in testing a new drug, researchers found that 10% 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 three will have this mild side effect
B. at least three will have this mild side effect
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OpenStudy (amistre64):
is 12 enough of a count?
OpenStudy (amistre64):
(12 C 3) .10^3 .90^9 maybe?
OpenStudy (amistre64):
at least 3 is 1-"2"
OpenStudy (amistre64):
if i hit the right buttons, id say A is .0852
OpenStudy (amistre64):
B might be .7699 if im remembering it right
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OpenStudy (amistre64):
there he is :)
OpenStudy (anonymous):
binomial this one
OpenStudy (amistre64):
yay!! i was right, basically
OpenStudy (dboyette):
exactly three is 0.85 and at least 3 would be 0.025? Am i right?
OpenStudy (anonymous):
\[P(X=3)=\dbinom{12}{3}(.1)^3(.9)^9\]
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OpenStudy (amistre64):
1 - exactly2
OpenStudy (amistre64):
.... or am i not hitting the right synapses
OpenStudy (amistre64):
1-P(1)-P(2) is what i got in me head
OpenStudy (anonymous):
at least three
compute
none
\[(.9)^{12}\]
1
\[12\times .1\times (.9)^{11}\] and 2
\[66(.1)^2(.9)^{10}\]
add them up and subtract the result from 1
OpenStudy (anonymous):
@ amistre you forgot
\[P(X=0)\] you need that one too
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