Answer:
0.31
Explanation:
From the information giving,
total number of diodes=30
total number of defective diodes=5,
total number of not defective diode=30-5=25
to determine the probability of at least one of the selected diodes is defective , we use the the equation blow
P(probability that at least one of the bulb is defective) = 1 - P(probability that none of the bulb is defective)
So we determine the probability of picking a none defective diode,
since we have 25 non-defective diode out of 30 diode, the probability of picking a none defective diode is
P(probability that none of the bulb is defective)=
[tex]Pr=\frac{C(25,2)}{C(30,2)}\\[/tex]
[tex]Pr=\frac{300}{435}\\ Pr=0.69\\[/tex]
Hence
P(probability that at least one of the bulb is defective) = 1 - P(probability that none of the bulb is defective)
P(probability that at least one of the bulb is defective)=1-0.69
P(probability that at least one of the bulb is defective)=0.31
