What is the Probability of Knee Problems Requiring Full Knee Replacement?

[pckofwolfs]

For b and c I have no idea where to go. I know for part b I need the probability of the full knee replacement, but that's all I know. I have no idea what needs to be done for part c. If someone could help that would be greatly apprecieted. I think part a is right. If not could it be explained why it is wrong.According to the American Academy of Orthopedic Surgeons, about 26% of orthopedic surgery
involves knee problems. More than two-thirds of the surgeries involve full knee replacements.
*Round your answer to the nearest percent if necessary*

Age of Adults Getting Knee Replacements
Age Percentage
18-44 2.8%
45-64 24.6%
65-74 43.3%
75-84 26.7%
85-older 2.9%

a) What is the probability that an orthopedic case selected at random involves knee
problems? [1]
The answer I put here was 26%.

b) Of those cases, from part (a), estimate the probability that the case requires full knee
replacement. [1]c) Compute the probability that an orthopedic case selected at random involves a knee
problem and requires a full knee replacement.

 

[I like Serena]

For b and c I have no idea where to go. I know for part b I need the probability of the full knee replacement, but that's all I know. I have no idea what needs to be done for part c. If someone could help that would be greatly apprecieted. I think part a is right. If not could it be explained why it is wrong.According to the American Academy of Orthopedic Surgeons, about 26% of orthopedic surgery
involves knee problems. More than two-thirds of the surgeries involve full knee replacements.
*Round your answer to the nearest percent if necessary*

Age of Adults Getting Knee Replacements
Age Percentage
18-44 2.8%
45-64 24.6%
65-74 43.3%
75-84 26.7%
85-older 2.9%

a) What is the probability that an orthopedic case selected at random involves knee
problems? [1]
The answer I put here was 26%.

b) Of those cases, from part (a), estimate the probability that the case requires full knee
replacement. [1]c) Compute the probability that an orthopedic case selected at random involves a knee
problem and requires a full knee replacement.

Hi pckofwolfs! Welcome to MHB! (Smile)

This looks to be a bit of a trick question.

So of all orthopedic surgeries about 26% involve knee problems.
And of those cases, more than two-thirds involve full knee replacements.
That looks to be the answer to b), doesn't it? (Wondering)

For c) we need to combine those two chances.
How can we combine them to get an answer that makes sense? (Wondering)

 

[pckofwolfs]

So for part b would I do something like 26 x (2/3) then subtract that answer from 26 to tell me how many are full knee?

 

[I like Serena]

So for part b would I do something like 26 x (2/3) then subtract that answer from 26 to tell me how many are full knee?

No, if we only look at the cases that involve knee surgery, we have over (2/3) knee replacement.
That means that the estimate for b) is simply (2/3).

 

[pckofwolfs]

so, for c I'd have (13/50) + (2/3) to be 139/150?

 

[I like Serena]

so, for c I'd have (13/50) + (2/3) to be 139/150?

That doesn't look right. (Shake)

The probability of a knee problem is 26%.
How can the probability of a knee problem AND a knee replacement be bigger than that? (Worried)

 

[pckofwolfs]

yeah, I have no idea how to do that then. I am taking the AND as a hint to use the and formulas I have say to do p(knee) * p(full). Which you said wouldn't be right.

 

[I like Serena]

yeah, I have no idea how to do that then. I am taking the AND as a hint to use the and formulas I have say to do p(knee) * p(full). Which you said wouldn't be right.

Actually, that is (more or less) right!
But that's not what you did...

Properly, it should be p(knee) * p(full given knee).

 

[pckofwolfs]

Therefore, 13/75 would be the answer?

 

[I like Serena]

Therefore, 13/75 would be the answer?

Yep. (Nod)