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[Step-by-Step] The length of life Y for fuses of a certain type is modeled by the exponential distribution, with

Question: The length of life \(Y\) for fuses of a certain type is modeled by the exponential distribution, with

\[f(y)=\left\{\begin{array}{lr} \left(\frac{1}{3}\right) e^{-\frac{y}{2}}, & y>0 \\ 0, & \text { otherwise } \end{array}\right.\]

(the measurements are in hundreds of hours).

  1. If two such fuses have independent lengths of life \(Y_{1}\) and \(Y_{2}\), find the joint pdf \(f\left(Y_{1}, Y_{2}\right)\).
  2. One fuse in part (a) is in a primary system, and the other is in a backup system that comes into use only if the primary system fails. The total effective length of life of the two fuses is then \(Y_{1}+Y_{2}\). Find \(P\left(Y_{1}+Y_{2} \leq 1\right)\).

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