Question

From past experience it is known that 25% of accounts in a large accounting population are in
error.
a) What is the probability that 5 accounts are audited before an account in error is found?
b) What is the probability that the will be three account in error in the first five accounts
audited?

Answer 1

a) This is a geometric distribution with p=0.75.

$P(X=6)=(1-p)^{6-1}p^1=0.75^5*0.25^1=0.0593.$

b) This is a binomial distribution with p=0.25, n=5.

$P(X=3)=C_5^3p^3(1-p)^2=10*0.75^2*0.25^3=0.0879.$