Consider a collection of 9 bears. There is a family of red bears consisting of o

Question

Consider a collection of 9 bears. There is a family of red bears consisting of one father bear, one mother bear, and one baby bear. There is a similar green bear family, and a similar blue bear family. We draw 5 consecutive times from this collection without replacement (i.e., not returning the bear after each draw). We keep track (in order) of the kind of bears that we get. Let X denote the number of red bears selected. For i=1,2,3,4,5, let Xi=1 if the ith bear selected is red, and Xi=0 otherwise. Refer to the red father bear as red bear #1, and the red mother bear as red bear #2, and the red baby bear as red bear #3. For i=1,2,3, let Yi=1 if the ith red bear is selected (at any time, i.e., on any of the five selections), and Yi=0 otherwise.

a. Find E(X2) using the probability mass function of X.

b. Find E(X2) in a different way, namely, using the fact that X=X1+?+X5 (the Xj's are dependent indicators). Expand (X1+?+X5)2 into 25 terms, where 20 of them will behave one way, and the other 5 will behave another way. You are expected to obtain the same answer as 12a.

c. Find E(X2) in a different way, namely, using the fact that X=Y1+Y2+Y3 (the Yj's are dependent indicators). Expand (Y1+Y2+Y3)2 into 9 terms, where 6 of them will behave one way, and the other 3 will behave another way. You are expected to obtain the same answer as 12a.

d. Find Var?(X)

Explanation / Answer

E(X2)=Pr(the second bear is red)

=Pr(the second bear is red|the first bear is red)Pr(the first bear be red)

+Pr(the second bear is red|the first bear is not red)Pr(the first bear is not red)

=2/8?3/9+3/8?6/9

=0.33

v(x)=E(X^2)-(E(X))^2

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