Answer
$0.1315$
Work Step by Step
The probability density $y$ for the failure of the first bulb is: $f(x)=\dfrac{e^{-x/800}}{800}$
and the probability that all three bulbs will fail within $1000$ hours is:
$$\int_{0}^{1000}\int_{0}^{1000} \int_{0}^{1000}f(x) f(y) f(z) dx \space dy \space dz=[\int_{0}^{1000} f(x)] dx]^3 \\=\int_{0}^{1000} \int_{0}^{1000-x} \int_{0}^{1000-x-y} e^{-(\dfrac{1}{800})^3 e^{-(x+y+z)/800}} dz dy dx \\=[\int_{0}^{1000} \dfrac{e^{-x/800}}{800} dx]^3\\=\dfrac{-1}{(800)^2} \times \int_0^{1000} [e^{-5/4} (1800-x) -800 e^{-x/800}]_0^{1000}\\=1-\dfrac{97 e^{-5/4}}{32} \\ \approx 0.1315$$
