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 9.3, Problem 106E

(a)

To determine

To find: The probability value in which the slot player get atleast one apple.

(a)

Expert Solution
Check Mark

Answer to Problem 106E

The probability that the slot player gets at least one apple in one pull of the lever is 0.67232.

Explanation of Solution

Given information:

The slot player gets at least one apple in one pull of the lever.

Formula used:

  P(At least 1 apple)=1P(No apple)

Calculation:

Obtain the probability that the slot player gets at least one apple in one pull of the lever:

It is given that, a slot machine has 5 wheels, and each wheel has 5 symbols (apple, grape, peach, pear and plum).

The wheel spins independently and the five symbols are equally likely to appear.

The probability that the slot player gets at least one apple in one pull of the lever is obtained as 0.67232 from the calculation given below:

  P(At least 1 apple)=1P(No apple)=1P(No apple on 5 wheels)=1[P(No apple on1st wheel)×....×P(No apple on5th wheel)]=1[45×45×45×45×45]=110243125=10.32768P(At least 1 apple)=0.67232

Therefore, the probability that the slot player gets at least one apple in one pull of the lever is 0.67232.

Conclusion:

The probability that the slot player gets at least one apple in one pull of the lever is 0.67232.

(b)

To determine

To find: The probability value based on the reader’s question.

(b)

Expert Solution
Check Mark

Answer to Problem 106E

The probability that the slot player gets at least one apple in five pulls of the lever is 0.996222.

Explanation of Solution

Given information:

In the readers view, the lever was pulled 5 times.

Formula used:

  P(At least 1 apple)=1P(No apple)

Calculation:

Compute the answer for reader’s question:

In the readers view, the lever was pulled 5 times.

The readers question is to compute the probability of getting at least 1 one apple in five pulls of the lever.

The probability that the slot player gets at least one apple in five pulls of the lever is obtained as 0.996222 from the calculation given below:

   P( At least 1 apple )=1P( No apple )

   =1P( No apple on 5 wheels for all teh 5 pulls )

   =1[ P( No apple on 1st wheel )×....×P( No apple on 5th wheel ) ]

   =1 [ 4 5 × 4 5 × 4 5 × 4 5 × 4 5 ] 5

   =1 ( 1024 3125 ) 5

   =1 ( 0.32768 ) 5

   =10.003778

   P( At least 1 apple )=0.67232

Therefore, the probability that the slot player gets at least one apple in five pulls of the lever is 0.996222.

Conclusion:

The probability that the slot player gets at least one apple in five pulls of the lever is 0.996222.

Chapter 9 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 9.1 - Prob. 6ECh. 9.1 - Prob. 7ECh. 9.1 - Prob. 8ECh. 9.1 - Prob. 9ECh. 9.1 - Prob. 10ECh. 9.1 - Prob. 11ECh. 9.1 - Prob. 12ECh. 9.1 - Prob. 13ECh. 9.1 - Prob. 14ECh. 9.1 - Prob. 15ECh. 9.1 - Prob. 16ECh. 9.1 - Prob. 17ECh. 9.1 - Prob. 18ECh. 9.1 - Prob. 19ECh. 9.1 - Prob. 20ECh. 9.1 - Prob. 21ECh. 9.1 - Prob. 22ECh. 9.1 - Prob. 23ECh. 9.1 - Prob. 24ECh. 9.1 - Prob. 25ECh. 9.1 - Prob. 26ECh. 9.1 - Prob. 27ECh. 9.1 - Prob. 28ECh. 9.1 - Prob. 29ECh. 9.1 - Prob. 30ECh. 9.1 - Prob. 31ECh. 9.1 - Prob. 32ECh. 9.2 - Prob. 1.1CYUCh. 9.2 - Prob. 2.1CYUCh. 9.2 - Prob. 3.1CYUCh. 9.2 - Prob. 33ECh. 9.2 - Prob. 34ECh. 9.2 - Prob. 35ECh. 9.2 - Prob. 36ECh. 9.2 - Prob. 37ECh. 9.2 - Prob. 38ECh. 9.2 - Prob. 39ECh. 9.2 - Prob. 40ECh. 9.2 - Prob. 41ECh. 9.2 - Prob. 42ECh. 9.2 - Prob. 43ECh. 9.2 - Prob. 44ECh. 9.2 - Prob. 45ECh. 9.2 - Prob. 46ECh. 9.2 - Prob. 47ECh. 9.2 - Prob. 48ECh. 9.2 - Prob. 49ECh. 9.2 - Prob. 50ECh. 9.2 - Prob. 51ECh. 9.2 - Prob. 52ECh. 9.2 - Prob. 53ECh. 9.2 - Prob. 54ECh. 9.2 - Prob. 55ECh. 9.2 - Prob. 56ECh. 9.2 - Prob. 57ECh. 9.2 - Prob. 58ECh. 9.2 - Prob. 59ECh. 9.2 - Prob. 60ECh. 9.2 - Prob. 61ECh. 9.2 - Prob. 62ECh. 9.3 - Prob. 1.1CYUCh. 9.3 - Prob. 1.2CYUCh. 9.3 - Prob. 1.3CYUCh. 9.3 - Prob. 2.1CYUCh. 9.3 - Prob. 3.1CYUCh. 9.3 - Prob. 3.2CYUCh. 9.3 - Prob. 63ECh. 9.3 - Prob. 64ECh. 9.3 - Prob. 65ECh. 9.3 - Prob. 66ECh. 9.3 - Prob. 67ECh. 9.3 - Prob. 68ECh. 9.3 - Prob. 69ECh. 9.3 - Prob. 70ECh. 9.3 - Prob. 71ECh. 9.3 - Prob. 72ECh. 9.3 - Prob. 73ECh. 9.3 - Prob. 74ECh. 9.3 - Prob. 75ECh. 9.3 - Prob. 76ECh. 9.3 - Prob. 77ECh. 9.3 - Prob. 78ECh. 9.3 - Prob. 79ECh. 9.3 - Prob. 80ECh. 9.3 - Prob. 81ECh. 9.3 - Prob. 82ECh. 9.3 - Prob. 83ECh. 9.3 - Prob. 84ECh. 9.3 - Prob. 85ECh. 9.3 - Prob. 86ECh. 9.3 - Prob. 87ECh. 9.3 - Prob. 88ECh. 9.3 - Prob. 89ECh. 9.3 - Prob. 90ECh. 9.3 - Prob. 91ECh. 9.3 - Prob. 92ECh. 9.3 - Prob. 93ECh. 9.3 - Prob. 94ECh. 9.3 - Prob. 95ECh. 9.3 - Prob. 96ECh. 9.3 - Prob. 97ECh. 9.3 - Prob. 98ECh. 9.3 - Prob. 99ECh. 9.3 - Prob. 100ECh. 9.3 - Prob. 101ECh. 9.3 - Prob. 102ECh. 9.3 - Prob. 103ECh. 9.3 - Prob. 104ECh. 9.3 - Prob. 105ECh. 9.3 - Prob. 106ECh. 9.3 - Prob. 107ECh. 9.3 - Prob. 108ECh. 9 - Prob. 1CRECh. 9 - Prob. 2CRECh. 9 - Prob. 3CRECh. 9 - Prob. 4CRECh. 9 - Prob. 5CRECh. 9 - Prob. 6CRECh. 9 - Prob. 7CRECh. 9 - Prob. 8CRECh. 9 - Prob. 9CRECh. 9 - Prob. 1PTCh. 9 - Prob. 2PTCh. 9 - Prob. 3PTCh. 9 - Prob. 4PTCh. 9 - Prob. 5PTCh. 9 - Prob. 6PTCh. 9 - Prob. 7PTCh. 9 - Prob. 8PTCh. 9 - Prob. 9PTCh. 9 - Prob. 10PTCh. 9 - Prob. 11PTCh. 9 - Prob. 12PTCh. 9 - Prob. 13PT
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