Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access)
Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access)
9th Edition
ISBN: 9781319126100
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution
Check Mark

Answer to Problem 18E

Solution: The probability of complementary event of A is 0.583.

Explanation of Solution

The nonoccurrence of event A will represent the complementary event of A denoted as A¯ and its probability is computed as

P(A)+P(A¯)=1

So

P(A¯)=1P(A)=10.417=0.583

So, the probability that A does not occur is 0.583.

(b)

To determine

To find: The probability that all four tosses result in the same outcome.

(b)

Expert Solution
Check Mark

Answer to Problem 18E

Solution: The probability of obtaining similar outcomes is 0.125, and is obtained by the help of addition rule of probability.

Explanation of Solution

Calculation:

If a coin is tossed 4 times, then there are a total of 24=16 possible outcomes in the sample space. The favorable outcomes to obtain same outcomes are {(HHHH),(TTTT)}. So, the probability that all the outcomes are same is computed by the addition theorem of probability as

P{(HHHH)(TTTT)}=P(HHHH)+P(TTTT)P{(HHHH)(TTTT)}

P{(HHHH)(TTTT)}=0 because the events are independent.

So

P{(HHHH)(TTTT)}=P(HHHH)+P(TTTT)=116+116=18=0.125

(c)

To determine

To find: The probability of obtaining at least one head and at least one tail.

(c)

Expert Solution
Check Mark

Answer to Problem 18E

Solution: The probability of at least one head and at least one tail is 0.875 and is obtained by complement rule of probability.

Explanation of Solution

Calculation:

The events favorable to getting at least 1 head and at least one tail is

{(HTTT),(THTT),(TTHT),(TTTH),(HHTT),(HTHT),(TTHH),(THTH),(HHHT),(HHTH),(HTHH),(THHH)}

The probability is obtained as

P[At least one head and one tail]=1P[0 Heads and 0 Tails]=1[P(HHHH)+P(TTTT)]=10.125=0.875

(d)

To determine

To find: If the events A and B are disjoint.

(d)

Expert Solution
Check Mark

Answer to Problem 18E

Solution: The provided events are not disjoint and the property of the probability theory that probability can never exceed one is used.

Explanation of Solution

Calculation:

The probability of events A and B are provided as

P(A)=0.4P(B)=0.8

Now, by the addition theorem of probability

P(AB)=P(A)+P(B)P(AB)

If events A and B are disjoint, then the probability of AB must have been zero.

So

P(AB)=P(A)+P(B)=0.4+0.8=1.2

Since the probability of any event can never be more than 1, hence A and B are not disjoint.

(e)

To determine

If the probability of an event A can be −0.04.

(e)

Expert Solution
Check Mark

Answer to Problem 18E

Solution: The provided scenario is not possible and the property of the probability theory that the probability can never be less than 0 is used.

Explanation of Solution

According to the provided information, the event A is very rare and its probability is 0.04. This is not possible because the probability of any event can never be negative. Probability always lies in between 0 and 1.

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Chapter 4 Solutions

Introduction to the Practice of Statistics 9E & LaunchPad for Introduction to the Practice of Statistics 9E (Twelve-Month Access)

Ch. 4.2 - Prob. 11UYKCh. 4.2 - Prob. 12UYKCh. 4.2 - Prob. 13UYKCh. 4.2 - Prob. 14UYKCh. 4.2 - Prob. 15UYKCh. 4.2 - Prob. 16UYKCh. 4.2 - Prob. 17ECh. 4.2 - Prob. 18ECh. 4.2 - Prob. 19ECh. 4.2 - Prob. 20ECh. 4.2 - Prob. 21ECh. 4.2 - Prob. 22ECh. 4.2 - Prob. 23ECh. 4.2 - Prob. 24ECh. 4.2 - Prob. 25ECh. 4.2 - Prob. 26ECh. 4.2 - Prob. 27ECh. 4.2 - Prob. 28ECh. 4.2 - Prob. 29ECh. 4.2 - Prob. 30ECh. 4.2 - Prob. 31ECh. 4.2 - Prob. 32ECh. 4.2 - Prob. 33ECh. 4.2 - Prob. 34ECh. 4.2 - Prob. 35ECh. 4.2 - Prob. 36ECh. 4.2 - Prob. 37ECh. 4.2 - Prob. 38ECh. 4.2 - Prob. 39ECh. 4.2 - Prob. 40ECh. 4.2 - Prob. 41ECh. 4.3 - Prob. 42UYKCh. 4.3 - Prob. 43UYKCh. 4.3 - Prob. 44UYKCh. 4.3 - Prob. 45ECh. 4.3 - Prob. 46ECh. 4.3 - Prob. 47ECh. 4.3 - Prob. 48ECh. 4.3 - Prob. 49ECh. 4.3 - Prob. 50ECh. 4.3 - Prob. 51ECh. 4.3 - Prob. 52ECh. 4.3 - Prob. 53ECh. 4.3 - Prob. 54ECh. 4.3 - Prob. 55ECh. 4.3 - Prob. 56ECh. 4.3 - Prob. 57ECh. 4.3 - Prob. 58ECh. 4.3 - Prob. 59ECh. 4.3 - Prob. 60ECh. 4.3 - Prob. 61ECh. 4.3 - Prob. 62ECh. 4.4 - Prob. 63UYKCh. 4.4 - Prob. 64UYKCh. 4.4 - Prob. 65UYKCh. 4.4 - Prob. 66UYKCh. 4.4 - Prob. 67UYKCh. 4.4 - Prob. 68ECh. 4.4 - Prob. 69ECh. 4.4 - Prob. 70ECh. 4.4 - Prob. 71ECh. 4.4 - Prob. 72ECh. 4.4 - Prob. 73ECh. 4.4 - Prob. 74ECh. 4.4 - Prob. 75ECh. 4.4 - Prob. 76ECh. 4.4 - Prob. 77ECh. 4.4 - Prob. 78ECh. 4.4 - Prob. 79ECh. 4.4 - Prob. 80ECh. 4.4 - Prob. 81ECh. 4.4 - Prob. 82ECh. 4.4 - Prob. 83ECh. 4.4 - Prob. 84ECh. 4.4 - Prob. 85ECh. 4.4 - Prob. 86ECh. 4.4 - Prob. 87ECh. 4.4 - Prob. 88ECh. 4.5 - Prob. 89UYKCh. 4.5 - Prob. 90UYKCh. 4.5 - Prob. 91UYKCh. 4.5 - Prob. 92UYKCh. 4.5 - Prob. 93UYKCh. 4.5 - Prob. 94UYKCh. 4.5 - Prob. 95UYKCh. 4.5 - Prob. 96ECh. 4.5 - Prob. 97ECh. 4.5 - Prob. 98ECh. 4.5 - Prob. 99ECh. 4.5 - Prob. 100ECh. 4.5 - Prob. 101ECh. 4.5 - Prob. 102ECh. 4.5 - Prob. 103ECh. 4.5 - Prob. 104ECh. 4.5 - Prob. 105ECh. 4.5 - Prob. 106ECh. 4.5 - Prob. 107ECh. 4.5 - Prob. 108ECh. 4.5 - Prob. 109ECh. 4.5 - Prob. 110ECh. 4.5 - Prob. 111ECh. 4.5 - Prob. 112ECh. 4.5 - Prob. 113ECh. 4.5 - Prob. 114ECh. 4.5 - Prob. 115ECh. 4.5 - Prob. 116ECh. 4.5 - Prob. 117ECh. 4.5 - Prob. 118ECh. 4.5 - Prob. 119ECh. 4.5 - Prob. 120ECh. 4.5 - Prob. 121ECh. 4.5 - Prob. 122ECh. 4.5 - Prob. 123ECh. 4 - Prob. 124ECh. 4 - Prob. 125ECh. 4 - Prob. 126ECh. 4 - Prob. 127ECh. 4 - Prob. 128ECh. 4 - Prob. 129ECh. 4 - Prob. 130ECh. 4 - Prob. 131ECh. 4 - Prob. 132ECh. 4 - Prob. 133ECh. 4 - Prob. 134ECh. 4 - Prob. 135ECh. 4 - Prob. 136ECh. 4 - Prob. 137ECh. 4 - Prob. 138ECh. 4 - Prob. 139E
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