Question 360068
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Hi,
Binomial Problem with n=13; p=0.20  Yes, you are on the right track there.

Note: The probability of x successes in n trials is: 

P = nCx* {{{p^x}}}*{{{q^(n-x)}}} where p and q are the probabilities of success and failure respectively.
In this case p(mild side effect)=.2 & q =.8 
nCx = {{{n!/(x!(n-x!))}}}

(A) exactly three will have this mild side effect
P(3 with mild side effects) = 13C3* {{{.2^3}}}*{{{.8^10}}}
13C3 = 13!/3!10! = 13*12*11/3*2 = 286
P(3 with mild side effects) = 286* {{{.2^3}}}*{{{.8^10}}}

(B) at least five will have this mild side effect. 
1-   P(0)   + P(1)       + P(2)             + P(3)         + P(4) 
1 - .8^13 + 13*.2^1.8^12 + 78*.2^2.8^11 + 286*.2^3.8^10 + 715*.2^4.8^9