Question

A box of 30 diodes is known to contain five defective ones. If two diodes are selected at random without replacement, what is the probability that at least one of these diodes is defective?

Answer 1

0.31

Step-by-step 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)=

`Pr=(C(25,2))/(C(30,2))\\`

`Pr=(300)/(435)\\ Pr=0.69\\`

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