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?

Expert's answer

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}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS