The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 11.1, Problem 26E

a.

To determine

To provide: The reason of Luis’ probability being incorrect.

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

5 six sides dice have been rolled simultaneously.

To obtain Yahtzee, all 5 dice must have same numbers.

According to Luis, the probability of getting a Yahtzee is (16)5 .

Calculation:

The following numbers could be obtained by players 1, 2, 3, 4, 5 and 6.

Here all the dice are independent.

Thus, the probability of getting a Yahtzee is calculated as:

  P(Yahtzee)=6×(16)5

b.

To determine

To explain: Whether Nasir should be surprised if he does not obtain Yahtzee even after rolls. Also, support the answer by computing appropriate probability.

b.

Expert Solution
Check Mark

Answer to Problem 26E

The probability is 0.9808.

Explanation of Solution

Calculation:

From above part, it is known that

  P(Yahtzee)=6×(16)5=0.0007716

  P(NoYahtzee)=16×(16)5=0.9992284

The probability of not getting Yahtzee even after 25 rolls is computed as:

  P(No Yahtzee)25=(0.9992254)25=0.9809=98.09%

Since, the probability is above 0.05. Thus, it could be concluded that the probability of not getting Yahtzee is likely. Thus, Nasir should not be surprised.

Chapter 11 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 11.1 - Prob. 3ECh. 11.1 - Prob. 4ECh. 11.1 - Prob. 5ECh. 11.1 - Prob. 6ECh. 11.1 - Prob. 7ECh. 11.1 - Prob. 8ECh. 11.1 - Prob. 9ECh. 11.1 - Prob. 10ECh. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.1 - Prob. 22ECh. 11.1 - Prob. 23ECh. 11.1 - Prob. 24ECh. 11.1 - Prob. 25ECh. 11.1 - Prob. 26ECh. 11.2 - Prob. 1.1CYUCh. 11.2 - Prob. 1.2CYUCh. 11.2 - Prob. 1.3CYUCh. 11.2 - Prob. 2.1CYUCh. 11.2 - Prob. 2.2CYUCh. 11.2 - Prob. 2.3CYUCh. 11.2 - Prob. 3.1CYUCh. 11.2 - Prob. 3.2CYUCh. 11.2 - Prob. 3.3CYUCh. 11.2 - Prob. 4.1CYUCh. 11.2 - Prob. 4.2CYUCh. 11.2 - Prob. 6.1CYUCh. 11.2 - Prob. 27ECh. 11.2 - Prob. 28ECh. 11.2 - Prob. 29ECh. 11.2 - Prob. 30ECh. 11.2 - Prob. 31ECh. 11.2 - Prob. 32ECh. 11.2 - Prob. 33ECh. 11.2 - Prob. 34ECh. 11.2 - Prob. 35ECh. 11.2 - Prob. 36ECh. 11.2 - Prob. 37ECh. 11.2 - Prob. 38ECh. 11.2 - Prob. 39ECh. 11.2 - Prob. 40ECh. 11.2 - Prob. 41ECh. 11.2 - Prob. 42ECh. 11.2 - Prob. 43ECh. 11.2 - Prob. 44ECh. 11.2 - Prob. 45ECh. 11.2 - Prob. 46ECh. 11.2 - Prob. 47ECh. 11.2 - Prob. 48ECh. 11.2 - Prob. 49ECh. 11.2 - Prob. 50ECh. 11.2 - Prob. 51ECh. 11.2 - Prob. 52ECh. 11.2 - Prob. 53ECh. 11.2 - Prob. 54ECh. 11.2 - Prob. 55ECh. 11.2 - Prob. 56ECh. 11.2 - Prob. 57ECh. 11.2 - Prob. 58ECh. 11.2 - Prob. 59ECh. 11.2 - Prob. 60ECh. 11.2 - Prob. 61ECh. 11.2 - Prob. 62ECh. 11.2 - Prob. 63ECh. 11.2 - Prob. 64ECh. 11 - Prob. 1CRECh. 11 - Prob. 2CRECh. 11 - Prob. 3CRECh. 11 - Prob. 4CRECh. 11 - Prob. 5CRECh. 11 - Prob. 6CRECh. 11 - Prob. 1PTCh. 11 - Prob. 2PTCh. 11 - Prob. 3PTCh. 11 - Prob. 4PTCh. 11 - Prob. 5PTCh. 11 - Prob. 6PTCh. 11 - Prob. 7PTCh. 11 - Prob. 8PTCh. 11 - Prob. 9PTCh. 11 - Prob. 10PTCh. 11 - Prob. 11PTCh. 11 - Prob. 12PTCh. 11 - Prob. 13PT
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