Question #286785

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
2022-01-28T09:24:33-0500
P(X<0.5,Y<0.5)=โˆซ00.5โˆซ00.532(x2+y2)dydxP(X<0.5, Y<0.5)=\displaystyle\int_{0}^{0.5}\displaystyle\int_{0}^{0.5}\dfrac{3}{2}(x^2+y^2)dydx=32โˆซ00.5[x2y+y33]dx0.50=\dfrac{3}{2}\displaystyle\int_{0}^{0.5}[x^2y+\dfrac{y^3}{3}]dx\begin{matrix} 0.5 \\ 0 \end{matrix}=32โˆซ00.5(12x2+124)dx=\dfrac{3}{2}\displaystyle\int_{0}^{0.5}(\dfrac{1}{2}x^2+\dfrac{1}{24})dx=32[16x3+124x]0.50=\dfrac{3}{2}[\dfrac{1}{6}x^3+\dfrac{1}{24}x]\begin{matrix} 0.5 \\ 0 \end{matrix}=32(148+148)=116=\dfrac{3}{2}(\dfrac{1}{48}+\dfrac{1}{48})=\dfrac{1}{16}P(X<0.5,Y<0.5)=116P(X<0.5, Y<0.5)=\dfrac{1}{16}

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!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS