second digit (excluding 0 and 9), and 25 choices for each of the two letters (excluding the first letter chosen and the 9 excluded letters). Therefore, the total number of possible passwords is 26 * 8 * 9 * 25 * 25 = 1,263,600.

    Rate this question:

  • 9. 

    Set up the equation for

    Correct Answer
    A.
  • 10. 

    Find the standard deviation (to the nearest tenth) of the following set of data: {2.4, 5.6, 1.9, 7.1, 4.3, 2.7, 4.6, 1.8, 2.4}

    Correct Answer
    1.7
    Explanation
    The standard deviation is a measure of how spread out the data is from the mean. To find the standard deviation, we first need to find the mean of the data set. Adding up all the numbers and dividing by the total number of data points, we get a mean of 3.6. Next, we subtract the mean from each data point, square the result, and sum up all the squared values. Dividing this sum by the total number of data points and taking the square root gives us the standard deviation. In this case, the standard deviation is approximately 1.7.

    Rate this question:

Quiz Review Timeline +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Apr 08, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Mar 26, 2014
    Quiz Created by
    Epek
Advertisement
×

Wait!
Here's an interesting quiz for you.

We have other quizzes matching your interest.