# Exponential Distribution Lifetime of a Lightbulb - Is my solution correct?

Can someone please check if I tackled this question correctly? I don't have answers to refer to. Thank you in advance!

The lifetime of a lightbulb expressed in days is exponentially distributed with paremeter 0.004.

a) What is the probability that the lightbulb lasts at least 300 days?

So we are looking for $$1-P([0,299])$$.

$$P([a,b]) = e^{-\lambda a}-e^{-\lambda b}$$

$$\lambda=0.004$$

$$P([0,299]) = e^{-0.004 *0}-e^{-0.004*299} = 1-0.3024=0.6975$$

From here follows that: $$P([300,∞)=1- 0.6975=0.3025$$

b) What is the probability that the lightbulb lasts at most one year?

Here we compute probability for the interval of $$[0,365]$$.

For $$a)$$ you have to regard that the random variable of the lifetime of a lightbulb is continuous. That means $$P(X\geq t)=1-P(X\leq t)$$
For b) you just insert the value for $$t=365$$ into the cdf. Or you integrate the pdf:
$$P(X\leq 365)=\int_0^{365} 0.004\cdot e^{-0.004x} \, dx$$