If the probability is 0.001 that a light bulb will fail on any given day, what is the probability that it will last at least 30 days?
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there are no new concepts from the other thread you posted. Can you see that the probability of it surviving 2 days is 0.999 * 0.999 And for 3 days its 0.999 * 0. 999 * 0.999 Can you generalise this to 30 days?
Originally Posted by ilikec If the probability is 0.001 that a light bulb will fail on any given day, what is the probability that it will last at least 30 days? Pr(at least 30 days) = Pr(30 days) + Pr(31 days) + Pr(32 days) + ..... where: Pr(30 days) = (0.999^30)(0.001) (why?) Pr(31 days) = (0.999^31)(0.001) (why?) etc. You now have an infinite geometric series, and the formula for finding such a sum is well known.
Originally Posted by mr fantastic Pr(at least 30 days) = Pr(30 days) + Pr(31 days) + Pr(32 days) + ..... where: Pr(30 days) = (0.999^30)(0.001) (why?) Pr(31 days) = (0.999^31)(0.001) (why?) etc. You now have an infinite geometric series, and the formula for finding such a sum is well known. The answer just comes out to , the same as by using SpringFan25's reasoning.
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