Question

Let X denote the diameter of an armored electric cable and Y denote the diameter of the ceramic mold that makes the cable. Bo

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Answer #1

1) Since total probability is 1 so

\int_{0}^{2}\int_0^{y} f(x,y)dxdy=1 \\ \Rightarrow \int_{0}^{2}\int_0^{y} Ky dxdy=1 \\ \Rightarrow K\int_{0}^{2}y^2 dy=1 \\ \Rightarrow K \left ( y^3\over 3 \right )_{0}^{2} =1 \\ \Rightarrow K={3\over 8}

2) The region is shown below

Hence the required probability is

p=\int_{0}^{0.25}\int_{0.5-x}^{2} f(x,y)dydx+ \int_{0.25}^{2}\int_{ x}^{2} f(x,y)dydx\\ \\~~ ~~~~~={3\over 8}\left ( \int_{0}^{0.25}\int_{0.5-x}^{2} y dydx+ \int_{0.25}^{2}\int_{ x}^{2} ydydx\right )\\\\ ~~~~~~~={3\over 8}\left ( \int_{0}^{0.25}\left [ y^2\over 2 \right ]_{0.5-x}^{2} dx+ \int_{0.25}^{2}\left [ y^2\over 2 \right ]_{ x}^{2} dx\right ) \\ \\ ~~~~~~~={3\over 16}\left ( \int_{0}^{0.25}[4-(0.5-x)^2]dx+ \int_{0.25}^{2}[4-x^2] dx\right ) \\ ~~~~~~~={3\over 16}[0.96354+4.33854] \\ ~~~~~~~\approx 0.99414

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