Question

A bin contains 78 light bulbs of which 9 are defective. If 3 light bulbs are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones. Round to the nearest thousandth if necessary.
0.692

0.002

0.721

0.885

Answer 1

Answer:

$P(3 Good) = (\frac{69}{78})^3 =0.692$

Step-by-step explanation:

For this case we have a total os light bulbs 78 and 9 of them are defective, that means 78-9 =69 are not defective. We can find the probability that a buld is defective or no like this:

$P(D) = \frac{9}{78}=0.115$

$P(ND) = \frac{69}{78}=0.885$

And for this case we want to find the probability that all the 3 bulbs selected would be good one, since the experiment is with replacement we can assume independence between the events of obtain a non defective bulb and we can find the probability like this:

$P(3 Good) = (\frac{69}{78})^3 =0.692$

Answer 2

With replacement:

P(GGG)=[(78-9)/78]^3

P(GGG)=(23/26)^3

P(GGG)=12167/17576

P(GGG)≈0.692