Question

Q1.

Theresa has many ICU patients in her care. Some of the patients are on feeding tubes while others are on IVs only. In addition, Theresa knows that some of the patients have diabetes while some don’t. The exact breakdown is as follows:

  (Use these data for this question only.)


                                  Feeding Tube         IV

Has Diabetes                   64                   50

No Diabetes                     5                   34

What is the probability of selecting someone who is on an IV, given that they do not have diabetes?

(If necessary, round answer to 4 decimal places)

Answer 1

We have given,

	Feeding Tube	IV	Total
Has Diabetes	64	50	114
No Diabetes	5	34	39
Total	69	84	153

We have compute,  probability of selecting someone who is on an IV, given that they do not have diabetes

That is P( someone who is on an IV | they do not have diabetes)

We know that

P(A|B)=P(A intersection B)/P(B)

Therefore,

P( someone who is on an IV | they do not have diabetes)=p(someone who is on an IV ,no diabetes) / p(no diabetes)

Here P(someone who is on an IV ,no diabetes)=34/153= 0.2222

P(no diabetes)=39/153= 0.2549

So P( someone who is on an IV | they do not have diabetes)= 0.2222/0.2549= 0.8717

probability of selecting someone who is on an IV, given that they do not have diabetes is 0.8717