1. ## failure time

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?

2. 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?

3. 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.

4. 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 $\displaystyle 0.999^{30}$, the same as by using SpringFan25's reasoning.