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)
Since there are [tex] 9 [/tex] defective bulbs, there are [tex] 63-9=54 [/tex] good ones.
So, the probability of choosing a good one with the first draw is [tex] \frac{54}{63} = \frac{6}{7} [/tex]. This holds for all draws, since the replacement resets the experiment each time.
