Stats: Modeling the World Nasta Edition Grades 9-12
Stats: Modeling the World Nasta Edition Grades 9-12
3rd Edition
ISBN: 9780131359581
Author: David E. Bock, Paul F. Velleman, Richard D. De Veaux
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 15E

(a)

To determine

To find: the probability that the 1st heart get is the 3rd card dealt.

(a)

Expert Solution
Check Mark

Answer to Problem 15E

0.1453

Explanation of Solution

Given:

Total number of cards = 52

Heart = 13

Other = 52-13= 39

Calculation:

Required to find the probability of getting first heart in the trial

  P(firs heart you get is the third card dealt)=P(first card is not a heart)×P(second card is not a heart)×P(Third card is a heart)=3952×3851×1350=19,266132,600=0.1453

(b)

To determine

To find: the probability that cards are all red.

(b)

Expert Solution
Check Mark

Answer to Problem 15E

0.1176

Explanation of Solution

Given:

Total number of cards = 52

Red = 26

Calculation:

Required probability that all three drawn cards are red

  (you cards are all red)=[P(first card is red)×P(Second card is red)×P(Third card is red)]=2652×2551×2450=15600132600=0.1176

The probability that all red is 0.1176

(c)

To determine

To find: the probability that get no spades.

(c)

Expert Solution
Check Mark

Answer to Problem 15E

0.4135

Explanation of Solution

Given:

Total number of cards = 52

Spades = 13

Other = 52-13 =39

Formula used:

Calculation:

  P(you get is no spades)=[P(first card is not a spade)×P(Second card is not a spade)×P(Third card is not a spade)]=3952×3851×3750=54834132600=0.4135

Probability that all cards dealt are not spade is 0.4135.

(d)

To determine

To find: the probability that at least one ace.

(d)

Expert Solution
Check Mark

Answer to Problem 15E

0.2174

Explanation of Solution

Given:

Total number of cards = 52

Aces= 4

Calculation:

  P(at least one ace)=1P(Noace)=1[P(first card is not an ace)×P(Second card is not an ace)×P(Third card is not an ace)]=1[4852×4751×4650]=1[103776132600]=10.7826=12174

The probability that at least one is Ace is 0.2174

Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Find number of persons in a part with 66 handshakes Combinations; Author: Anil Kumar;https://www.youtube.com/watch?v=33TgLi-wp3E;License: Standard YouTube License, CC-BY
Discrete Math 6.3.1 Permutations and Combinations; Author: Kimberly Brehm;https://www.youtube.com/watch?v=J1m9sB5XZQc;License: Standard YouTube License, CC-BY
How to use permutations and combinations; Author: Mario's Math Tutoring;https://www.youtube.com/watch?v=NEGxh_D7yKU;License: Standard YouTube License, CC-BY
Permutations and Combinations | Counting | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=0NAASclUm4k;License: Standard Youtube License
Permutations and Combinations Tutorial; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=XJnIdRXUi7A;License: Standard YouTube License, CC-BY