Question

You pick two marbles at random out of a box that has 5 red and 5 blue marbles. If they are the same color, you win $1. If they are of different colors, you lose $1 (i.e. your payout is -$1). What is the expected value of your payout

Answer 1

Answer:

$E(X) =sum_{i=1}^n X_i P(X_i)$

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:

$E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11$

$E(X^2) =sum_{i=1}^n X^2_i P(X_i)$

$E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1$

And the variance is defined as:

$Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}$

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:

$p = \frac{Possible}{ Total}= \frac{2C1 * (5C2)}{10C2}= \frac{2*10}{45}= \frac{4}{9}$

So then the probability of no win would be given by the complement:

$q = 1-p = 1- \frac{4}{9}= \frac{5}{9}$

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:

$E(X) =sum_{i=1}^n X_i P(X_i)$

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:

$E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11$

We can calculate the second monet like this:

$E(X^2) =sum_{i=1}^n X^2_i P(X_i)$

$E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1$

And the variance is defined as:

$Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}$

Answer 2

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