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)=(xn)pxqn−x
a) p(X=1)
=(16)0.0510.955
=0.2321
b) P(X≥1)
=∑x=160.05x0.956−x
=0.2649
c)P(X>1∣X≥1)=p(X≥1p(x>1)
P(X>1)=∑x=260.05x0.956−x
=0.0327
P(X>1∣X≥1)=0.26490.0327
=0.1237
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