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2 years ago

So, if you had 25 video games and 1/5 were sports games, how many of your games would be sports?

Mathematics

1 answer:

blsea [12.9K]2 years ago

5 0

Answer:

5 sport games

Step-by-step explanation:

25 total video games * (1 sport game / 5 video games) = 5 sport games

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Step-by-step explanation:

Let's start defining the random variables for this exercise :

$X_{1}:$ '' The proportion of the particulates that are removed by the first pass ''

$X_{2}:$ '' The proportion of what remains after the first pass that is removed by the second pass ''

$Y:$ '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

$Y=(1-X_{1})(1-X_{2})$

We also assume that $X_{1}$ and $X_{2}$ are independent random variables with common pdf.

The probability density function for both variables is $f(x)=4x^{3}$ for $0$ and $f(x)=0$ otherwise.

The first step to solve this exercise is to find the expected value for $X_{1}$ and $X_{2}$ .

Because the variables have the same pdf we write :

$E(X_{1})= E(X_{2})=E(X)$

Using the pdf to calculate the expected value we write :

$E(X)=\int\limits^a_b {xf(x)} \, dx$

Where $a=$ ∞ and $b=$ - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between $0$ and $1$ ⇒

Using the pdf we calculate the expected value :

$E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}$

⇒ $E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}$

Now we need to use some expected value properties in the expression of $Y$ ⇒

$Y=(1-X_{1})(1-X_{2})$ ⇒

$Y=1-X_{2}-X_{1}+X_{1}X_{2}$

Applying the expected value properties (linearity and expected value of a constant) ⇒

$E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})$

Using that $X_{1}$ and $X_{2}$ have the same expected value $E(X)$ and given that $X_{1}$ and $X_{2}$ are independent random variables we can write $E(X_{1}X_{2})=E(X_{1})E(X_{2})$ ⇒