Question

A bin contains 63 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.

Answer 1

Since there are $9$ defective bulbs, there are $63-9=54$ good ones.

So, the probability of choosing a good one with the first draw is $\frac{54}{63} = \frac{6}{7}$. This holds for all draws, since the replacement resets the experiment each time.

So, the answer is $\left( \frac{6}{7} \right)^3 \approx 0.630$

Answer 2

With replacement, probability of selecting a good bulb remains

(63-9)/63

=6/7

If the bulbs are selected WITH replacement three times, then

P(all good)

= (6/7)^3 (by the multiplication rule)

= 216/343

= 0.630 (to three decimal places)