# Events and Sets

#### nickar1172

##### New member
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

##### Member
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

##### New member
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

##### Member
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%

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

##### Well-known member
MHB Math Helper
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.