Question

Suppose I have an urn with 9 balls: 4 green, 3 yellow and 2 white ones. I draw a ball from the urn repeatedly with replacement. (a) Suppose I draw n times. Let X., be the number of times I saw a green ball followed by a yellow ball. Calculate the expectation Ex, (b) Let y be the number of times I drew a green ball before the first white draw. Calculate E[Y]. Can you give an intuitive explanation for your answer

Answer 1

`E(X_n)=(2(n-1))/(27)`

`E(y)=(14)/(9)`

Explanation:

From the question we are told that:

Sample size n=9

Number of Green
`g=4`

Number of yellow
`y=4`

Number of white
`w=4`

Probability of Green Followed by yellow P(GY) ball

`P(GY)=(4)/(9)*(3)/(9)`

`P(GY)=(4)/(27)`

Generally the equations for when n is even is mathematically given by

`Probability of success P(S)=(4)/(27)`

`Probability of Failure P(F)=(27-4)/(27)`

`Probability of Failure P(F)=(23)/(27)`

Therefore

`E(X_n)=(n)/(2)*P`

`E(X_n)=(n)/(2)*(4)/(27)`

`E(X_n)=(2n)/(27)`

Generally the equations for when n is odd is mathematically given by

`(n-1)/(2)`

`E(X_n)=(n-1)/(2)*(4)/(27)`

`E(X_n)=(2(n-1))/(27)`

b)

Probability of drawing white ball

`P(w)=(2)/(9)`

Therefore

`E(w)=(1)/(p)`

`E(w)=(1)/((2)/(9))`

`E(w)=(9)/(2)`

Therefore

`E(y)=[E(w)-1](4)/(9)`

`E(y)=[(9)/(2)-1](4)/(9)`

`E(y)=(14)/(9)`