A company manufactures fuses. The percentage of non-defective fuses is 95.4%. A sample of 10 fuse was selected. Calculate the probability of selecting at least 3 defective fuses.

Given that the the percentage of non-defective fuses is 95.4%, the probability of a defective fuses is

P(defective)= 1- P(non-defective)

= 100-95.4

= 4.6%

This implies that the percentage of defective fuses, p = 0.046

Each fuse can either be defective or non-defective, which are the only two possible outcomes. Therefore, we will use the binomial distribution which gives the probability of exactly x successes on n repeated trials and is given by:

P(X=x) = (n!/(x!(n-x)!)).px.(1-p)n-x

given that 10 fuses are samples, it implies that n=10

The probability of selecting at least 3 defective fuses:

P(X≥3) = 1 - P(x <3)

where

P(X<3) = P(X=0) + P(X=1) + P(X=2)

By using the formula

P(X=x) = (n!/(x!(n-x)!)).px .(1-p)n-x

P(x=0) = (10!/(0!(10-0)!)) x (0.046)0 .(0.954)10 = 0.6244

P(x=1) = (10!/(1!(10-1)!)).(0.046)1 .(0.954)9 = 0.3011

P(x=2) = (10!/(2!(10-2)!)).(0.046)2 .(0.954)8 = 0.0653

Thus

P(X<3) = 0.6244+0.3011+0.0653 = 0.9908

and

P(X≥3) = 1 - P(x <3)

= 1 - 0.9908

= 0.0092

= 0.92%

Answer: There is 0.92% probability of selecting at least 3 defective fuses