The probability of nonoccurrence of event A .

Question

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Answer 1

Question

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution

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¯)=1−P(A)=1−0.417=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

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

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

(d)

To determine

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

(d)

Expert Solution

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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

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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Answer 2

Question

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution

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¯)=1−P(A)=1−0.417=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

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

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

(d)

To determine

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

(d)

Expert Solution

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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

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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Answer 3

Question

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution

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¯)=1−P(A)=1−0.417=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

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

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

(d)

To determine

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

(d)

Expert Solution

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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

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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Answer 4

Question

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution

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¯)=1−P(A)=1−0.417=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

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

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

(d)

To determine

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

(d)

Expert Solution

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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

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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Answer 5

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution

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¯)=1−P(A)=1−0.417=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

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

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

(d)

To determine

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

(d)

Expert Solution

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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

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.

Answer 6

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

(a)

Expert Solution

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¯)=1−P(A)=1−0.417=0.583

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

Answer 7

Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event A.

Answer 8

(a)

Expert Solution

Answer to Problem 18E

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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¯)=1−P(A)=1−0.417=0.583

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

Answer 13

(b)

To determine

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

(b)

Expert Solution

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

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

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.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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

Answer 20

(c)

To determine

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

(c)

Expert Solution

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

Answer 21

(c)

To determine

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

Answer 22

(c)

Expert Solution

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.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

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]=1−P[0 Heads and 0 Tails]=1−[P(HHHH)+P(TTTT)]=1−0.125=0.875

Answer 27

(d)

To determine

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

(d)

Expert Solution

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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.

Answer 28

(d)

To determine

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

Answer 29

(d)

Expert Solution

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.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

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(A∪B)=P(A)+P(B)−P(A∩B)

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

So

P(A∪B)=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.

Answer 34

(e)

To determine

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

(e)

Expert Solution

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.

Answer 35

(e)

To determine

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

Answer 36

(e)

Expert Solution

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.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

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.