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The lengths of life Y for a type of fuse has an
Chapter 5, Problem 151SE(choose chapter or problem)
The lengths of life \(Y\) for a type of fuse has an exponential distribution with a density function given by
\(f(y)=\left\{\begin{array}{ll}
(1 / \beta) e^{-y / \beta}, & y \geq 0 \\
0, & \text { elsewhere }
\end{array}\right.\)
a If two such fuses have independent life lengths \(Y_{1} \text { and } Y_{2}\), find their joint probability density function.
b One fuse from 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 life length of the two fuses, therefore, is \(Y_{1}+Y_{2}\). Find \(P\left(Y_{1}+Y_{2} \leq a\right)\), where \(a>0\).
Equation Transcription:
Text Transcription:
Y
f(y)={ (1/\beta)e_-y/\beta, y\geq 0 0, elsewhere
Y_1 and Y_2
Y_1+Y_2
P(Y_1+Y_2 \geq a)
a>0
Questions & Answers
QUESTION:
The lengths of life \(Y\) for a type of fuse has an exponential distribution with a density function given by
\(f(y)=\left\{\begin{array}{ll}
(1 / \beta) e^{-y / \beta}, & y \geq 0 \\
0, & \text { elsewhere }
\end{array}\right.\)
a If two such fuses have independent life lengths \(Y_{1} \text { and } Y_{2}\), find their joint probability density function.
b One fuse from 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 life length of the two fuses, therefore, is \(Y_{1}+Y_{2}\). Find \(P\left(Y_{1}+Y_{2} \leq a\right)\), where \(a>0\).
Equation Transcription:
Text Transcription:
Y
f(y)={ (1/\beta)e_-y/\beta, y\geq 0 0, elsewhere
Y_1 and Y_2
Y_1+Y_2
P(Y_1+Y_2 \geq a)
a>0
ANSWER:
Solution:
Step 1 of 2:
The life lengths, Y, for a type of fuse has an exponential distribution with a density function
We have to find,
- The joint probability density function.
-
P(Y1+Y2
, where a>0.