A number of lightbulbs have lifetimes X that are iid and exponentially distributed with a mean of 1/4. As a lightbulb fails, it is replaced with another until the bulbs run out. Using Chebyshev Inequality, estimate the number of lightbulbs we will need such that the probability that we have a functioning bulb at t = 10 is at least 0.8.
My attempt so far:
I have a random variable Y which is a sum of N X's. This is because as they burn out I replace them and so the mean of Y is N/4 and since these are iid the variance of Y is N/16.
Now I try to use Chebyshev Inequality, since I'm interested in the point y = 10 I have $\ |Y-mu|=|10-N/4| $, and the standard form is $\ P(|Y-mu|>=k)<=var(Y)/k^2 $. This implies that the inequality puts an upper bound but I'm looking for "at least 0.8", a lower bound. My question is can I then do the following to transform it into an lower bound, $\ 1-P(|Y-mu|<=k)>=var(Y)/k^2 $?