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The lifetimes of three different types of computer components follow exponential distributions with parameters 1,2, and 3 (in particular, that means that the average lifetimes for the three types are 1,2, and 3 years) i.e. $$f_1(x_1) = e^{-x_1}, f_2(x_2) = \frac{1}{2}e^{-\frac{x_2}{2}}, f_3(x_3) = \frac{1}{3}e^{-\frac{x_3}{3}} $$ with support sets x_i = 0, i = 1,2,3. One component from each type is randomly sampled and the corresponding lifetimes recorded. Find $$P(X_1 > 5X_2>10X_3)$$

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You need to enclose subscripts and superscripts that don't consist of a single token in braces to form a group. –  joriki Dec 6 '12 at 21:40
    
thanks joriki. I think Fab got to it before me haha so wherever you are thank you Fab. –  Max Dec 6 '12 at 21:57
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I get $1/374$ when I evaluate $$ P\left(X_1 > 5 X_2 > 10 X_3\right) = \int_0^\infty dx_3 \ f_3\left(x_3\right) \int_{2 x_3}^\infty dx_2 \ f_2\left(x_2\right) \int_{5 x_2}^\infty dx_1 \ f_1\left(x_1\right). $$

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The basic solution technique would be something as follows...

If the failure of the first element happens at time $x$, then the failure of the second element will have to happen before a time $5x$ or greater, and event that happens with probability

$$\int_0^{5x} \frac{1}{2}e^{-\frac{t}{2}}dt = \left.-e^{-\frac{t}{2}}\right|_0^{5x} = 1-e^{-\frac{5x}{2}}$$

If you integrate the above function for all possible values of $x$ weighted by their relative probability (the first probability density function) you get

$$\int_0^\infty e^{-x} (1-e^{-\frac{5x}{2}}) dx = \int_0^\infty (e^{-x}-e^{-\frac{7x}{2}}) dx = 1 - \frac{2}{7} = \frac{5}{7}$$

So $P(X_1 > 5X_2) = \frac{5}{7}$.

Your problem is slightly more complicated, but should be solvable using the same principle.

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