Question

We have three urns: the first urn has 6 red balls and 4 green balls; the...

We have three urns: the first urn has 6 red balls and 4 green balls; the second urn has 15 red balls and 5 green balls and the third urn has 20 red balls and 10 green balls. We pick 4 balls from the first urn (sampling with replacement); we select 5 balls from the second urn (sampling with replacement) and we select 10 balls from the third urn (sampling with replacement). Let X1 denote the number of red balls which are obtained from the first urn; let X2 denote the number of red balls which are obtained from the second urn and let X3 denote the number of red balls which are obtained from the third urn.

a) Find the expected value and variance of X1.

b) Find the expected value and the variance of X2.

c) Find the expected value and the variance X3.

d) Find the expected value of the total number of balls which are obtained from the 3 urns.

e) FInd the variance of the total number of red balls which are obtained form teh 3 urns.

f) If we define the "payoff" to be 5 times the total number of red balls from the three urns, then what is the expeted value of the payoff?

g) Referring to the preceeding exercise, what is the standard deviation of the payoff?

a)

Total number of balls: 6+4=10

Since sampling are done with replacement so probability of getting red ball in each draw is

p = 4 / 10 =0.40

Here X1 has binomial distribution with parameters n=4 and p=0.40.

The expected value and variance is:

E(X1) = np = 4 * 0.40 = 1.6

Var(X1) = np(1-p) = 4 *0.40 * 0.60 = 0.96

b)

Total number of balls: 15+5 =20

Since sampling are done with replacement so probability of getting red ball in each draw is

p = 15 / 20 =0.75

Here X2 has binomial distribution with parameters n=5 and p=0.75.

The expected value and variance is:

E(X2) = np = 5 * 0.75 =3.75

Var(X2) = np(1-p) = 5*0.75 * 0.25 = 0.9375

c)

Total number of balls: 20+10=30

Since sampling are done with replacement so probability of getting red ball in each draw is

p = 20 / 30 = 2/3

Here X3 has binomial distribution with parameters n=10 and p=2/3.

The expected value and variance is:

E(X3) = np = 10 * (2/3) =6.667

Var(X3) = np(1-p) = 10 * 2/3 * 1/3 = 2.222

d)

Let T shows the total number of red balls. The expected value of the total number of balls which are obtained from the 3 urns is

E(T) = E(X1)+ E(X2)+ E(X3) = 1.6+3.75 + 6.667 = 12.017

e)

Var(T) =Var(X1)+Var(X2)+Var(X3) = 0.96 + 0.9375 + 2.222 = 4.1195

f)

Let P shows the pay off. So,

E(P) = E(5T)= 5E(T) = 5 * 12.017 = 60.085

g)

The standard deviation of the payoff is

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