Question #286782

A production facility contains two machines that are used to rework items that are initially defective. Let ๐‘‹ be the number of hours that the first machine is in use and let ๐‘Œ be the number of hours that the second machine is in use, on a randomly chosen day. Assume that ๐‘‹ and ๐‘Œ have a joint probability density function given by ๐‘“(๐‘ฅ) = { 3 2 (๐‘ฅ 2 + ๐‘ฆ 2 ) 0 < ๐‘ฅ < 1 ๐‘Ž๐‘›๐‘‘ 0 < ๐‘ฆ < 1 0 ๐‘œ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘ค๐‘–๐‘ ๐‘’. a. What is the probability that both machines are in operation for less than half an hour?

1
Expert's answer
2022-01-26T17:09:32-0500

To find the probability that both machines are in operation for less than half an hour, we determine the probability,

p(0<x<0.5,0<y<0.5)=โˆซ00.5โˆซ00.5f(x,y)dydx=โˆซ00.5โˆซ00.532(x2+y2)dydx=32โˆซ00.5โˆซ00.5(x2+y2)dydx=32โˆซ00.5(x2y+y33)โˆฃ00.5dx=32โˆซ00.5(0.5x2+124)dx=32(16x3+x24)โˆฃ00.5=32(148+148)=116p(0\lt x\lt 0.5,0\lt y\lt 0.5)=\displaystyle\int^{0.5}_0\displaystyle\int^{0.5}_0 f(x,y)dydx\\ =\displaystyle\int^{0.5}_0\displaystyle\int^{0.5}_0 {3\over2}(x^2+y^2)dydx\\ ={3\over2}\displaystyle\int^{0.5}_0\displaystyle\int^{0.5}_0 (x^2+y^2)dydx ={3\over2}\displaystyle\int^{0.5}_0(x^2y+{y^3\over3})|^{0.5}_0dx\\ ={3\over2}\displaystyle\int^{0.5}_0(0.5x^2+{1\over24})dx={3\over2}({1\over6}x^3+{x\over24})|^{0.5}_0={3\over2}({1\over48}+{1\over48})={1\over16}

Therefore, the probability that both machines are in operation for less than half an hour is 116{1\over16}


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