Question

A manufacturer is interested in reliability of the light bulbs it produces. Light bulbs
fail if they are not functioning when they are switched on. Suppose that the probability of
failure each time the light bulb is switched on is p and the performance is independent
form one trial to another.
(a) The manufacturer is interested in the probability that the light bulb functions at least
50 times. For what values of p is this probability at least 90%?

Answer 1

This would be a geometric distribution. You can think of it as 50 is the number of "failures" until the first "success> Although, here the success is defined as the light bulb not functioning. ("Failure" and "Success" here are just mathematical terms to define the geometric distribution... where you have an "event" occurring multiple times in a row, until the "opposite event" FIRST occurs). And the probability of success here is \(p\).

So, let \(X\) be the number of times the light bulb functions before the first time it stops functioning (i.e. X is the number of failures before the 1st success).

Then \(X\sim\text{GEO}(p)\) with pdf:

\(P(X=x)=p(1-p)^x, \,\,\, x=0,1,2,3,...\)

Now, to find the probability that it functions at least 50 times, then you are interested in \(X\) being 50 or more, hence: \(P(X\ge 50)\)

Which you can calculate as:
\[P(X\ge 50)=\sum_{x=50}^{\infty}p(1-p)^x=p\sum_{x=50}^{\infty}(1-p)^x=p\frac{(1-p)^{50}}{1-(1-p)}=(1-p)^{50}\]
You can do this sum since it's a geometric series, and since \(p\) is a probability, then \(0<p<1\implies 0<(1-p)<1\)
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Now, you get this probability to be 90% , we set it up to be:
\[P(X\ge 50)=0.90\]

Now, we found \(P(X\ge 50)=(1-p)^{50}\) Hence,
\((1-p)^{50}=0.9\)
\(\left((1-p)^{50}\right)^{1/50}=0.9^{1/50}\\
1-p=0.9^{1/50}\\
p=1-0.9^{1/50}\approx0.002105\)

Answer 2

At times, when the bulbs fail to function, they are repaired and they are tested again.
What is the expected number of trials before a given light bulb fails for the second
time?