A bin contains 71 light bulbs of which 9 are defective. If 6 light bulbs are randomly selected from the bin with replacement, how do you find the probability that all the bulbs selected are good ones?
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"Find the sum of the integers between 2 and 100 which are divisible by 3 ?"
#(62/71)^6# But to avoid the curse of the lazy-monster, you must earn the treasure hidden in the cave (the answer fully simplified)... by reading the explanation.
If 9 of the 71 bulbs are defective, then 62 are good. That gives you a #62/71# chance of getting a good one once. You said that they were replaced, so that means that each time you get a bulb, it is independent from the previous draws, because taking a bulb out and putting it back in doesn't affect the ratio of good to defective.
So, the second time you draw, it will still be a #62/71# chance of a good bulb. However, if you find the chances of getting 2 in a row, you'll see that the 2nd draw is a #62/71# chance after getting a #62/71# chance. So, it's actually #62/71# of #62/71#, or #(62/71 * 62/71)#.
Repeat this for 6 times, because you draw six times, and you get #(62/71)^6# which equates to #0.4434# or #44.34# percent.
