Respuesta :
Answer:
[tex] E(X) =sum_{i=1}^n X_i P(X_i)[/tex]
And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:
[tex] E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11[/tex]
[tex] E(X^2) =sum_{i=1}^n X^2_i P(X_i)[/tex]
[tex] E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1[/tex]
And the variance is defined as:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}[/tex]
Step-by-step explanation:
For this case we know that we can win if we select 2 balls of the same color, so we can find the probability of win like this:
[tex] p = \frac{Possible}{ Total}= \frac{2C1 * (5C2)}{10C2}= \frac{2*10}{45}= \frac{4}{9}[/tex]
So then the probability of no win would be given by the complement:
[tex] q = 1-p = 1- \frac{4}{9}= \frac{5}{9}[/tex]
We can define the random variable X who represent the amount of money that we can win.
And we can use the definition of expected value given by:
[tex] E(X) =sum_{i=1}^n X_i P(X_i)[/tex]
And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:
[tex] E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11[/tex]
We can calculate the second monet like this:
[tex] E(X^2) =sum_{i=1}^n X^2_i P(X_i)[/tex]
[tex] E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1[/tex]
And the variance is defined as:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}[/tex]
A part of the question is missing and it says;
b) Calculate the variance of the amount you win.
Answer:
A) Expected value of Pay out;
(E(Y)) = - 0.11
B) Variance of the amount won; Var(Y) = 0.9879
Step-by-step explanation:
A) From the question, we have;
X ∈ {0,1,2} and by the nature of the question, X has a hypergeometric distribution as;
P(X =i) = [(5,i) (5, 2 - i)] / (10,2)
Furthermore, when we consider the random variable Y that marks the amount of money that we win, we'll get a function of X as;
Y = 1X [x∈{0,2}] - [1X(x=1)]
If we now use the linearity of expectation, we'll get;
E(Y) = E [X(x∈{0,2}) ] - [E(X(x=1))]
= P(X = 0) + P(X = 2) - P(X = 1)
For 2 balls with probability of a win, P = (possible outcome) /(total outcome) =
{(2C1) (5C2)}/(10C2) = (2 x 10)/45 = 20/45
While, for probability of no win;
P = 1 - (20/25) = 25/25
So, E(Y) =20/45 - 25/45 = - 5/45 =
- 0.11
B) Now let's calculate for the variance;
Var(Y) = E(Y(^2)) - E(Y)^(2)
Now, from question a, using the equation of Y, we can say;
(Y)^(2) = (1^2)X [x∈{0,2}] - [(1^2)X(x=1)]
And so;
E(Y^(2)) = P(X = 0) + P(X = 2) + P(X = 1) = (1^2)(20/45) + (-1^2)(25/45) = 45/45 = 1
So, Var(Y) = 1 - (-0.11)^2 = 0.9879
