anonymous
  • anonymous
You have a torn tendon and are facing arthroscopic surgery to fix it. The surgeon explains the risks of the surgery. Infection occurs in 2% of all cases and the repair fails in 11% of the cases. 0.5% of the time the repair fails and infection occurs. What is the probability that the operation is successful and infection-free?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I have a tree diagram started|dw:1439489950303:dw|
kropot72
  • kropot72
|dw:1439491456952:dw| The first step is to calculate the probability of failure, given that there is infection. \[\large P(F \cap I)=0.005\] \[\large P(F|I)=\frac{P(F \cap I)}{P(I)}=\frac{0.005}{0.02}=0.25\] Then the probability of success, given infection occurs is \[\large P(S|I)=1-P(F|I)=1-0.25=0.75\] Are you able to follow so far?
anonymous
  • anonymous
Yes, but does it not give you the probability of failure and infection? I thought that's what the '0.5% of the time the repair fails and infection occurs,' meant....

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kropot72
  • kropot72
The probability of success and infection is given by \[\large P(S \cap I)=P(I) \times P(S|I)=0.02\times0.75=0.015\] We are given that the repair fails in 11% of all cases, therefore the overall probability of success is 1 - 0.11 = 0.89 We have found that the probability of success and infection is 0.015. Therefore the probability of success and no infection is 0.89 - 0.015 = 0.875 A corrected drawing follows. |dw:1439493603377:dw|
anonymous
  • anonymous
That makes sense..
kropot72
  • kropot72
That's good :)
anonymous
  • anonymous
but at the same time, I can't quite wrap my brain around it. I see what you're saying, and it mathematically makes sense, but I don't quite understand why it is what it is.
kropot72
  • kropot72
The first step is to find the probability of success in the presence of infection. We are given that the probability of failure in the presence of infection is 0.005 (5%). So we need to use conditional probability theory to find the probability of success in the presence of infection, which is found to be 0.015. It is also effectively given that the overall probability of success (including with and without infection) is 0.89 (1 - 0.11), being 1 minus the overall probability of failure. Now we can separate out the probability of success and no infection, by simply subtracting 0.015 from 0.89. Giving 0.875.
anonymous
  • anonymous
Just to see if I followed this properly, the missing number at the end, which would be infection+success would be 0.105|dw:1439495124441:dw|
kropot72
  • kropot72
No so. The probability that the operation is successful and infection-free is already on the probability tree, being 0.875.
anonymous
  • anonymous
Isn't the decimal of a percent such as 5%, 0.05 rather than 0.005 ??
anonymous
  • anonymous
Isn't 0.005 0.5%?
kropot72
  • kropot72
And the probability of infection plus success is also on the tree, being 0.015. The probability of failure and no infection was not on my diagram. the reason being that it is not asked for. "Isn't 0.005 0.5%". Yes it is, that is the conversion from a percentage to a probability.
anonymous
  • anonymous
Okay, thank you for your help. (:
kropot72
  • kropot72
|dw:1439495695074:dw|
kropot72
  • kropot72
You're welcome :)

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