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

Answer:

$E(X_n)=\frac{2(n-1)}{27}$

$E(y)=\frac{14}{9}$

Step-by-step 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

 $P(GY)=\frac{4}{9}*\frac{3}{9}$

 $P(GY)=\frac{4}{27}$

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

 $Probability of success P(S)=\frac{4}{27}$

 $Probability of Failure P(F)=\frac{27-4}{27}$

 $Probability of Failure P(F)=\frac{23}{27}$

Therefore

 $E(X_n)=\frac{n}{2}*P$

 $E(X_n)=\frac{n}{2}*\frac{4}{27}$

 $E(X_n)=\frac{2n}{27}$

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

 $\frac{n-1}{2}$

 $E(X_n)=\frac{n-1}{2}*\frac{4}{27}$

 $E(X_n)=\frac{2(n-1)}{27}$

b)

Probability of drawing white ball

 $P(w)=\frac{2}{9}$

Therefore

 $E(w)=\frac{1}{p}$

 $E(w)=\frac{1}{\frac{2}{9}}$

 $E(w)=\frac{9}{2}$

Therefore

 $E(y)=[E(w)-1]\frac{4}{9}$

 $E(y)=[\frac{9}{2}-1]\frac{4}{9}$

 $E(y)=\frac{14}{9}$