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Suppose that you have torn a tendon and are fac- ing surgery to repair it. The orthopedic surgeon explains the risks to you. Infection occurs in 3% of such operations, the repair fails in 14%, and both infection and failure occur together 1% of the time. What is the probability that the operation is successful for someone who has an operation that is free from infection?(a) 0.8342 (b) 0.8400 (c) 0.8600 (d) 0.8660 (e) 0.9900

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Answer:

D) 0.8660

Step-by-step explanation:

Infection occurs in 3% of operations.

Repair fails in 14% of operations.

Both failure and infection occurs together in 1%

P(infection) = 0.03

P(repair fails) = 0.14

P(infection and repair fails) = 0.01

From general addition rule

P(A or B) = P(A) + P(B) - P(A and B)

P(infection or repair fail) = P(infection) + P(repair fails) - P(infection and repair fails)

= 0.03 + 0.14 - 0.01

= 0.17 - 0.01

= 0.16

Using complement rule

P(A') = 1 - P(A)

P(no infection and repair successful) = 1 - P(infection or repair fail)

= 1 - 0.16

= 0.84

P(no infection) = 1 - P(infection)

= 1 - 0.03

= 0.97

Using conditional probability

P(B|A) = P(A and B) / P(A)

P(repair successful | no infection) = P(no infection and repair successful) / P(no infection)

= 0.84/0.97

= 0.8660(D)

lhmarianateixeira

In this exercise we have to use the knowledge of probability and thus calculate the success of each given operation, in this way we will find that:

Letter D

From the probability values ​​informed in the text we found that:

  • Infection occurs in 3% of operations.
  • Repair fails in 14% of operations.
  • Both failure and infection occurs together in 1%

Using the probability formula for one of the cases previously reported, we have:

  • [tex]P(infection) = 0.03[/tex]
  • [tex]P(repair fails) = 0.14[/tex]
  • [tex]P(infection\ and \ repair \ fails) = 0.01[/tex]

From general addition rule:

[tex]P(A or B) = P(A) + P(B) - P(A and B)[/tex]

Substituting the equations already known and read previously, and we find that:

[tex]P(infection \ or \ repair \ fail) = P(infection) + P(repair fails) - P(infection \ and \ repair \ fails)[/tex]

Putting the known values ​​in the formula above, we have:

[tex]= 0.03 + 0.14 - 0.01\\= 0.17 - 0.01\\= 0.16[/tex]

Using complement rule:

[tex]P(A') = 1 - P(A)[/tex]

Substituting the equations already known and read previously, and we find that:

[tex]P(no\ infection\ and\ repair\ successful) = 1 - P( \ infection\ or \ repair\ fail)\\= 1 - 0.16\\= 0.84\\P(no infection) = 1 - P(infection)\\= 1 - 0.03\\= 0.97[/tex]

Using conditional probability:

[tex]P(B|A) = P(A and B) / P(A)[/tex]

Putting the known values ​​in the formula above, we have:

[tex]P(repair\ successful | no\ infection) = P(no\ infection\ and \ repair\ successful) / P(no \ infection)\\= 0.84/0.97\\= 0.8660[/tex]

See more about probability at brainly.com/question/795909

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