Respuesta :
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.
[tex]P(ancient)=0.86[/tex]
(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 [tex]P(ancient)=0.86[/tex] and the probability that the second computer is ancient is [tex]P(ancient)=0.86[/tex]
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".
[tex]P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396[/tex]
(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"
[tex]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[/tex]
(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".
[tex]P(C)=1-P(A)=1-0.3479=0.6520[/tex]
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.
Using the binomial distribution, it is found that:
a) 0.7396 = 73.96% probability that both computers are ancient.
b) 0.3479 = 34.79% probability that all seven computers are ancient.
c) 0.6521 = 65.21% probability that at least one of seven randomly selected computers is cutting dash edge. Since this probability is greater than 5%, it would not be unusual that at least one of seven randomly selected computers is cutting dash edge.
---------------------------------------
For each computer, there are only two possible outcomes. Either it is ancient, or it is not. The probability of a computer being classified as ancient is independent of any other computer, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
It is the probability of exactly x successes on n repeated trials, given by:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameter p is the probability of a success on a single trial.
- For this problem, 86% of the computers are classified as ancient, thus [tex]p = 0.86[/tex].
Item a:
- 2 computers, thus [tex]n = 2[/tex].
- The probability is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.86)^{2}.(0.14)^{0} = 0.7396[/tex]
0.7396 = 73.96% probability that both computers are ancient.
Item b:
- 7 computers, thus [tex]n = 7[/tex].
- The probability is P(X = 7).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{7,7}.(0.86)^{7}.(0.14)^{0} = 0.3479[/tex]
0.3479 = 34.79% probability that all seven computers are ancient.
Item c:
The probability of at least one cutting dash edge is the probability that not all area ancient, that is:
[tex]P(X < 7) = 1 - P(X = 7) = 1 - 0.3479 = 0.6521[/tex]
0.6521 = 65.21% probability that at least one of seven randomly selected computers is cutting dash edge. Since this probability is greater than 5%, it would not be unusual that at least one of seven randomly selected computers is cutting dash edge.
A similar problem is given at https://brainly.com/question/13836999
