Respuesta :
Answer:
a) Marginal probability of X
[tex]f_x(x)=\frac{2}{3}(x+1)[/tex]
b) Marginal probability of Y
[tex]f_y(y)=\frac{2}{3}(1+2y)[/tex]
c) The probability that the drive-through facility is busy less than one-half of the time is P=0.5.
Step-by-step explanation:
The joint density function of probability is:
[tex]f(x, y) = (2/3) (x + 2y)\\\\0 \leq x \leq 1, \,0 \leq y \leq 1[/tex]
a) The marginal density of X is
[tex]f_x(x)=\int_0^1 f(x,y)dy=(2/3)(x+2\int_0^1 y\,dy)=(2/3)*(x+y^2)\\\\f_x(x)=(2/3)*(x+(1-0))\\\\f_x(x)=(2/3)*(x+1)[/tex]
b) The marginal density of Y is:
[tex]f_y(y)=\int_0^1 f(x,y)dx=(2/3)*(\int_0^1x\,dx+2y)=(2/3)*(1^2-0^2+2y)\\\\f_y(y)=(2/3)*(1+2y)[/tex]
c) Probability that the drive-through facility is busy less than one-half of the time. This is P(X<0.5).
[tex]P(X<0.5)=\int_0^{0.5}\,(2/3)(x+1)dx=(2/3)[(0.5^2+0.5)-(0^2+0)]\\\\P(X<0.5)=(2/3)*(0.75-0)=0.5[/tex]
a) Marginal probability of X
[tex]\int_x(x)=\frac{2}{3} (x+1)[/tex]
b) Marginal probability of Y
[tex]\int_y(y)=\frac{2}{3} (1+2y)[/tex]
c) The probability that the drive-through facility is busy less than one-half of the time is P=0.5.
Let's start the question:
The joint density function of probability is:
[tex]\int(x,y)=(2/3)(x+2y)[/tex]
[tex]0\leq x\leq 1,0\leq y\leq 1[/tex]
a) The marginal density of X is
[tex]\int_x(x)=\int\limits^0_1 {(x,y)} \, dy=(2/3)(x+2)\int\limits^0_1 {y} \, dxy=(2/3)*(x+y^2)\\\\ \int_x(x)=(2/3)*(x+(1-0))\\\\ \int_x(x)=(2/3)*(x+1)[/tex]
b) The marginal density of Y is:
[tex]\int_y(y)=\int\limits^0_1 {(x,y)} \, dx=(2/3)*(\int\limits^0_1 {x} \, dx +2y)=(2/3)*(1^2-0^2+2y)\\\\\int_y(y)=(2/3)*(1+2y)[/tex]
c) Probability that the drive-through facility is busy less than one-half of the time.
This is P(X<0.5).
Learn more:
brainly.com/question/22751827
