Question

A package of 6 fuses are tested where the probability an individual fuse is defective is
0.05. (That is, 5% of all fuses manufactured are defective). What is the probability
that more than one fuse will be defective, given that at least one is defective?

Answer 1

$P(X>1|X\ge1)=\frac {P(X>1) \bigcap P(X\ge 1)}{P(X\ge 1)}$

=$\frac {P(X>1)}{P(X\ge 1)}$

For a binomial probability,

$P(X=x) ={n\choose x} *p^x *q^{n-x}$

P(X>1)=$\sum_{i=2}^6{6\choose x} *0.05^x *0.95^{6-x}$

=0.0328

$P(X\ge 1)=\sum_{i=1}^6{6\choose x} *0.05^x *0.95^{6-x}$

=0.2649

Then

$\frac {0.0328}{0.2649}$

=0.1238