Answer:
[tex]E(Y)=\frac{1}{25}[/tex]
Step-by-step explanation:
Let's start defining the random variables for this exercise :
[tex]X_{1}:[/tex] '' The proportion of the particulates that are removed by the first pass ''
[tex]X_{2}:[/tex] '' The proportion of what remains after the first pass that is removed by the second pass ''
[tex]Y:[/tex] '' The proportion of the original particulates that remain in the sample after two passes ''
We know the relation between the random variables :
[tex]Y=(1-X_{1})(1-X_{2})[/tex]
We also assume that [tex]X_{1}[/tex] and [tex]X_{2}[/tex] are independent random variables with common pdf.
The probability density function for both variables is [tex]f(x)=4x^{3}[/tex] for [tex]0<x<1[/tex] and [tex]f(x)=0[/tex] otherwise.
The first step to solve this exercise is to find the expected value for [tex]X_{1}[/tex] and [tex]X_{2}[/tex].
Because the variables have the same pdf we write :
[tex]E(X_{1})= E(X_{2})=E(X)[/tex]
Using the pdf to calculate the expected value we write :
[tex]E(X)=\int\limits^a_b {xf(x)} \, dx[/tex]
Where [tex]a=[/tex] ∞ and [tex]b=[/tex] - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between [tex]0[/tex] and [tex]1[/tex] ⇒
Using the pdf we calculate the expected value :
[tex]E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}[/tex]
⇒ [tex]E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}[/tex]
Now we need to use some expected value properties in the expression of [tex]Y[/tex] ⇒
[tex]Y=(1-X_{1})(1-X_{2})[/tex] ⇒
[tex]Y=1-X_{2}-X_{1}+X_{1}X_{2}[/tex]
Applying the expected value properties (linearity and expected value of a constant) ⇒
[tex]E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})[/tex]
Using that [tex]X_{1}[/tex] and [tex]X_{2}[/tex] have the same expected value [tex]E(X)[/tex] and given that [tex]X_{1}[/tex] and [tex]X_{2}[/tex] are independent random variables we can write [tex]E(X_{1}X_{2})=E(X_{1})E(X_{2})[/tex] ⇒
[tex]E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2})[/tex] ⇒
[tex]E(Y)=E(1)-2E(X)+[E(X)]^{2}[/tex]
Using the value of [tex]E(X)[/tex] calculated :
[tex]E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}[/tex]
[tex]E(Y)=\frac{1}{25}[/tex]
We find that the expected value of the variable [tex]Y[/tex] is [tex]E(Y)=\frac{1}{25}[/tex]
