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ASQ CSSBB Exam - Topic 6 Question 124 Discussion

Posted by Anthony on Apr 12, 2026 at 05:21 PM

A medicine with efficacy of .52 is given to five patients. Find the approximate probability that at least one of the patients is cured. (Hint: Use the binomial formula.)

Suggested Answer: A

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20 comments · newest first
JO
Jovita 1 month ago
I’m leaning towards option A, .975. It feels right based on my calculations!
KE
Kenneth 1 month ago
Exactly! That gives us the probability of at least one being cured.
GL
Glynda 2 months ago
So, we do (0.48)^5 for none being cured, then subtract from 1?
YU
Yun 2 months ago
Right! If the efficacy is .52, then the chance of not being cured is .48.
EL
Eloisa 2 months ago
Yeah, I feel like I need to calculate the probability of none being cured first.
IS
Isaiah 2 months ago
This question is tricky! I think it's about using the binomial formula.
ST
Stephen 2 months ago
Binomial formula is key here, definitely use it!
AR
Arthur 2 months ago
Totally agree, .975 makes sense with those odds!
LE
Leah 3 months ago
I don’t think it’s that high, sounds off.
KA
Katina 3 months ago
Wait, how do you get that number?
LI
Lina 3 months ago
The probability of at least one cured is about .975!
SO
Sommer 3 months ago
Efficacy of .52 means we should expect some cures.
NE
Nelida 3 months ago
Totally agree with .975!
RO
Rodrigo 4 months ago
Wait, how can it be that high?
DO
Douglass 5 months ago
I think it’s closer to .531.
YU
Yuki 5 months ago
The binomial formula is key here!
CA
Candra 6 months ago
I’m a bit confused about the numbers. Is .52 the probability of success or failure? I hope I remember this correctly!
TA
Tawna 6 months ago
I feel like I might have seen a similar question where we calculated the probability of success and failure.
PE
Peter 6 months ago
I think we need to find the probability of none being cured first, then subtract from 1. That sounds familiar from practice problems.
LA
Lavonna 6 months ago
I remember we talked about the binomial formula in class, but I'm not sure how to apply it here.
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