This is a binomial distribution case where p=0.05,n=6, q=1-0.05=0.95
Let X denote the random variable for the number of fuses.
"P(X=x) ={n\\choose x} p^x q^{n-x}"
a) p(X=1)
"={6\\choose 1}{0.05^1}{0.95^5}"
=0.2321
b) "P(X \\geq 1)"
"={\\sum_{x=1}^6} 0.05^x 0.95^{6-x}"
=0.2649
c)"P(X>1|X\\geq 1)=\\frac {p(x>1)}{p(X\\geq 1}"
P(X>1)"={\\sum_{x=2}^6} 0.05^x 0.95^{6-x}"
=0.0327
"P(X>1|X\\geq 1)=\\frac{0.0327}{0.2649}"
=0.1237
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