Suppose you have a flashlight that uses 2 batteries and a package of 3 new batteries. Suppose that each battery has an independent lifetime (in hours) with exponential$\displaystyle (\lambda)$ distribution. Find the distribution of the number of hours T that you can use the flashlight.

My thoughts are:

P(T = t) = P(2 of the batteries have lifetimes of t hours)

= P(X = t, Y = t) + P(X = t, Z = t) + P(Y = t, Z = t)

= $\displaystyle \lambda e^{-\lambda t}\lambda e^{-\lambda t} + \lambda e^{-\lambda t}\lambda e^{-\lambda t} + \lambda e^{-\lambda t}\lambda e^{-\lambda t}$

= 3$\displaystyle \lambda^2 e^{-2\lambda t}$

Could someone tell me if this is correct?