Clinical Probability Question (HY Behavioral Science)

[cfcrml]

Can someone clarify this for me? I'm REALLY bad at math.

on p.112 on Fadem's HY Behavioral Science (2nd edition):

20% of patients who take a new drug for hypertension develop nausea. If two patients (patients A and B) take the drug, calculate the probability that at least one patient (i.e. either A or B or both A and B) will experience nausea.

The explanation says that you should add the probability of patient A experiencing nausea to the probability of B experiencing nausea and then subtract the probability of both A and B experiencing nauseas, so
0.20 + 0.20 -0.04 = 0.36
(0.04 is the probability that both A and B will experience nausea).

My question is, why do you subtract 0.04? Don't you add it to 0.2 and 0.2? For the life of me, I can't figure this one out. Thanks for your help!!!

---

Members don't see this ad.

 

[bob sacamano]

Can someone clarify this for me? I'm REALLY bad at math.

on p.112 on Fadem's HY Behavioral Science (2nd edition):

20% of patients who take a new drug for hypertension develop nausea. If two patients (patients A and B) take the drug, calculate the probability that at least one patient (i.e. either A or B or both A and B) will experience nausea.

The explanation says that you should add the probability of patient A experiencing nausea to the probability of B experiencing nausea and then subtract the probability of both A and B experiencing nauseas, so
0.20 + 0.20 -0.04 = 0.36
(0.04 is the probability that both A and B will experience nausea).

My question is, why do you subtract 0.04? Don't you add it to 0.2 and 0.2? For the life of me, I can't figure this one out. Thanks for your help!!!

the easiest way for me to conceptualize this problem is to calculate the probability that NEITHER patient gets nausea, then subtract that from 1. that will leave all other scenarios possible (pt A only gets nausea, pt B only, both get nausea). the probability of one not getting nausea is 0.8, so the prob of BOTH patients NOT having nausea is 0.8 x 0.8 = 0.64. subtract that from 1 to get 0.36

 

[cfcrml]

Thanks, that helps! But I still don't understand why you would SUBTRACT 0.04 from 0.2 and 0.2.

statistically speaking, the best way to solve this problem is to calculate the probability that NEITHER patient gets nausea, then subtract that from 1. that will leave all other scenarios possible (pt A only gets nausea, pt B only, both get nausea). the probability of one not getting nausea is 0.8, so the prob of BOTH patients NOT having nausea is 0.8 x 0.8 = 0.64. subtract that from 1 to get 0.36

 

[bry2781]

i hope this makes some sense... The 0.2 accounts for any time that someone is affected (be this alone or along with a second person). so, the 0.2 plus 0.2 accounts for the situation where either of the two people in question are affected. The answer to the question can't be anymore than this value. you cant be affected a second time. thus, you have to subtract the instances that BOTH A and B are affect (0.2 x 0.2), because those instances are already accounted for in the 0.2 + 0.2 figure you arrived at earlier. did that make sense? hope so (for your sake and mine... )

 

[cfcrml]

Yes! Thank you!!

i hope this makes some sense... The 0.2 accounts for any time that someone is affected (be this alone or along with a second person). so, the 0.2 plus 0.2 accounts for the situation where either of the two people in question are affected. The answer to the question can't be anymore than this value. you cant be affected a second time. thus, you have to subtract the instances that BOTH A and B are affect (0.2 x 0.2), because those instances are already accounted for in the 0.2 + 0.2 figure you arrived at earlier. did that make sense? hope so (for your sake and mine... )

 

[AlternateSome1]

I always tend to think of ven diagrams. You can picture the two .2 circles overlapping and then take the surface area of the entire figure rather than either individual circle's area.