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2 votes
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.

2 Answers

4 votes

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
(54)/(63) = (6)/(7). This holds for all draws, since the replacement resets the experiment each time.


So, the answer is
\left( (6)/(7) \right)^3 \approx 0.630
User Ritesh Adulkar
by
8.8k points
6 votes

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)

User Samuel Zhang
by
7.7k points
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