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xxMikexx [17]
2 years ago
8

A computer can be classified as either cutting dash edge or ancient. Suppose that 86​% of computers are classified as ancient. ​

(a) Two computers are chosen at random. What is the probability that both computers are ancient​? ​(b) Seven computers are chosen at random. What is the probability that all seven computers are ancient​? ​(c) What is the probability that at least one of seven randomly selected computers is cutting dash edge​? Would it be unusual that at least one of seven randomly selected computers is cutting dash edge​?
Mathematics
1 answer:
Leno4ka [110]2 years ago
5 0

Answer:

(a) The probability that both computers are ancient is 0.7396

(b) The probability that all seven computers are ancient is 0.3479

(c) The probability that at least one of seven randomly selected computers is cutting dash edge is 0.6520.

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

Step-by-step explanation:

We know that 86​% of computers are classified as ancient. This means, if one computer is chosen at random, there is an 86% chance that it will be classified as ancient.

P(ancient)=0.86

(a) To find the probability that two computers are chosen at random and both are ancient​ you must,

The probability that the first computer is ancient is P(ancient)=0.86 and the probability that the second computer is ancient is P(ancient)=0.86

These events are independent; the selection of one computer does not affect the selection of another computer.

When calculating the probability that multiple independent events will all occur, the probabilities are multiplied, this is known as the rule of product.

Let A be the event "the first computer is ancient" and B the event "the second computer is ancient".

P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396

(b) To find the probability that seven computers are chosen at random and all are ancient​ you must,

Following the same logic in part (a) we have

Let A be the event "the first computer is ancient",

B the event "the second computer is ancient",

C the event "the third computer is ancient",

D the event "the fourth computer is ancient",

E the event "the fifth computer is ancient",

F the event "the sixth computer is ancient", and

G the event "the seventh computer is ancient"

P(A\:and \:B\:and \:C\:and \:D\:and \:E\:and \:F\:and \:G)=\\P(A)\cdot P(B)\cdot P(C)\cdot P(D)\cdot P(E)\cdot P(F)\cdot P(G) =(0.86)^7=0.3479

(c) To find the probability that at least one of seven randomly selected computers is cutting dash edge​ you must

Use the concept of complement. The complement of an event is the subset of outcomes in the sample space that are not in the event.

Let C the event "the computer is cutting dash edge".

Let A the event "the seven computers are ancient".

P(C)=1-P(A)=1-0.3479=0.6520

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

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White raven [17]

Answer:

a. 1.0833

Absolute Error = 0.416667

Relative Error = 1.250002

b. 0.0011070

Absolute Error = 0.0011070

Relative Error = 0.003321

Step-by-step explanation:

Given

1/3 = 0.333333

3/4 = 0.75

100/301 = 0.332226

a.

1/3 + 3/4

= 0.333333 + 0.75

= 1.083333

= 1.0833 ------ Approximated to 5 significant digits

Absolute Error = |Real Value - Estimated Value|

Relative Error = Absolute Error/Real Value

Assume 1/3 to be the real value and 3/4 to be the estimated value

Absolute Error = |0.333333 - 0.75|

Absolute Error = |-0.416667|

Absolute Error = 0.416667

Relative Error = 0.416667/0.333333

Relative Error = 1.250002

b.

1/3 - 100/301

= 0.333333 - 0.332226

= 0.001107

= 0.0011070 ----- Approximated to 5 significant digits

Assume 1/3 to be real value and 100/301 to be estimated value

Absolute Error = 0.333333 - 0.332226

Absolute Error = 0.0011070

Relative Error = 0.0011070/0.333333

Relative Error = 0.003321

Absolute and relative errors are approximation errors and they are due to the discrepancy between an exact value and some approximation to them.

The absolute error is the magnitude of the difference between the exact value and the approximation. The relative error is the absolute error divided by the magnitude of the exact value

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Determine the range of the following graph:
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Answer:

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Step-by-step explanation:

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Find the value of x for which ABCD must be a parallelogram.<br> (4x - 1)<br> (x + 38)°
mart [117]

Answer:

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Step-by-step explanation:

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3 years ago
For a certain weekend, the weatherman predicts that it will rain with a 40% probability on Saturday and a 50% probability on Sun
melomori [17]

Answer:

70%

Step-by-step explanation:

The probability it rains on at least one of the days is:

P = P(Sat & not Sun) + P(Sun & not Sat) + P(Sat & Sun)

P = (0.40)(1−0.50) + (0.50)(1−0.40) + (0.40)(0.50)

P = 0.20 + 0.30 + 0.20

P = 0.70

Another way to look at it is 1 − the probability that it doesn't rain on either day.

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P = 1 − (1−0.40)(1−0.50)

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There is a 70% probability that it will rain on at least one day.

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