Probability of High BP & Heart Trouble in Hospital Patients

[nickar1172]

I am not positive if this question has a formula behind it because that's how terrible my math is,

1. A hospital Survey indicates that 35% of patients admitted have high blood pressure, 53% have heart trouble, and 22% have both. Find the probability that a patient admitted to this hospital has:

a) neither high blood pressure nor heart trouble

I did 35% + 53% = 88%, 88% - 100% = 12%

b) high blood pressure or heart trouble, but not both

13% + 31% = 44%

c) high blood pressure given that he has heart trouble

I did not no how to solve this question and guessed 35%

d)heart trouble given that he does not have high blood pressure

same as c) I just guess and put 53%

 

[Barioth]

I'll give you the general formula, Try and see if you can how you can use them.

we'll put
A: The patient has high blood preasure
B: The patient has heart problem

From what you have writen

Can you write down

$P(A), P(B), P(A \cap B)$

Here are some general formula:

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

With those two formula you should be able to solve your problem.

 

[nickar1172]

dude I am bugging out right now can you just please give me the answers to these or plug in the individual formulas for each

 

[Barioth]

dude I am bugging out right now can you just please give me the answers to these or plug in the individual formulas for each

The answer to A- 44%, B- 22%, C- 41.5%

I don't think these answer on their own will help you...

as I said earlier, can you find out the value of

[FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]∩[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main])[/FONT]

 

[HallsofIvy]

Another way to do this: First, to avoid percentages, imagine there are 100 patients. 35 of them have high blood pressure, 53 have heart trouble, and 22 have both. Since the "35" who have high blood pressure includes the "22" who have both, 35- 22= 13 have high blood pressure only. Similarly, 53- 22= 31 have heart trouble only.

So: of 100 patients, 13 have high blood pressure only, 31 gave heart trouble only, and 22 have both. 100- (13+ 31+ 22)= 100- 66= 34 have neither.

Now you can answer:
a) neither high blood pressure nor heart trouble

b) high blood pressure or heart trouble, but not both

c) high blood pressure given that he has heart trouble

d)heart trouble given that he does not have high blood pressure

By taking the total number who fit the description and divide by 100. For example, since 34 have neither high blood pressure nor heart trouble, the answer to (a) is 34/100= 34%. Your "12%" is wrong because your "35+ 53" is the wrong calculation. The 35 and 53 include the 22 who have both and so are counted twice. You could do that calculation as 35+ 53- 22.