word problem involving percents

E

eric beans

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Approximately 1 in 14 men older than the age of 50 has prostate cancer. The level of 'prostate specific antigen' (PSA) is used as a preliminary screening test for prostate cancer. 7% of men with prostate cancer do not have a high level of PSA. These results are known as 'false negatives'. 75% of those men with a high level of PSA do not have cancer. These results are known as 'false positives'. If a man over 50 has a normal level of PSA, what are the chances that he has prostate cancer?

A 7%
B 25%
C 5%
D 0.7%
E 0.5%

Answer D

why?

What am I misunderstanding here?
1/14*.07 = .005 = .5%
 
eric beans said:
Approximately 1 in 14 men older than the age of 50 has prostate cancer. The level of 'prostate specific antigen' (PSA) is used as a preliminary screening test for prostate cancer. 7% of men with prostate cancer do not have a high level of PSA. These results are known as 'false negatives'. 75% of those men with a high level of PSA do not have cancer. These results are known as 'false positives'. If a man over 50 has a normal level of PSA, what are the chances that he has prostate cancer?

A 7%
B 25%
C 5%
D 0.7%
E 0.5%

Answer D

why?

What am I misunderstanding here?
1/14*.07 = .005 = .5%
Click to expand...
7% of men with prostate cancer do not have a high level of PSA--that is given. But why are you multiplying 1/ 14 by 7% (or .07)? What is your logic. Please be clear.
 
7% with "normal" levels have cancer. Question is asking if you have normal levels, what are your chances of having cancer.

What I don't understand is how did they get 0.7% for the answer.
 
eric beans said:
7% with "normal" levels have cancer. Question is asking if you have normal levels, what are your chances of having cancer.

What I don't understand is how did they get 0.7% for the answer.
Click to expand...
Please think about it. Why do you think that it is wrong?
 
I don't know.

How would you answer this question? What's your thought behind it.
 
What you found is the probability of having cancer AND normal level, not of having cancer GIVEN normal level.
 
So how would you go about finding it for a normal level? So it would be just straight 1/14?
 
THINK, and tell us your thoughts.

Have you learned about Bayes Theorem?

Have you tried making a table like this?

[MATH]\begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & & & 1/14 \\ \text{not C} \\ \hline \text{total}& & &{1} \end{array}[/MATH]
 
Oh. maybe that's the problem. I haven't had bayes theorem yet. Please explain. All I've been taught/learned is simple probability calculations using "and" "or" to multiple or add different odds.
 
Please tell us what you HAVE learned and are expected to use here. I didn't say Bayes is necessary here, just that this sort of problem is often taught in that context.
 
basic high school probability stuff. i didn't think this would involve a PhD in math to solve.
 
" If a man over 50 has a normal level of PSA, what are the chances that he has prostate cancer? "
"7% of men with prostate cancer do not have a high level of PSA."
....and 1/14 have prostate cancer.

sooo of the people who have cancer, 7% of them have normal levels of PSA....so 1/14*7% was my thinking.
 
7% of men with prostate cancer do not have a high level of PSA---translate this into a probability of some event.
75% of those men with a high level of PSA do not have cancer --translate
1 in 14 men older than the age of 50 has prostate cancer-----translate
 
1 in 14 men older than the age of 50 has prostate cancer-----of the total population say out of 1000 people, only 1/14 have prostate cancer....so only 70 people would have this condition.
7% of men with prostate cancer do not have a high level of PSA--- out of a population of 1000 people, the 70 who do have cancer, of those 7% didn't show high levels of PSA. so (1/14*7%=.005) 5 people showed no signs of high psa but do have cancer.
75% of those men with a high level of PSA do not have cancer -- of the total population say out of 1000 people, however many of them have high psa, 75% of that group do not have cancer. so that must mean that 25% of those with high levels must account for the other (70-5 = 65) 65 people who did have high psa AND have cancer.

....so that must mean, 65*4 =260 people tested with high psa levels total.

that must mean out of a 1000 population, 1000-260 = 740 didn't test high for PSA...but had "normal" levels.

looking back at the question..." If a man over 50 has a normal level of PSA, what are the chances that he has prostate cancer? " , well i know that 740 have normal level PSA. ...now what? Oh i got it! 5/740 = .0067 rounded off to .7%!

the problem was in my initial calculations wasn't using the "normal level" population. I was using the WHOLE population both high and low psa population total. right?

holy crap! that's a ALOT of calculations! how does this relate to beyes? what is beyes??? what portion of what i did referred to as beyes???


what's the faster way of solving this???
 
Last edited:
1000/14 is not 70. Note that 14*70=980
Without rounding off until the very end I got .0068 so your answer looks good.
Don't you think it was better doing it yourself?
I think filling in Dr Perterson's table is the best way to do this problem, but not necessarily the fastest.
 
Last edited:
Here's how I do it, using the table:

Approximately 1 in 14 men older than the age of 50 has prostate cancer: fill in total with cancer, C = 1/14 = 0.071428.
7% of men with prostate cancer do not have a high level of PSA: fill in "C and not P" with 0.07*0.071428 = 0.005.

[MATH]\displaystyle \begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & & 0.005 & 0.071428 \\ \text{not C} \\ \hline \text{total}& & &{1} \end{array}[/MATH]
Now subtract to fill in "C and P": 0.071428 - 0.005 = 0.066428.
75% of those men with a high level of PSA do not have cancer: "P and not C" is 1/4 of P, so P = 4*0.066428 = 0.265714.
Subtract to get "P and not C": 0.265714 - 0.066428 = 0.199285.

[MATH]\displaystyle \begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & 0.066428 & 0.005 & 0.071428 \\ \text{not C} & 0.199285 \\ \hline \text{total}& 0.265714 & &{1} \end{array}[/MATH]
Now subtract to fill in the missing totals: 1 - 0.265714 = 0.734286; 1 - 0.071428 = 0.928572.

[MATH]\displaystyle \begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & 0.066428 & 0.005 & 0.071428 \\ \text{not C} & 0.199285 & & 0.928572 \\ \hline \text{total}& 0.265714 & 0.734286 &{1} \end{array}[/MATH]
We could fill in the missing space, but don't need to. P(C | not P) = P(C and not P)/P(not P) = 0.005/0.734286 = 0.0068, which is our answer.

Yes it takes a lot of calculations; you just have to be organized to keep track of it.
 
Dr.Peterson said:
Here's how I do it, using the table:

Approximately 1 in 14 men older than the age of 50 has prostate cancer: fill in total with cancer, C = 1/14 = 0.071428.
7% of men with prostate cancer do not have a high level of PSA: fill in "C and not P" with 0.07*0.071428 = 0.005.

[MATH]\displaystyle \begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & & 0.005 & 0.071428 \\ \text{not C} \\ \hline \text{total}& & &{1} \end{array}[/MATH]
Now subtract to fill in "C and P": 0.071428 - 0.005 = 0.066428.
75% of those men with a high level of PSA do not have cancer: "P and not C" is 1/4 of P, so P = 4*0.066428 = 0.265714.
Subtract to get "P and not C": 0.265714 - 0.066428 = 0.199285.

[MATH]\displaystyle \begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & 0.066428 & 0.005 & 0.071428 \\ \text{not C} & 0.199285 \\ \hline \text{total}& 0.265714 & &{1} \end{array}[/MATH]
Now subtract to fill in the missing totals: 1 - 0.265714 = 0.734286; 1 - 0.071428 = 0.928572.

[MATH]\displaystyle \begin{array}{*{20}{c}} \text { }&\text{P}&\text{not P}&\text{total} \\ \hline \text {C} & 0.066428 & 0.005 & 0.071428 \\ \text{not C} & 0.199285 & & 0.928572 \\ \hline \text{total}& 0.265714 & 0.734286 &{1} \end{array}[/MATH]
We could fill in the missing space, but don't need to. P(C | not P) = P(C and not P)/P(not P) = 0.005/0.734286 = 0.0068, which is our answer.

Yes it takes a lot of calculations; you just have to be organized to keep track of it.
Click to expand...
so what exactly is beyes theory in a nut shell? subtracting percentages from a whole 1?
 
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