Asked by monica
n=18,x=14,p=0.75 using binomial probablility
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Answered by
MathMate
Given X~Bin(n=18,p=0.75)
and let C(n,r)=n!/(r!(n-r)!)
P(X=14)=C(n,X)*p^X*(1-p)^(n-X)
=(n!/X!(n-X)!)*p^X*(1-p)^(n-X)\
=(18!/(14!(4!)) * 0.75^14 * 0.25^4
=0.2 approx.
and let C(n,r)=n!/(r!(n-r)!)
P(X=14)=C(n,X)*p^X*(1-p)^(n-X)
=(n!/X!(n-X)!)*p^X*(1-p)^(n-X)\
=(18!/(14!(4!)) * 0.75^14 * 0.25^4
=0.2 approx.
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