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Question 656790: In testing a new drug, reseachers found that 5% of all patients using it will have a mild side effect. A random sample of 12 patients using the drug is selected. Find the probability that:
(A) EXACTLY TWO will have this mild side effect.
(B) AT LEAST two will have this mild side effect.
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This is my work ------>
n= 12
p=0.05
mean=n*p
mean= 12*0.05
result is: mean=0.6 am I on the right track or off the tracks? with explaination so I can learn this and quit bugging you guys. Please help!!!!
Found 2 solutions by solver91311, ewatrrr: Answer by solver91311(24713) (Show Source): Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!
Hi,
Re: TY, w/o capability of using Your own calculator, the only option would be using an on-line site like the following,
which is relatively uncomplicated calculator to enter the information given:
http://stattrek.com/online-calculator/binomial.aspx
P(x = 2) = .0988 (rounding to 4 decimal places)
P(x≥ 2) = .1184
Important thing to understand is with only 5% having a mild side effct...
Probability of even 2 out of 12 people having a side effect is very small.
This is a case of Bionomial Probability Distribution
Note: The probability of x successes in n trials is:
P = nCx* where p and q are the probabilities of success and failure respectively.
this case p = .05 & q = .95 and n = 12
nCx = 
P(x = 2) = 12C2(.05)^2(.95)^10 0r binompdf(12,.05,2) using TI83
P(x ≥ 2) = 1- P(x <2) = 1-P(0) - P(1)
0r 1 - binomcdf(12,.05,1) using TI83
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