Question 368260
First we find the probability that in all three consecutive test 

fuses are not defective.



probability of first fuse is not defective = 5/10  (in total 10, 5 no defetive)


as now there are 4 non-defective fuse remain in total 9.

probability of second fuse is not defective = 4/9 (in total 9, 4 non defective)

probability of third fuse is not defective = 3/8  (in total 8, 3 non defective)

thus probability = 5/10 * 4 /9 * 3/8 = 1/12


we know p' = 1-p

thus the probability of not happening it = 1 - 1/12 =  11/12



probability that at least one of 3 test will fail = 11/12