EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
6th Edition
ISBN: 9781319287573
Author: Starnes
Publisher: MPS PUB
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Question
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Chapter 5.2, Problem 42E

(a)

To determine

Probability for getting non − hearts.

(a)

Expert Solution
Check Mark

Answer to Problem 42E

Probability,

  P(Hc)=0.75

Explanation of Solution

Given information:

Two − way table represents the sample space for this chance process:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.2, Problem 42E , additional homework tip  1

Calculations:

Complement rule:

  P(Ac)=P(notA)=1P(A)

We know that

H: getting a heart

F: getting a face card

Look at the bottom left corner of the table,

There are total 52 cards in a standard deck.

Thus,

The number of possible outcomes is 52.

Note that

13 of the 52 cards are Heart.

Thus,

The number of favorable outcomes is 13.

If we divide the number of favorable outcomes by the number of possible outcome, we get the probability.

  P(H)=NumberoffavorableoutcomesNumberofpossibleoutcomes=1352

Apply the complement rule:

  P(Hc)=1P(H)=11352=3952=34=0.75

Thus,

Probability for getting non − hearts is 0.75.

(b)

To determine

Probability for getting non − hearts and face card.

(b)

Expert Solution
Check Mark

Answer to Problem 42E

Probability,

  P(HcandF)0.1731

Explanation of Solution

Given information:

Two − way table represents the sample space for this chance process:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.2, Problem 42E , additional homework tip  2

Calculations:

We know that

H: getting a heart

F: getting a face card

Hc: getting a non - heart

Look at the bottom left corner of the table,

There are total 52 cards in a standard deck.

Thus,

The number of possible outcomes is 52.

Note that

9 of the 52 cards are Face cards and non − Heart.

Thus,

The number of favorable outcomes is 9.

If we divide the number of favorable outcomes by the number of possible outcome, we get the probability.

  P(HcandF)=NumberoffavorableoutcomesNumberofpossibleoutcomes=9520.1731=17.31%

Thus,

Around 17.31% of all cards are non − Heart and Face cards and the probability is approx. 0.1731.

(c)

To determine

Probability for getting non − heart or face card.

(c)

Expert Solution
Check Mark

Answer to Problem 42E

Probability,

  P(HcorF)0.8077

Explanation of Solution

Given information:

Two − way table represents the sample space for this chance process:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.2, Problem 42E , additional homework tip  3

Calculations:

Complement rule:

  P(Ac)=P(notA)=1P(A)

General addition rule:

For any two events,

  P(AorB)=P(A)+P(B)P(AandB)

From Part (a) and Part (b) results,

We have

Probability for getting non − Heart,

  P(Hc)=3952

Probability for getting non − Heart and Face card,

  P(HcandF)=952

Now,

Note that

12 of the 52 cards are Face cards.

Thus,

In this case, the number of favorable outcomes is 12 and number of possible outcomes in 52.

  P(F)=NumberoffavorableoutcomesNumberofpossibleoutcomes=1252

Apply general addition rule:

  P(HcorF)=P(Hc)+P(F)P(HcandF)=3952+1252952=4252=21260.8077=80.77%

Thus,

Around 80.77% of all cards are non − Heart or Face cards and the probability is approx. 0.8077.

Chapter 5 Solutions

EBK PRACTICE OF STAT.F/AP EXAM,UPDATED

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.2 - Prob. 43ECh. 5.2 - Prob. 44ECh. 5.2 - Prob. 45ECh. 5.2 - Prob. 46ECh. 5.2 - Prob. 47ECh. 5.2 - Prob. 48ECh. 5.2 - Prob. 49ECh. 5.2 - Prob. 50ECh. 5.2 - Prob. 51ECh. 5.2 - Prob. 52ECh. 5.2 - Prob. 53ECh. 5.2 - Prob. 54ECh. 5.2 - Prob. 55ECh. 5.2 - Prob. 56ECh. 5.2 - Prob. 57ECh. 5.2 - Prob. 58ECh. 5.2 - Prob. 59ECh. 5.2 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5.3 - Prob. 84ECh. 5.3 - Prob. 85ECh. 5.3 - Prob. 86ECh. 5.3 - Prob. 87ECh. 5.3 - Prob. 88ECh. 5.3 - Prob. 89ECh. 5.3 - Prob. 90ECh. 5.3 - Prob. 91ECh. 5.3 - Prob. 92ECh. 5.3 - Prob. 93ECh. 5.3 - Prob. 94ECh. 5.3 - Prob. 95ECh. 5.3 - Prob. 96ECh. 5.3 - Prob. 97ECh. 5.3 - Prob. 98ECh. 5.3 - Prob. 99ECh. 5.3 - Prob. 100ECh. 5.3 - Prob. 101ECh. 5.3 - Prob. 102ECh. 5.3 - Prob. 103ECh. 5.3 - Prob. 104ECh. 5.3 - Prob. 105ECh. 5.3 - Prob. 106ECh. 5.3 - Prob. 107ECh. 5.3 - Prob. 108ECh. 5 - Prob. R5.1RECh. 5 - Prob. R5.2RECh. 5 - Prob. R5.3RECh. 5 - Prob. R5.4RECh. 5 - Prob. R5.5RECh. 5 - Prob. R5.6RECh. 5 - Prob. R5.7RECh. 5 - Prob. R5.8RECh. 5 - Prob. T5.1SPTCh. 5 - Prob. T5.2SPTCh. 5 - Prob. T5.3SPTCh. 5 - Prob. T5.4SPTCh. 5 - Prob. T5.5SPTCh. 5 - Prob. T5.6SPTCh. 5 - Prob. T5.7SPTCh. 5 - Prob. T5.8SPTCh. 5 - Prob. T5.9SPTCh. 5 - Prob. T5.10SPTCh. 5 - Prob. T5.11SPTCh. 5 - Prob. T5.12SPTCh. 5 - Prob. T5.13SPTCh. 5 - Prob. T5.14SPT
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