Probability light bulb exponential distribution

Question

I came across a question and am a little confused by the correct answer and the answer I calculated.

Assume you have a light bult that will burnout after $t$ hours where $t$ ranges from $0$ to infinity with an exponential density

$f(t)= Le^{-Lt}$ where $L= .01$. What is the probability that the bulb will NOT burn out before $T$ hours.

I did it the following way: I integrated the function: so i got $-e^{-.01t}$ and thentook the complement and so I got $1+e^{-.01t}$ but the correct answer is evidently just $e^{-.0t}$ and I don't understand how we arrive at this conclusion

Answer 1

You did things almost fully correctly. The probability that the lifetime of the bulb is less than $T$ is $$\int_0^T \lambda e^{-\lambda t}\,dt.$$

An antiderivative is $-e^{-\lambda t}$. Plug in $T$ and take away the result of plugging in $0$. We get $1-e^{-\lambda T}$. Take this away from $1$.

Another way: We can calculate the required probability of survival to at least time $T$ (death at $T$ or after) as $$\int_T^\infty \lambda e^{-\lambda T}\,dt.$$

Remark: Forgetting about plugging in $0$ is a common error. After all, we can safely forget when we are integrating a polynomial.

Answer 2

You integrated wrong. Probabilities can't be negative or greater than $1$. I suspect you forgot to evaluate at $t=0$.

Answer 3

If $T$ has density $f_T(t)=L\mathrm{e}^{-Lt}\mathbf{1}_{t>0}$, then $$ F_T(t)=P(T\leq t)=\int_0^ t L\mathrm{e}^{-Ls}\,\mathrm ds=1-\mathrm{e}^{-Lt},\quad t>0. $$ Thus $$ P(T>t)=\mathrm{e}^{-Lt}. $$

Answer 4

The cumulative distribution function of the exponential distribution is $F(x) = 1-e^{-\lambda x}$, so the right answer is indeed $e^{-0.1 T}$. To derive this we integrate \begin{equation} P(X > T) = \int_{T}^{\infty}\lambda e^{-\lambda x} dx = \left[-e^{-\lambda x}\right]_{T}^{\infty} = \lim_{R\rightarrow\infty} - e^{-\lambda R} + e^{-\lambda T} = 0 + e^{-\lambda T} = e^{-\lambda T}. \end{equation}